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### Course: Integrated math 1>Unit 6

Lesson 3: Equivalent systems of equations

# Equivalent systems of equations review

Two systems of equations are equivalent if they have the same solution(s). This article reviews how to tell if two systems are equivalent.
Systems of equations that have the same solution are called equivalent systems.
Given a system of two equations, we can produce an equivalent system by replacing one equation by the sum of the two equations, or by replacing an equation by a multiple of itself.
In contrast, we can be sure that two systems of equations are not equivalent if we know that a solution of the one is not a solution of the other.
Note: This idea of equivalent systems of equations pops up again in linear algebra. However, the examples and explanations in this article are geared to a high school algebra 1 class.

## Example 1

We're given two systems of equations and asked if they're equivalent.
System ASystem B
$\begin{array}{r}-12x+9y=7\\ \\ 9x-12y=6\end{array}$$\begin{array}{r}-12x+9y=7\\ \\ 3x-4y=2\end{array}$
If we multiply the second equation in System B by $3$, we get:
$\begin{array}{rl}3x-4y& =2\\ \\ 3\left(3x-4y\right)& =3\left(2\right)\\ \\ 9x-12y& =6\end{array}$
Replacing the second equation of System B with this new equation, we get an equivalent system:
$\begin{array}{r}-12x+9y=7\\ \\ 9x-12y=6\end{array}$
Whoa! Look at that! This system is the same as System A, which means system A is equivalent to System B.

## Example 2

We're given two systems of equations and asked if they're equivalent.
System ASystem B
$\begin{array}{rl}-9x-4y& =5\\ \\ 2x+5y& =-4\end{array}$$\begin{array}{rl}-7x+y& =1\\ \\ 2x+5y& =-4\end{array}$
Interestingly, if we sum the equations in System A, we get:
Replacing the first equation in System A with this new equation, we get a system that's equivalent to System A:
$\begin{array}{rl}-7x+y& =1\\ \\ 2x+5y& =-4\end{array}$
Lo and behold! This is System B, which means that System A is equivalent to System B.

## Example 3

We're given two systems and asked to prove that they aren't equivalent by finding a solution of one that is not a solution of the other.
System ASystem B
$\begin{array}{rl}-4x+10y& =1\\ \\ -1x-2y& =-3\end{array}$$\begin{array}{rl}-9x-y& =8\\ \\ -1x-2y& =4\end{array}$
Notice how the coefficients for $x$ and $y$ in the second equations of both systems are the same. However, the constant terms in the two equations are different!
Whichever pair of values for $x$ and $y$ that makes System A true will make System B false, and vice versa.
For example, $x=1$, $y=1$ is a solution to the second equation in System A, but it's not a solution to the second equation in System B.
System A and System B are not equivalent.

## Practice

Problem 1
Elsa and Olaf's teacher gave them a system of linear equations to solve. They each took a few steps that led to the systems shown in the table below.
Teacher
$5x+3y=-1$
$4x-9y=8$
ElsaOlaf
$4x-9y=8$$15x+9y=-3$
$9x-6y=7$$4x-9y=-5$
Which of them obtained a system that is equivalent to the teacher's system?
Remember that two linear systems are "equivalent" if they have the same solution.

Want more practice? Check out this exercise.

## Want to join the conversation?

• What if it had fractions?
• change fractions into whole numbers. remember, whenever you do something to one-side, you will have to do to another-side to maintain the balance (equivalent).

For example, is A,B equivalent to C,D?
----------------------------------------
A) -4x + 10y = 1
B) -5x - 11y = 7
----------------------------------------
C) -9x -y = 8
D) -5/10x +5/4y = 125/1000
----------------------------------------
like Khan always say, try to solve this yourself.
the answer is yes, these 2 systems are equivalent.
C is equivalent to A&B combined.
D times 8 is equivalent to A.

Hope it helps!

**I'm not a pro in Math, in fact, I've just started learning Math about 2 months ago. So, please elaborate if I said something wrong, thanks!**
• When we subtract or add one equation to the second equation in a system of equations, the sum or difference produces an entirely new line, a line with a new slope and y-intercept. This is contrary to my instincts because we are adding or subtracting equivalent values when we add or subtract the second equation from the first. Could you explain to me how we get an entirely unique line as a result? How is it that adding and subtracting equivalent values produces something new?
• The process in which you determine which system of equations are equivalent is confusing. In the examples they tell you which systems are equivalent but it doesn't really explain how you can figure that out.
• Equivalent systems have either: the same lines (obviously); multiples of the original lines; or combinations of the original lines (adding or subtracting two original lines).
• what do you do if you click the wrong one? :(
• It's not a test so I just clicked the right one and then clicked check again. If you do that it doesn't even show you ever got it wrong.
• Why, when you add/subtract one equation from the other on, does it make an equation that is considered to be equal?
(1 vote)
• The properties of equality allow us to add/subtract the same value to both sides of an equation and still have an equivalent equation.

The 2 sides of an equation are equivalent. So, when you add/subtract one equation to another, you are adding/subtracting equal values (same values) to both sides of the equation. The values just don't look the same.

Hope this helps.
• "Replacing the first equation in System A with this new equation, we get a system that's equivalent to System A:"
This sentence from example 2 makes no sense.
• I think he meant "equivalent to System B"
• How can you tell whether or not an elimination process only replaces one side or both?
• You simply do the same thing for both sides. If there is an equal sign, then both sides should be equal.
• Elsa and olaf!! On a math example?!?!
• ayye....can any one help me wth some steps to this...OH and my name is xxxevolution
• XXevolution,

This is the end of the unit. Unless you can ask about specific steps, at this point, I would suggest that you go back through the videos and exercises.

If you have a specific step on which you need help, please explain which step it is.
• Does anyone else also get the sense that the substitution method is a lot nicer (i.e. faster and takes less space on paper), but that unless they give one of the equations in point-intercept form (thus signaling that you should be using substitution over elimination), it just devolves into a mess of fractions that turn out to be uglier and harder to solve than elimination would've been in the first place?

Far's I can tell if all the terms in at least one of the equations don't have a common denominator, it's definitely gonna devolve into a mess of fractions (coz you have to divide all the terms by the coefficient of one of the variables to isolate said variable, so if it doesn't evenly divide into the other variable AND ALSO into the constant, there'll obviously be fractions).

Have I missed something? Has anyone else noticed this, and have you found a way around it, back into sweet substitution bliss? What're your thoughts?