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8th grade (Eureka Math/EngageNY)
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, equals, 2
5, equals, 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 left parenthesis, start color #11accd, 5, end color #11accd, comma, start color #1fab54, 1, end color #1fab54, right parenthesis.
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, equals, 2 into one of the original equations, and solve for x.
So our solution is left parenthesis, 3, comma, 2, right parenthesis.
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, equals, 4 into one of the original equations, and solve for y.
So our solution is left parenthesis, 4, comma, minus, 2, right parenthesis.
Let's practice!
Challenge problem
Want to join the conversation?
- 6x=10y-10
X+y+7= 0
Solve by substitution(4 votes) - Solve the system of equation and find the largest value of x+y?
problem:
x+6y=5,
xy=1(2 votes) - (x+9/4)+(y-1/3)=1
2x-y=11
solve by elimination(2 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)