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Mathway: Math Problem Solver 4+

Homework scanner & calculator, chegg, inc..

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Mathway is the world’s smartest math calculator for algebra, graphing, calculus and more! Mathway gives you unlimited access to math solutions that can help you understand complex concepts. Simply point your camera and snap a photo or type your math homework question for step-by-step answers. If a premium subscription option is selected: • Payment will be charged to iTunes Account at confirmation of purchase • Subscription automatically renews unless auto-renew is turned off at least 24-hours before the end of the current period • Account will be charged for renewal within 24-hours prior to the end of the current period, at the same monthly or annual rate selected at the beginning of the subscription • Subscriptions may be managed by the user and auto-renewal may be turned off by going to the user's Account Settings after purchase Terms of Use: https://www.mathway.com/terms Privacy Policy: https://www.mathway.com/privacy

Version 5.5.3

This update includes bug fixes and performance improvements. Powered by Chegg, Mathway is your one-stop shop for homework help. Is your equation too complex to type? Save time with our snap and solve feature! Just take a photo and instantly see the solution. Our interactive calculator also allows you to select your class subject (now including physics!) to generate all the functions you need to solve that equation. Need more than an answer? Upgrade your subscription and get unlimited access to detailed equation breakdowns, helping you learn how to solve it one step at a time.

Ratings and Reviews

418.2K Ratings

Recommend!!!!!!?

Once I was doing my homework I couldn’t figure it out I asked my mom to help me and I showed her and she fell asleep so I kind of started panicking Went downstairs to the table I started struggling and thinking and thinking so I went to the App Store looked and looked and looked and looked left to right for helping your homeworkIs the best I really do recommend it it is the best it will save your life and when I mean save your life I mean really save your life nobody really understands when they say that they saved their lifeThey never understand how much that person means they don’t even know how serious the persons like saying I really do recommend this app messed up in the world five star review amazing helpful save your life get it get it now get it now it’s the best app ever students allowed teacher teachers not allowed can students really need this app so don’t get mad at them if they have this app it’s the best time with them so hope you get it

All it does is give the answer

Seriously saying, this app is literally just useless unless if you know the answer to math problems and want to check your answers. Otherwise, paying wouldn’t be a consideration for me. For those who are on a tight budget, and have a strong desire to reach their future career, I’d say paying wouldn’t be worth it, especially if you are just doing a living only as a student with no career and if you do not have enough financial support. I’ve lived a long way with this app studying all day and all night and yet there is high demand for paying just to get the work shown to get the idea why a question is answered in a certain way. This is why nowadays I read reviews from various books titled for math and read books that give a full explanation of how an answer is earned. As a life lesson it’s just good to say the fact that spending is not worth it for this app since there are various places and apps to ask questions related to learning math (at least in my area). This app should be like a substitute teacher or tutor for those who are having trouble, but the fact that there is such huge fee just ruins the reason why I should install this app. Teachers ask for showing work, and yet I’m stuck, having no idea what to do to get the right answers. This also goes to understanding how an answer is caught as well.

It works but the full version hasn’t for me

I have been using this a lot this semester so I decided to pay for the monthly subscription then I could cancel the membership at the end of the semester, but it doesn’t even work. The subscription shows in my subscriptions list in my settings but I still can’t view the steps, and when I attempt to see how the problem is solved it brings me to the menu to subscribe to the paid part of the app. Honestly though, other than that. This app works pretty great, solves problems the right way and you can select the method of how you want the problem solved. You can change what kind of math you are trying to solve, and the answers are pretty spot on. The predictive text is even really good on this, it’ll help solve word problems and you kind of don’t em have to type anything in sometimes. However, the keyboard is very buggy and you can’t type very fast. It’s also kind of hard to navigate as there are like 4 different keyboard menus to choose from, I always forget where to find things and it kind of takes a while to load the text sometimes. But it does solve the problems.

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Percentage Table

WHAT is P% of X?

Ex: What is 25% of 150?

Y is 25% of 150

Equation: Y = P% * X

Solving our equation for Y

Y = 25% * 150

Converting percent to decimal:

p = 25%/100 = 0.25

Y = 0.25 * 150

Y is WHAT Percent of X?

35 is what percent of 175?

35 is P% of 175

Solving our equation for P

P% = 35/175

Convert decimal to percent:

P% = 0.2 * 100 = 20%

Y is P% of WHAT?

35 is 12% of what?

35 is 12% of X

Solving our equation for X

p = 12%/100 = 0.12

X = 35/0.12

X = 291.6667

WHAT % of X is Y?

Ex: What % of 2400 is 360?

P% of 2400 is 360

Equation: P% * X = Y

P% = 360/2400

P% = 0.15 * 100 = 15%

P% of What is Y Calculator?

35% of what is 560?

35% of X is 560

X = 560/35%

p = 35%/100 = 0.35

X = 560/0.35

Ex: P% of X is What Calculator?

32% of 560 is what?

32% of 560 is Y

Y = 32% * 560

p = 32%/100 = 0.32

Y = 0.32 * 560

Ex: Y out of What is P% Calculator?

350 out of what is 15%?

350 out of X is 15%

Equation: Y/X = P%

X = 350/15%

p = 15%/100 = 0.15

X = 350/0.15

X = 2333.3333

Ex: What out of X is P% Calculator?

What out of 450 is 12%?

Y out of 450 is 12%

Y = 12% * 450

Y = 0.12 * 450

Ex: Y out of X is WHAT Percent Calculator?

570 out of 850 is what %?

570 out of 850 is P%

P% = 570/850

P% = 0.6706 * 100 = 67.06%

X + (X × P%) = Y

Ex: X plus P% is WHAT?

350 plus 12% is what?

350 plus 12% is Y

X(1 + P%) = Y

Y = X(1 + P%)

Y = 350(1 + 12%)

Y = 350(1 + 0.12)

Y = 350(1.12)

Ex: X plus WHAT Percent is Y?

420 plus what % is 560?

