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### Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Lesson 4: Topic D: Systems of linear equations and their solutions- Systems of equations: trolls, tolls (1 of 2)
- Systems of equations: trolls, tolls (2 of 2)
- Testing a solution to a system of equations
- Solutions of systems of equations
- Systems of equations with graphing
- Systems of equations with graphing
- Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: chores
- Systems of equations with graphing
- Systems of equations with elimination: 3t+4g=6 & -6t+g=6
- Systems of equations with elimination
- Systems of equations with elimination: x+2y=6 & 4x-2y=14
- Systems of equations with elimination: -3y+4x=11 & y+2x=13
- Systems of equations with elimination: 2x-y=14 & -6x+3y=-42
- Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
- Systems of equations with elimination challenge
- Systems of equations with substitution: 2y=x+7 & x=y-4
- Systems of equations with substitution
- Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution: 9x+3y=15 & y-x=5
- Systems of equations with substitution
- Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
- Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
- Systems of equations with elimination: apples and oranges
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: King's cupcakes
- Systems of equations with elimination: Sum/difference of numbers
- Systems of equations with elimination: potato chips
- Systems of equations with elimination: coffee and croissants
- Systems of equations with substitution: coins
- Systems of equations with substitution: potato chips
- Systems of equations with substitution: shelves
- Systems of equations word problems
- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- Solutions to systems of equations: consistent vs. inconsistent
- Systems of equations number of solutions: fruit prices (1 of 2)
- Systems of equations number of solutions: fruit prices (2 of 2)
- Systems of equations number of solutions: y=3x+1 & 2y+4=6x
- Solutions to systems of equations: dependent vs. independent
- Number of solutions to a system of equations graphically
- Number of solutions to a system of equations graphically
- Forming systems of equations with different numbers of solutions
- Number of solutions to a system of equations algebraically
- Comparing Celsius and Fahrenheit temperature scales
- Converting Fahrenheit to Celsius

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# Systems of equations with elimination

Walk through examples of solving systems of equations with elimination.

In this article, we're going to be solving systems of linear equations using a strategy called elimination. First, we need to understand that it's okay to add equations together.

**Key idea:**Anytime we have two true equations, we can add or subtract them to create another true equation.

For example, here are two very basic true equations:

$2=2$ $5=5$

We can add these equations together to create another true equation:

Or we can subtract these equations to create another true equation:

Here's another example with more complicated equations:

Great, now that we see that it's okay to add or subtract equations, we can move onto solving a system of equations using elimination.

## Solving a system of equations using elimination

We'll solve this system of equations as an example:

The hard thing about solving is that there are two variables $x$ and $y$ . If only we could get rid of one of the variables...

Here's an idea! Let's add the two equations together to cancel out the $y$ variable:

Brilliant! Now we have an equation with just the $x$ variable that we know how to solve:

Baller! Let's use the first equation to find $y$ when $x$ equals $5$ :

Sweet! So the solution to the system of equations is $({5},{1})$ .

## Multiplying one of the equations by a constant, then using elimination

That last example worked out great because the $y$ variable was eliminated when we added the equations. Sometimes it isn't quite that easy.

Take this system of equation as an example:

If we add these equations, neither the $x$ or $y$ variable will be eliminated, so that won't work. Here are the steps for problems like this:

**Step 1: Multiply one of the equations by a constant so that when we add it to the other equation, one of the variables is eliminated.**

**Step 2: Add the new equation to the equation we didn't use in step 1 in order to eliminate one of the variables.**

**Step 3: Solve for**$y$ .

**Step 4: Substitute**$y=2$ into one of the original equations, and solve for $x$ .

So our solution is $(3,2)$ .

## Multiplying *both* of the equations by a constant, then using elimination

Sometimes we'll need to multiply both equations by a constant when using elimination.

For example, consider this system of equations:

Here are the steps to solve a system of equations like this one:

**Step 1: Multiply each equation by a constant so we can eliminate one variable.**

**Step 2: Combine the new equations to eliminate one variable.**

**Step 3: Substitute**$x=4$ into one of the original equations, and solve for $y$ .

So our solution is $(4,-2)$ .

## Let's practice!

## Challenge problem

## Want to join the conversation?

- 6x=10y-10

X+y+7= 0

Solve by substitution(5 votes) - Solve the system of equation and find the largest value of x+y?

problem:

x+6y=5,

xy=1(3 votes) - (x+9/4)+(y-1/3)=1

2x-y=11

solve by elimination(3 votes) - Ok when it comes to elimination my multiplying both equations so that we can can cancel out variable, it is not making sense `to me because as it shows on the article the variables that are being canceled are both positive. I am not understanding how they both cancel out.(2 votes)
- Check out this video to help you get started. https://www.khanacademy.org/math/arithmetic/arith-review-negative-numbers/arith-review-number-opposites/v/opposite-of-a-number(1 vote)

- Honestly, I cannot stand graphing. I consider myself to be great at math but when it comes to graphing, no matter how hard I try I just can't remember how to do them.(2 votes)
- find all the values of x such that y=0. y=2[2x-3(x-4)]-5(x-4)(2 votes)
- how do we solve equations with the process of elimination? please help asap(1 vote)
- Elimination works best when you can make the coefficients (numbers) of x or y match but be opposite in sign.

Suppose your problem is

2x+4y=8

3x+5y=12

Since, 2x ≠ 3x or 4y≠5y, (1) pick your variable to "eliminate" and (2) scale them up with multiplication to equal but opposite values.

Like 2x(3)=6x and 3x(-2)=-6x.

Multiply all of the equation by your factor.

[2x+4y=8]•3 and [3x+5y=12]•-2 to get

6x+12y=24 and -6x-10y=-24.

Since we are dealing with equality, left sides of both equations can be added together, and right sides can be added together.

(6x+12y) + (-6x-10y) = 24+(-24)

2y=0

So, y=0.

Once you know y=0, go back to one of your (original) equations. Substitute y=0 into the equation and solve for x.

2x+4(0)=8

2x=8

x=4(2 votes)

- hi this is defeinetly a question sure it is please dont ban me or suspend this acount(1 vote)
- Ok so the equation is 2x+2y=28 and x*y=40 and I made it 40/x=-1x+14 (might be wrong) and now I have no idea what to do(1 vote)
- So far, you're right! Now, you substitute y for x+14.(1 vote)

- (y-3)(2y+4)+8=(y-3)(y-2)(1 vote)