420 plus P% is 560

P% = Y/X - 1

P% = 560/420 - 1

P% = 0.3333 * 100 = 33.33%

Ex: What plus P% is Y?

What plus 25% is 450?

X plus 25% is 450

X = Y/(1 + P%)

X = 450/(1 + 25%)

X = 450/(1 + 0.25)

X = 450/1.25

X - (X × P%) = Y

Ex: X minus P% is WHAT?

450 minus 12% is what?

450 minus 12% is Y

X(1 - P%) = Y

Y = X(1 - P%)

Y = 450(1 - 12%)

Y = 450(1 - 0.12)

Y = 450(0.88)

Ex: X minus WHAT Percent is Y?

650 minus what % is 325?

650 minus P% is 325

P% = 1 - Y/X

P% = 1 - 325/650

P% = 0.5 * 100 = 50%

Ex: WHAT minus P% is Y?

What minus 2% is 490?

X minus 2% is 490

X = Y/(1 - P%)

X = 490/(1 - 2%)

p = 2%/100 = 0.02

X = 490/(1 - 0.02)

X = 490/0.98

Percentage Change Formula

Percentage Change = [ (New_Value - Original_Value) / |Original_value| ] * 100

Ex: Percentage Change from 250 to 350?

Solution: New_Value = 350, Original_Value = 250

Substitute the given values in the formula.

Percentage Change = (350 - 250) / |250| * 100

Percentage Change = (100/250) * 100 = (2/5) * 100 = 0.4 * 100

Percentage Change = 40%

∴ Percentage Change from 250 to 350 is 40%

Percent Error Formula

percent error = [ (AV - OV) / | AV | ] * 100%

where: AV is the actual value, and OV is the observed value.

Ex: Percent Error from 200 to 250?

Solution: actual value = 200, observed value = 250

Percent Error = [ (AV - OV) / | AV | ] * 100%

Percent Error = (250 - 200) / |200| * 100

Percent Error = (50/200) * 100 = (1/4) * 100 = 0.25 * 100

Percent Error = 25%

∴ Percent Error from 200 to 250 is 25%

Percentage Increase Formula

Percentage Increase = [ (FV - SV) / | SV | ] * 100%

where: FV is the final value, and SV is the starting value.

Ex: Percent Increase from 90 to 108?

Solution: Starting Value = 90, Final Value = 108

Percentage Increase = (108 - 90) / |90| * 100

Percentage Increase = (18/90) * 100 = (1/5) * 100 = 0.2 * 100

Percentage Increase = 20%

∴ Percent Increase from 90 to 108 is 20%

Percentage Decrease Formula

Percentage Decrease = [ (SV - FV) / | SV | ] * 100%

Ex: Percent Decrease from 80 to 68?

Solution: Starting Value = 80, Final Value = 68

Percentage Increase = [ (SV - FV) / | SV | ] * 100%

Per. Decrease = (80 - 68) / |80| * 100

Per. Decrease = (12/80) * 100 = (3/20) * 100 = 0.15 * 100

Per. Decrease =

∴ Percent Decrease from 80 to 68 is 15%

Percentage Difference Formula

Percentage Difference = 100 * |a - b| / ((a + b) / 2)

Ex: Percent Difference from 300 to 200?

Solution: a = 300, b = 200

Percentage Difference = 100 * |a - b| / ((a + b) / 2)%

Percentage Difference = 100 * |300 - 200| / ((300+200) / 2)%

P.D = 100 * |100| / (500/2)

P.D = (10000 / 250)%

∴ Percent Difference from 300 to 200 is 40%

Percentage of a Percentage Formula

Percentage of a Percentage = (pct1 ÷ 100) × (pct2 ÷ 100) × 100

where: pct1 is the percentage 1, and pct2 is the percentage 2.

Ex: Find 25% of 60%

Solution: pct1 = 300, pct2 = 200

p = (pct1 ÷ 100) × (pct2 ÷ 100) × 100

p = (25 ÷ 100) × (60 ÷ 100) × 100

p = .25 × .6 × 100

p = .15 × 100

∴ 25% of 60% is 15%

Doubling Time Formula

doubling time = log(2) / log(1 + r)

Where: r = a constant growth rate

Convert Fraction to Percentage

Simple Formula to Convert Fraction to Percent

(n / d) * 100 = P

where n is the numerator, d is the denominator, and p is the percentage.

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Example (Click to try)

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  • Type in your equation like y=2x+1 (If you have a second equation use a semicolon like y=2x+1 ; y=x+3)
  • Press Calculate it to graph!

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Order of Operations

The order of operations is a set of rules that is to be followed in a particular sequence while solving an expression. In mathematics with the word operations we mean, the process of evaluating any mathematical expression, involving arithmetic operations such as division, multiplication, addition, and subtraction. Let us learn in detail about the order of operations rules and how well we can remember the rules using short tricks.

What is the Order of Operations?

The order of Operations is the rule in math that states we evaluate the parentheses/brackets first, the exponents/the orders second, division or multiplication third (from left to right, whichever comes first), and the addition or subtraction at the last (from left to right, whichever comes first). In math, there might be several operations to be done while evaluating an expression, and simplification at the end yields different results. However, we can only have one correct answer for any sort of expression. To identify the correct answer we simplify any given mathematical expression using a certain set of rules. These rules revolve around all the basic operators used in maths. Operators such as addition (+), subtraction (-), division (÷), and multiplication (×). Look at the given image to get a glimpse of how the order of operations exactly looks like.

order of operations

Order of Operations Definition

As we discussed above Order of operations can be defined as, a set of basic rules of precedence we use while solving any mathematical expression, involving multiple operations. When a subexpression appears between two operators, the operator that comes first according to the list given below should be applied first. The order of operations, rules are expressed here:

  • Brackets ( ), { }, [ ]
  • Division (÷) and Multiplication (×)
  • Addition (+) and Subtraction (-)

The above-mentioned set of rules always varies according to the respective given mathematical expressions.

Order of Operations Rules

While performing any sort of an operation on the respective numbers present in the expression we will follow the given basic rules in the particular sequence.

Order of Operations Rule 1: Observe the expression. The first rule is to solve the numbers present inside the parentheses or brackets. We solve inside to out grouping operations. Note the pattern of brackets present in the expression, there is a particular order to solve the parentheses, i.e., [ { ( ) } ]. First, solve the round brackets ( ) → curly brackets { } → box brackets [ ]. Inside the parantheses the order of operations are to be followed. Order of Operations Rule 2: After solving the numbers in the parentheses, look for any term present in the form of exponents and solve it. Order of Operations Rule 3: Now we are left with the basic four operators. Look for the numbers with the operation of multiplication or division, solve them from left to right. Order of Operations Rule 4: Lastly, look for the terms with addition or subtraction and solve them from left to right.

These rules have a specific acronym name. We call them PEMDAS or BODMAS . Let us learn now what exactly PEMDAS or BODMAS is.

Order of Operations - PEMDAS vs BODMAS

The PEMDAS or BODMAS is the two different acronym names given to learn the rules. These two names state the order in which the operations in an expression should be followed. Here is the detailed term for each letter used in the mentioned acronyms. First, we will discuss the PEMDAS.

Order of Operations PEMDAS

  • P stands for Parentheses ( ), { }, [ ]
  • E stands for Exponents (a 2 ) (For example, here, a is a number with exponent 2 )
  • M stands for Multiplication (×)
  • D stands for Division (÷)
  • A stands for Addition (+)
  • S stands for Subtraction (-)

Order of Operations BODMAS

  • B stands for Brackets ( ), { }, [ ]
  • O stands for Order

With the help of the above denotations, we can easily solve the mathematical expressions and get the correct answer.

How to Use Order of Operations?

Let us look at the different examples mentioned below to understand the accuracy of the rules used in order of operations.

1) For solving parentheses in order of operations :

Expression: 4 × (5 + 2) Solution: 4 × ( 7 ) = 28 (Correct (✔). This is a correct way to solve the parentheses) Let us look at another approach for the same expression. 4 × ( 5 + 2) = 20 + 2 = 22 (Incorrect (✘). This is an incorrect way to solve the parentheses)

2) For solving exponents in order of operations

Expression: 4 × (5 2 ) Solution: 4 × ( 25 ) = 100 (Correct (✔). This is a correct way to solve the exponents) Let us look at another approach for the same expression. 4 × ( 5 2 ) = 20 2 = 400 ((Incorrect (✘). This is an incorrect way to solve the exponents)

3) For multiplication or division and addition or subtraction

Expression: 3 + 5 × 2 Solution: 3 + 5 × 2 = 3 + 10 = 13 (Correct (✔). This is a correct way.) Let us look at another approach for the same expression. 3 + 5 × 2 = 8 × 2 = 16 (Incorrect (✘). This is an incorrect way.)

Expression: 3 - 6 ÷ 2 Solution: 3 - 6 ÷ 2 = 3 - 3 = 0 (Correct (✔). This is a correct way.) Let us look at another approach for the same expression. 3 - 6 ÷ 2 = (-3) ÷ 2 = -3/2 (Incorrect (✘). This is an incorrect way.)

Always remember while following the rules of order of operations do multiplication or division before addition or subtraction

Ways to Remember Order of Operations

We just read about the two different words PEMDAS and BODMAS. This is the best way to remember the order of operations. PEMDAS can be remembered by the phrase "Please Excuse My Dear Aunt Sally". In the order of operations, it means "Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction". Here multiplication and division, addition and subtraction are together. Similarly, we can remember the order of operations with the word BODMAS (Brackets, Orders, Division, Multiplication, Addition, and Subtraction.).

The easiest way to learn the order of operation is to perform the given steps:

  • Start simplifying terms within the brackets first
  • Solve the exponential terms.
  • Perform division or multiplication.
  • Perform addition or subtraction.

Note: While performing the order of operations on any given expression, we must observe the pattern of operators.

Real-Life Applications of Order of Operations

A lot of activities in our life require some sort of order of operation to perform it well. Let us take an everyday problem. Suppose you went to purchase five pepperoni pizzas that cost $20 each, and you want to split the total cost among 5 people evenly. To find out how much each person needs to pay let's use the order of operations here.

Total number of people = 5 Total number of pizzas = 5 Cost of one pizza = $20 Let us frame an expression using PEMDAS: Expression: (20 + 20 + 20 + 20 + 20) ÷ 5 or (5 × 20) ÷ 5 Solution: According to PEMDAS or BODMAS we will first solve the parentheses. (100) ÷ 5 = 20 According to the order of operations, each person needs to pay $20.

Similar to the above-mentioned problem, we have many day-to-day real-life instances where we use order of operations to deal with our problems.

☛Related Articles on Order of Operations

Check out the interesting articles below and learn more about the topic Order of Operations and its applications in detail.

  • Subtraction
  • Multiplication
  • Order of Operations Worksheets 5th Grade

Order of Operations Examples

Example 1: Help Jack in solving the following problem with the help of order of operations rules. a) 18 ÷ (9 - 2 × 3)

Solution : Given expression: 18 ÷ (9 - 2 × 3) According to the order of operations rule, we have to solve parentheses first. Please note here inside the parentheses we have two operations present, multiplication, and subtraction. First, multiply 2 × 3 = 6 18 ÷ (9 - 6) Now subtract 6 from 9, 18 ÷ (3) Now divide 18 ÷ 3 = 6

Example 2: Simplify the given expression using the order of operations rules. ​(6 × 2 - 6 - 1) × 2 2

Solution: We know that the order of operation follows either PEMDAS or BODMAS . Let us follow the order of operations rules and simplify the given expression.

Step 1: First, we need to solve the numbers within the parentheses. Multiply 6 by 2 in the given expression, ​​​​​(6 × 2 - 6 - 1) × 2 2 , we get, (12 - 6 - 1) × 2 2 . Step 2- Now, we need to subtract 6 from 12 inside the bracket, so, we get, (6 - 1) × 2 2 . Step 3- Remove parentheses after subtracting 6 - 1, we get, 5 × 2 2 . Step 4- Solve exponent, i.e 2 2 = 4. Step 5- Multiply 5 by 4 to get the final answer, which is, 5 × 4 = 20. ∴ ​(6 × 2 - 6 - 1) × 2 2 = 20 .

Example 3: Evaluate the expression using the order of operations: (1 + 20 − 9 ÷ 3 2 ) ÷ ((2 + 1) 2 + 16 ÷ 2)

Solution: Let us see how we can apply the rules of the order of operations in solving the given expression. Step 1: First, we need to simplify the innermost bracket, (1 + 20 − 9 ÷ 3 2 ) ÷ (3 2 + 16 ÷ 2) Step 2: Now we have to evaluate exponents, (1 + 20 − 9 ÷ 9) ÷ (9 + 16 ÷ 2) Step 3: Now, we need to divide 9 by 9 and 16 by 2 inside the brackets, and we get, (1 + 20 − 1) ÷ (9 + 8) Step 4: Adding 1 and 20 we get 21. Now subtract 1 from 21 we get 20. Now, (20) ÷ (9 + 8) Step 5: Add 9 + 8 and divide the result by 20. 20 ÷ 17 = 20/17 Step 6: ∴ (1 + 20 − 9 ÷ 3 2 ) ÷ ((2 + 1) 2 + 16 ÷ 2) = 20/17

Example 4: Solve the statement problem using the order of operations. If 72 is divided by the sum of 4 and 5, then subtracted from 10, what will be the final answer?

Solution: Let us first write the given statement into mathematical form. 10 - [72 ÷ (4 + 5)] Using the order of operations rules this expression can be simplified as: = 10 - [72 ÷ (4 + 5)] = 10 - [72 ÷ 9] = 10 - 8 = 2 ∴ 10 - [72 ÷ (4 + 5)] = 2

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solve the following problem 8 212 × $19 =

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Practice Questions on Order of Operations

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FAQs on Order of Operations

What is the order of operations in math.

The order of operation in math is a set of rules revolving around 4 major operators. According to the order of operations, there is a particular sequence which we need to follow on each operator while solving the given mathematical expression.

How to Solve Order of Operations?

To solve the order of operations first observe the expression and note what pattern it exactly follows. Now start using either PEMDAS or BODMAS to solve the given expression. As per the rules of order of operations do look for parentheses first, then exponents, then move towards multiplication or division and addition or subtraction from left to right.

How to Do Order of Operations with Integers?

We know that integers are positive numbers and negative numbers. We can easily perform the order of operations with integers by following the given steps:

  • Look for the integers present inside the parentheses or brackets and solve them.
  • After solving the integers in the parentheses, look for any integer term present in the form of exponents and solve it.
  • Now we are left with the basic four operators to be performed on integers. Look for the integers with the operation of multiplication or division and solve them from the left-hand side to the right-hand side.
  • Lastly, look for the integers with addition or subtraction and solve them.
  • In the case of integers, we need to make sure we are properly multiplying the signs. Such that, (-) × (-) = + and (+) × (-) = -

How to Remember Order of Operations?

To remember the order of operations we use two famous acronyms, i.e. PEMDAS and BODMAS. We use either of the two according to the rules of order of operations. PEMDAS or BODMAS helps in remembering the process of solving any order of operation for any n number of expressions.

How to do Order of Operations with Exponents?

According to the PEMDAS, the letter E stands for exponents which come as the second step in order of operations. Let us look at the given example to clearly understand how to do the order of operations with exponents. Expression: 7 × (2 2 ) Solution: 7 × ( 4 ) = 28 (Correct (✔). This is a correct way to solve the exponents) Let us look at another approach for the same expression. 7 × ( 2 2 ) = 14 2 = 196 ((Incorrect (✘). This is an incorrect way to do order of operations with exponents)

What is the Correct Order of Operations?

The correct order of operations can be easily expressed by using the word PEMDAS or BODMAS. The two words can be described as PEMDAS (Parentheses, Exponents, Multiplication or Division, and Addition or Subtraction). Similarly, for BODMAS (Brackets, Orders, Division, Multiplication, Addition, and Subtraction.)

What is the Order of Operations without Parentheses?

Going by the rules of the order of operations if we remove the parentheses, then we are left with EMDAS. EMDAS stands for (Exponents, Multiplication or Division, and Addition or Subtraction). If in the expression we don't have any exponential term then we need to perform multiplication or division first and moving forward we proceed with addition or subtraction. The situation may vary according to the operators present in the given expression

When Do We Use Order of Operations?

A lot of instances in our life pass through some sort of order of operations to perform it well. Every day we encounter such a scenario. For example, going to the grocery market and purchasing things we quickly perform the order of operations in our head. This helps us in reducing the turnaround time at the billing counter.

What Operation Is Completed First In the Order of Operations?

In the above sections, we read about two acronyms BODMAS and PEMDAS. According to both the acronyms, in order of operations, we simplify parentheses or the brackets first.

What is the Use of Order of Operation Calculator?

Order of operations calculator is an online tool and the fastest method with which we can evaluate any given numerical expression keeping the order of operations rules in mind. To use the order of operation calculator we need to enter the numerical expression in the correct format. Try Cuemath's order of operations calculator and solve the expressions quickly within a seconds.

☛Also Check:

For more practice try these:

  • Order of Operations With Exponents Worksheets
  • Advanced Order of Operations Worksheets
  • Order of Operations Worksheets

What Are the 4 Order of Operations?

The 4 major order of operations are:

  • Parentheses.
  • An exponential term.
  • Multiplication or division .
  • At the end addition or subtraction.

The four order of operations can be easily recalled at any given point in time by learning the acronyms PEMD AS or B ODMAS .

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  • x+y+z=25,\:5x+3y+2z=0,\:y-z=6
  • x+2y=2x-5,\:x-y=3
  • 5x+3y=7,\:3x-5y=-23
  • x^2+y=5,\:x^2+y^2=7
  • xy+x-4y=11,\:xy-x-4y=4
  • 3-x^2=y,\:x+1=y
  • xy=10,\:2x+y=1
  • How do you solve a system of equations by substitution?
  • To solve a system of equations by substitution, solve one of the equations for one of the variables, and substitute this expression into the other equation. Then, solve the resulting equation for the remaining variable and substitute this value back into the original equation to find the value of the other variable.
  • How do you solve a system of equations by graphing?
  • To solve a system of equations by graphing, graph both equations on the same set of axes and find the points at which the graphs intersect. Those points are the solutions.
  • How do you solve a system of equations by elimination?
  • To solve a system of equations by elimination, write the system of equations in standard form: ax + by = c, and multiply one or both of the equations by a constant so that the coefficients of one of the variables are opposite. Then, add or subtract the two equations to eliminate one of the variables. Solve the resulting equation for the remaining variable.
  • What are the solving methods of a system of equations?
  • There are several methods for solving a system of equations, including substitution, elimination, and graphing.
  • What is a system of linear equations?
  • A system of linear equations is a system of equations in which all the equations are linear and in the form ax + by = c, where a, b, and c are constants and x and y are variables.

system-of-equations-calculator

  • High School Math Solutions – Systems of Equations Calculator, Nonlinear In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how... Read More

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  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
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  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Evaluate  2 × 2  determinants.
  • Use Cramer’s Rule to solve a system of equations in two variables.
  • Evaluate  3 × 3  determinants.
  • Use Cramer’s Rule to solve a system of three equations in three variables.
  • Know the properties of determinants.

We have learned how to solve systems of equations in two variables and three variables, and by multiple methods: substitution, addition, Gaussian elimination, using the inverse of a matrix, and graphing. Some of these methods are easier to apply than others and are more appropriate in certain situations. In this section, we will study two more strategies for solving systems of equations.

Evaluating the Determinant of a 2×2 Matrix

A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an invertible matrix and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section.

Find the Determinant of a 2 × 2 Matrix

The determinant of a 2 × 2 2 × 2 matrix, given

is defined as

Notice the change in notation. There are several ways to indicate the determinant, including det ( A ) det ( A ) and replacing the brackets in a matrix with straight lines, | A | . | A | .

Finding the Determinant of a 2 × 2 Matrix

Find the determinant of the given matrix.

Using Cramer’s Rule to Solve a System of Two Equations in Two Variables

We will now introduce a final method for solving systems of equations that uses determinants. Known as Cramer’s Rule , this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704-1752), who introduced it in 1750 in Introduction à l'Analyse des lignes Courbes algébriques . Cramer’s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns.

Cramer’s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used.

To understand Cramer’s Rule, let’s look closely at how we solve systems of linear equations using basic row operations. Consider a system of two equations in two variables.

We eliminate one variable using row operations and solve for the other. Say that we wish to solve for x . x . If equation (2) is multiplied by the opposite of the coefficient of y y in equation (1), equation (1) is multiplied by the coefficient of y y in equation (2), and we add the two equations, the variable y y will be eliminated.

Now, solve for x . x .

Similarly, to solve for y , y , we will eliminate x . x .

Solving for y y gives

Notice that the denominator for both x x and y y is the determinant of the coefficient matrix.

We can use these formulas to solve for x x and y , y , but Cramer’s Rule also introduces new notation:

  • D : D : determinant of the coefficient matrix
  • D x : D x : determinant of the numerator in the solution of x x x = D x D x = D x D
  • D y : D y : determinant of the numerator in the solution of y y y = D y D y = D y D

The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express x x and y y as a quotient of two determinants.

Cramer’s Rule for 2×2 Systems

Cramer’s Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables.

Consider a system of two linear equations in two variables.

The solution using Cramer’s Rule is given as

If we are solving for x , x , the x x column is replaced with the constant column. If we are solving for y , y , the y y column is replaced with the constant column.

Using Cramer’s Rule to Solve a 2 × 2 System

Solve the following 2 × 2 2 × 2 system using Cramer’s Rule.

Solve for x . x .

Solve for y . y .

The solution is ( 2 , −3 ) . ( 2 , −3 ) .

Use Cramer’s Rule to solve the 2 × 2 system of equations.

Evaluating the Determinant of a 3 × 3 Matrix

Finding the determinant of a 2×2 matrix is straightforward, but finding the determinant of a 3×3 matrix is more complicated. One method is to augment the 3×3 matrix with a repetition of the first two columns, giving a 3×5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example.

Find the determinant of the 3×3 matrix.

  • Augment A A with the first two columns. det ( A ) = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | a 1 a 2 a 3 b 1 b 2 b 3 | det ( A ) = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | a 1 a 2 a 3 b 1 b 2 b 3 |
  • From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal.
  • From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal.

The algebra is as follows:

Finding the Determinant of a 3 × 3 Matrix

Find the determinant of the 3 × 3 matrix given

Augment the matrix with the first two columns and then follow the formula. Thus,

Find the determinant of the 3 × 3 matrix.

Can we use the same method to find the determinant of a larger matrix?

No, this method only works for 2 × 2 2 × 2 and 3 × 3 3 × 3 matrices. For larger matrices it is best to use a graphing utility or computer software.

Using Cramer’s Rule to Solve a System of Three Equations in Three Variables

Now that we can find the determinant of a 3 × 3 matrix, we can apply Cramer’s Rule to solve a system of three equations in three variables . Cramer’s Rule is straightforward, following a pattern consistent with Cramer’s Rule for 2 × 2 matrices. As the order of the matrix increases to 3 × 3, however, there are many more calculations required.

When we calculate the determinant to be zero, Cramer’s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system.

Consider a 3 × 3 system of equations.

If we are writing the determinant D x , D x , we replace the x x column with the constant column. If we are writing the determinant D y , D y , we replace the y y column with the constant column. If we are writing the determinant D z , D z , we replace the z z column with the constant column. Always check the answer.

Solving a 3 × 3 System Using Cramer’s Rule

Find the solution to the given 3 × 3 system using Cramer’s Rule.

Use Cramer’s Rule.

The solution is ( 1 , 3 , −2 ) . ( 1 , 3 , −2 ) .

Use Cramer’s Rule to solve the 3 × 3 matrix.

Using Cramer’s Rule to Solve an Inconsistent System

Solve the system of equations using Cramer’s Rule.

We begin by finding the determinants D , D x , and  D y . D , D x , and  D y .

We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables.

  • Multiply equation (1) by −2. −2.
  • Add the result to equation ( 2 ) . ( 2 ) .

We obtain the equation 0 = −8 , 0 = −8 , which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 1 .

Use Cramer’s Rule to Solve a Dependent System

Solve the system with an infinite number of solutions.

Let’s find the determinant first. Set up a matrix augmented by the first two columns.

As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out.

  • Multiply equation (1) by −2 −2 and add the result to equation (3): − 2 x + 4 y − 6 z = 0 2 x − 4 y + 6 z = 0 0 = 0 − 2 x + 4 y − 6 z = 0 2 x − 4 y + 6 z = 0 0 = 0
  • Obtaining an answer of 0 = 0 , 0 = 0 , a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See Figure 2 .

Understanding Properties of Determinants

There are many properties of determinants . Listed here are some properties that may be helpful in calculating the determinant of a matrix.

Properties of Determinants

  • If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal.
  • When two rows are interchanged, the determinant changes sign.
  • If either two rows or two columns are identical, the determinant equals zero.
  • If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero.
  • The determinant of an inverse matrix A − 1 A − 1 is the reciprocal of the determinant of the matrix A . A .
  • If any row or column is multiplied by a constant, the determinant is multiplied by the same factor.

Illustrating Properties of Determinants

Illustrate each of the properties of determinants.

Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal.

Augment A A with the first two columns.

Property 2 states that interchanging rows changes the sign. Given

Property 3 states that if two rows or two columns are identical, the determinant equals zero.

Property 4 states that if a row or column equals zero, the determinant equals zero. Thus,

Property 5 states that the determinant of an inverse matrix A − 1 A − 1 is the reciprocal of the determinant A . A . Thus,

Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus,

Using Cramer’s Rule and Determinant Properties to Solve a System

Find the solution to the given 3 × 3 system.

Using Cramer’s Rule , we have

Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out.

  • Multiply equation (3) by –2 and add the result to equation (1). − 2 x − 4 y − 4 x = − 8    2 x + 4 y + 4 z = 2 0 = − 6 − 2 x − 4 y − 4 x = − 8    2 x + 4 y + 4 z = 2 0 = − 6

Obtaining a statement that is a contradiction means that the system has no solution.

Access these online resources for additional instruction and practice with Cramer’s Rule.

  • Solve a System of Two Equations Using Cramer's Rule
  • Solve a Systems of Three Equations using Cramer's Rule

7.8 Section Exercises

Explain why we can always evaluate the determinant of a square matrix.

Examining Cramer’s Rule, explain why there is no unique solution to the system when the determinant of your matrix is 0. For simplicity, use a 2 × 2 2 × 2 matrix.

Explain what it means in terms of an inverse for a matrix to have a 0 determinant.

The determinant of 2 × 2 2 × 2 matrix A A is 3. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer.

For the following exercises, find the determinant.

| 1 2 3 4 | | 1 2 3 4 |

| − 1 2 3 − 4 | | − 1 2 3 − 4 |

| 2 − 5 − 1 6 | | 2 − 5 − 1 6 |

| − 8 4 − 1 5 | | − 8 4 − 1 5 |

| 1 0 3 − 4 | | 1 0 3 − 4 |

| 10 20 0 − 10 | | 10 20 0 − 10 |

| 10 0.2 5 0.1 | | 10 0.2 5 0.1 |

| 6 − 3 8 4 | | 6 − 3 8 4 |

| − 2 − 3 3.1 4 , 000 | | − 2 − 3 3.1 4 , 000 |

| − 1.1 0.6 7.2 − 0.5 | | − 1.1 0.6 7.2 − 0.5 |

| − 1 0 0 0 1 0 0 0 − 3 | | − 1 0 0 0 1 0 0 0 − 3 |

| − 1 4 0 0 2 3 0 0 − 3 | | − 1 4 0 0 2 3 0 0 − 3 |

| 1 0 1 0 1 0 1 0 0 | | 1 0 1 0 1 0 1 0 0 |

| 2 − 3 1 3 − 4 1 − 5 6 1 | | 2 − 3 1 3 − 4 1 − 5 6 1 |

| − 2 1 4 − 4 2 − 8 2 − 8 − 3 | | − 2 1 4 − 4 2 − 8 2 − 8 − 3 |

| 6 − 1 2 − 4 − 3 5 1 9 − 1 | | 6 − 1 2 − 4 − 3 5 1 9 − 1 |

| 5 1 − 1 2 3 1 3 − 6 − 3 | | 5 1 − 1 2 3 1 3 − 6 − 3 |

| 1.1 2 − 1 − 4 0 0 4.1 − 0.4 2.5 | | 1.1 2 − 1 − 4 0 0 4.1 − 0.4 2.5 |

| 2 − 1.6 3.1 1.1 3 − 8 − 9.3 0 2 | | 2 − 1.6 3.1 1.1 3 − 8 − 9.3 0 2 |

| − 1 2 1 3 1 4 1 5 − 1 6 1 7 0 0 1 8 | | − 1 2 1 3 1 4 1 5 − 1 6 1 7 0 0 1 8 |

For the following exercises, solve the system of linear equations using Cramer’s Rule.

2 x − 3 y = −1 4 x + 5 y = 9 2 x − 3 y = −1 4 x + 5 y = 9

5 x − 4 y = 2 − 4 x + 7 y = 6 5 x − 4 y = 2 − 4 x + 7 y = 6

6 x − 3 y = 2 − 8 x + 9 y = −1 6 x − 3 y = 2 − 8 x + 9 y = −1

2 x + 6 y = 12 5 x − 2 y = 13 2 x + 6 y = 12 5 x − 2 y = 13

4 x + 3 y = 23 2 x − y = −1 4 x + 3 y = 23 2 x − y = −1

10 x − 6 y = 2 − 5 x + 8 y = −1 10 x − 6 y = 2 − 5 x + 8 y = −1

4 x − 3 y = −3 2 x + 6 y = −4 4 x − 3 y = −3 2 x + 6 y = −4

4 x − 5 y = 7 − 3 x + 9 y = 0 4 x − 5 y = 7 − 3 x + 9 y = 0

4 x + 10 y = 180 − 3 x − 5 y = −105 4 x + 10 y = 180 − 3 x − 5 y = −105

8 x − 2 y = −3 − 4 x + 6 y = 4 8 x − 2 y = −3 − 4 x + 6 y = 4

x + 2 y − 4 z = − 1 7 x + 3 y + 5 z = 26 − 2 x − 6 y + 7 z = − 6 x + 2 y − 4 z = − 1 7 x + 3 y + 5 z = 26 − 2 x − 6 y + 7 z = − 6

− 5 x + 2 y − 4 z = − 47 4 x − 3 y − z = − 94 3 x − 3 y + 2 z = 94 − 5 x + 2 y − 4 z = − 47 4 x − 3 y − z = − 94 3 x − 3 y + 2 z = 94

4 x + 5 y − z = −7 −2 x − 9 y + 2 z = 8 5 y + 7 z = 21 4 x + 5 y − z = −7 −2 x − 9 y + 2 z = 8 5 y + 7 z = 21

4 x − 3 y + 4 z = 10 5 x − 2 z = − 2 3 x + 2 y − 5 z = − 9 4 x − 3 y + 4 z = 10 5 x − 2 z = − 2 3 x + 2 y − 5 z = − 9

4 x − 2 y + 3 z = 6 − 6 x + y = − 2 2 x + 7 y + 8 z = 24 4 x − 2 y + 3 z = 6 − 6 x + y = − 2 2 x + 7 y + 8 z = 24

5 x + 2 y − z = 1 − 7 x − 8 y + 3 z = 1.5 6 x − 12 y + z = 7 5 x + 2 y − z = 1 − 7 x − 8 y + 3 z = 1.5 6 x − 12 y + z = 7

13 x − 17 y + 16 z = 73 − 11 x + 15 y + 17 z = 61 46 x + 10 y − 30 z = − 18 13 x − 17 y + 16 z = 73 − 11 x + 15 y + 17 z = 61 46 x + 10 y − 30 z = − 18

− 4 x − 3 y − 8 z = − 7 2 x − 9 y + 5 z = 0.5 5 x − 6 y − 5 z = − 2 − 4 x − 3 y − 8 z = − 7 2 x − 9 y + 5 z = 0.5 5 x − 6 y − 5 z = − 2

4 x − 6 y + 8 z = 10 − 2 x + 3 y − 4 z = − 5 x + y + z = 1 4 x − 6 y + 8 z = 10 − 2 x + 3 y − 4 z = − 5 x + y + z = 1

4 x − 6 y + 8 z = 10 − 2 x + 3 y − 4 z = − 5 12 x + 18 y − 24 z = − 30 4 x − 6 y + 8 z = 10 − 2 x + 3 y − 4 z = − 5 12 x + 18 y − 24 z = − 30

For the following exercises, use the determinant function on a graphing utility.

| 1 0 8 9 0 2 1 0 1 0 3 0 0 2 4 3 | | 1 0 8 9 0 2 1 0 1 0 3 0 0 2 4 3 |

| 1 0 2 1 0 −9 1 3 3 0 −2 −1 0 1 1 −2 | | 1 0 2 1 0 −9 1 3 3 0 −2 −1 0 1 1 −2 |

| 1 2 1 7 4 0 1 2 100 5 0 0 2 2,000 0 0 0 2 | | 1 2 1 7 4 0 1 2 100 5 0 0 2 2,000 0 0 0 2 |

| 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 0 | | 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 0 |

Real-World Applications

For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution.

Two numbers add up to 56. One number is 20 less than the other.

Two numbers add up to 104. If you add two times the first number plus two times the second number, your total is 208

Three numbers add up to 106. The first number is 3 less than the second number. The third number is 4 more than the first number.

Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined.

For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule.

You invest $10,000 into two accounts, which receive 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account?

You invest $80,000 into two accounts, $22,000 in one account, and $58,000 in the other account. At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts?

A theater needs to know how many adult tickets and children tickets were sold out of the 1,200 total tickets. If children’s tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many children’s tickets and adult tickets were sold?

A concert venue sells single tickets for $40 each and couple’s tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many couple’s tickets were sold?

You decide to paint your kitchen green. You create the color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix?

You sold two types of scarves at a farmers’ market and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold?

Your garden produced two types of tomatoes, one green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have?

At a market, the three most popular vegetables make up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share?

At the same market, the three most popular fruits make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold.

Three artists performed at a concert venue. The first one charged $15 per ticket, the second artist charged $45 per ticket, and the final one charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band?

A movie theatre sold tickets to three movies. The tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie were sold?

For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate-covered cashews. The nutritional information for these items is shown in Table 1 .

For the special “low-carb”trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix?

For the “hiking” mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix?

For the “energy-booster” mix, there are 1,000 pieces in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix?

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Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
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  • Publisher/website: OpenStax
  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Section URL: https://openstax.org/books/college-algebra-2e/pages/7-8-solving-systems-with-cramers-rule

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Algebra: Ticket Word Problems

Related Pages Word Problems Involving Age Coin Word Problems More Algebra Lessons

Ticket problems are algebra word problems that involve various tickets of different values.

Ticket problems are word problems similar to coin problems and stamp problems as tickets may be denominated in specific values.

Be careful to distinguish between the value of the items and the quantity of the items.

A table can be useful for distinguishing between quantity and value in this type of word problems.

Example: The cost of tickets for a play is $3.00 for adults and $2.00 for children. 350 tickets were sold and $950 was collected. How many tickets of each type were sold?

Solution: Step 1: Set up a table with quantity and value.

Step 2: Fill in the table with information from the question.

The cost of tickets for a play is $ 3.00 for adults and $ 2.00 for children. 350 tickets were sold and $ 950 was collected. How many tickets of each type were sold?

Let x = number of $3 tickets Let y = number of $2 tickets Total = quantity × value

Step 3: Add down each column to get the equations

x + y = 350                     (equation 1) 3x + 2y = 950                 (equation 2)

Use Substitution Method Isolate variable x in equation 1

x = 350 – y                      (equation 3)

Substitute equation 3 into equation 2

3(350 – y) + 2y = 950 1050 – 3y + 2y = 950 3y – 2y = 1050 – 950 y = 100

Substitute y = 100 into equation 1

x + 100 = 350 x = 250

Answer: 250 $3 tickets and 100 $2 tickets were sold.

Ticket Word Problems

Example: Jim sold 120 tickets to a game. Adult tickets are $24 each and children tickets are $13 each. Sales are $2110. How many of each was sold?

Word Problems With Theater Attendance

Example: 500 people see a play. Children are charged $15 and adults $25. Ticket proceeds are $11,250. How many children attended the play?

Algebra - Word Problems: With Movie Tickets

Example: For a theater showing 202 tickets were sold. A child’s ticket costs $6.00 and an adult’s ticket costs $10.00. How many tickets of each were sold if the receipts totaled $1708.00?

Solve Ticket Word Problems In Three Variables

Example: The Goonies sold 785 concert tickets for a total of $17,650. If good tickets cost $40 and cheap tickets cost $15, how many of each type of ticket did they sell?

Example: At the arcade, Sammy won two blue tickets, 1 yellow ticket and 3 red tickets for a total of 1,500 points. Jamal won 1 blue, 2 yellow and 2 red for 1225 points. Yvonne won 2 blue, 3 yellow and 1 red for 1200 points. How much is each ticket worth?

Word Problem - Mixture Of Ticket Sales

Example: For opening night 328 tickets were sold. Students paid $2 each while non-students paid $4 each. If a total of $910 were collected, how many students and how many non-students attended?

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Solving Percent Problems

Learning Objective(s)

·          Identify the amount, the base, and the percent in a percent problem.

·          Find the unknown in a percent problem.

Introduction

Percents are a ratio of a number and 100. So they are easier to compare than fractions, as they always have the same denominator, 100. A store may have a 10% off sale. The amount saved is always the same portion or fraction of the price, but a higher price means more money is taken off. Interest rates on a saving account work in the same way. The more money you put in your account, the more money you get in interest. It’s helpful to understand how these percents are calculated.

Parts of a Percent Problem

Jeff has a coupon at the Guitar Store for 15% off any purchase of $100 or more. He wants to buy a used guitar that has a price tag of $220 on it. Jeff wonders how much money the coupon will take off the original $220 price.

Problems involving percents have any three quantities to work with: the percent , the amount , and the base .

The percent has the percent symbol (%) or the word “percent.” In the problem above, 15% is the percent off the purchase price.

The base is the whole amount. In the problem above, the whole price of the guitar is $220, which is the base.

The amount is the number that relates to the percent. It is always part of the whole. In the problem above, the amount is unknown. Since the percent is the percent off , the amount will be the amount off of the price .

You will return to this problem a bit later. The following examples show how to identify the three parts, the percent, the base, and the amount.

The previous problem states that 30 is a portion of another number. That means 30 is the amount. Note that this problem could be rewritten: 20% of what number is 30?

Solving with Equations

Percent problems can be solved by writing equations. An equation uses an equal sign (= ) to show that two mathematical expressions have the same value.

Percents are fractions, and just like fractions, when finding a percent (or fraction, or portion) of another amount, you multiply.

The percent of the base is the amount.

Percent of the Base is the Amount.

Percent · Base = Amount

Once you have an equation, you can solve it and find the unknown value. To do this, think about the relationship between multiplication and division. Look at the pairs of multiplication and division facts below, and look for a pattern in each row.

Multiplication and division are inverse operations. What one does to a number, the other “undoes.”

When you have an equation such as 20% · n = 30, you can divide 30 by 20% to find the unknown: n =  30 ÷ 20%.

You can solve this by writing the percent as a decimal or fraction and then dividing.

n = 30 ÷ 20% =  30 ÷ 0.20 = 150

You can estimate to see if the answer is reasonable. Use 10% and 20%, numbers close to 12.5%, to see if they get you close to the answer.

10% of 72 = 0.1 · 72 = 7.2

20% of 72 = 0.2 · 72 = 14.4

Notice that 9 is between 7.2 and 14.4, so 12.5% is reasonable since it is between 10% and 20%.

This problem is a little easier to estimate. 100% of 24 is 24. And 110% is a little bit more than 24. So, 26.4 is a reasonable answer.

Using Proportions to Solve Percent Problems

Let’s go back to the problem that was posed at the beginning. You can now solve this problem as shown in the following example.

You can estimate to see if the answer is reasonable. Since 15% is half way between 10% and 20%, find these numbers.

10% of 220 = 0.1 · 220 = 22

20% of 220 = 0.2 · 220 = 44

The answer, 33, is between 22 and 44. So $33 seems reasonable.

There are many other situations that involve percents. Below are just a few.

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  21. Ticket Word Problems (video lessons, examples, solutions)

    Solution: Step 1: Set up a table with quantity and value. Step 2: Fill in the table with information from the question. The cost of tickets for a play is $ 3.00 for adults and $ 2.00 for children. 350 tickets were sold and $ 950 was collected. How many tickets of each type were sold? Let x = number of $3 tickets Let y = number of $2 tickets

  22. Solving Percent Problems

    Move the decimal point two places to the right to write the decimal as a percent. Answer. 12.5% of 72 is 9. You can estimate to see if the answer is reasonable. Use 10% and 20%, numbers close to 12.5%, to see if they get you close to the answer. 10% of 72 = 0.1 · 72 = 7.2. 20% of 72 = 0.2 · 72 = 14.4.

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