Solve Systems of Two Linear Equations: CCSS.Math.Content.8.EE.C.8b - 8th Grade Math
Card 0 of 96
Solve the following system of equations:


Solve the following system of equations:
Set the two equations equal to one another:
2x - 2 = 3x + 6
Solve for x:
x = -8
Plug this value of x into either equation to solve for y. We'll use the top equation, but either will work.
y = 2 * (-8) - 2
y = -18
Set the two equations equal to one another:
2x - 2 = 3x + 6
Solve for x:
x = -8
Plug this value of x into either equation to solve for y. We'll use the top equation, but either will work.
y = 2 * (-8) - 2
y = -18
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Solve the system for
and
.
Solve the system for and
.
The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply
by
to get
.
Then, we can add
to this equation to yield
, so
.
We can plug that value into either of the original equations; for example,
.
So,
as well.
The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply by
to get
.
Then, we can add to this equation to yield
, so
.
We can plug that value into either of the original equations; for example, .
So, as well.
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What is the solution to the following system of equations:


What is the solution to the following system of equations:
By solving one equation for
, and replacing
in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.
By solving one equation for , and replacing
in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.
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Find the solution to the following system of equations.


Find the solution to the following system of equations.


To solve this system of equations, use substitution. First, convert the second equation to isolate
.

Then, substitute
into the first equation for
.


Combine terms and solve for
.


Now that we know the value of
, we can solve for
using our previous substitution equation.

To solve this system of equations, use substitution. First, convert the second equation to isolate .
Then, substitute into the first equation for
.
Combine terms and solve for .
Now that we know the value of , we can solve for
using our previous substitution equation.
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Find a solution for the following system of equations:

Find a solution for the following system of equations:
When we add the two equations, the
and
variables cancel leaving us with:
which means there is no solution for this system.
When we add the two equations, the and
variables cancel leaving us with:
which means there is no solution for this system.
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Solve the following system of equations:

Solve the following system of equations:
When we add the two equations, the
variables cancel leaving us with:

Solving for
we get:


We can then substitute our value for
into one of the original equations and solve for
:

When we add the two equations, the variables cancel leaving us with:
Solving for we get:
We can then substitute our value for into one of the original equations and solve for
:
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Solve this system of equations for
:


Solve this system of equations for :
Multiply the bottom equation by 5, then add to the top equation:






Multiply the bottom equation by 5, then add to the top equation:
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Solve this system of equations for
:


Solve this system of equations for :
Multiply the top equation by
:



Now add:





Multiply the top equation by :
Now add:
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Solve this system of equations for
:


Solve this system of equations for :
Multiply the bottom equation by
, then add to the top equation:





Divide both sides by 


Multiply the bottom equation by , then add to the top equation:
Divide both sides by
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Solve this system of equations for
:


Solve this system of equations for :
Multiply the top equation by
:



Now add:





Multiply the top equation by :
Now add:
Compare your answer with the correct one above
Solve the following system of equations.


Solve the following system of equations.
We are given


We can solve this by using the substitution method. Notice that you can plug
from the first equation into the second equation and then get

Add
to both sides


Add 9 to both sides


Divide both sides by 5


So
. We can use this value to find y by using either equation. In this case, I'll use
.



So the solution is

We are given
We can solve this by using the substitution method. Notice that you can plug from the first equation into the second equation and then get
Add to both sides
Add 9 to both sides
Divide both sides by 5
So . We can use this value to find y by using either equation. In this case, I'll use
.
So the solution is
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Solve the set of equations:


Solve the set of equations:
Solve the first equation for
:



Substitute into the second equation:


Multiply the entire equation by 2 to eliminate the fraction:




Using the value of
, solve for
:




Therefore, the solution is 
Solve the first equation for :
Substitute into the second equation:
Multiply the entire equation by 2 to eliminate the fraction:
Using the value of , solve for
:
Therefore, the solution is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can add our equations together to cancel out the 
![\frac{\begin{array}[b]{r}-7x-y=4\ +\ 4x+y=-7\end{array}}{ -3x=-3}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953479/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to subtract
from both sides to isolate the 
![\frac{\begin{array}[b]{r}4+y=-7\ \ -4\ \ \ \ \ \ \ \ -4\end{array}}{\\y=-11}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953493/gif.latex)
Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can add our equations together to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to subtract from both sides to isolate the
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can subtract our equations to cancel out the 
![\frac{\begin{array}[b]{r}4x-4y=5\ -\ 9x-4y=-15\end{array}}{ -5x=20}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953833/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to add
to both sides to isolate the 
![\frac{\begin{array}[b]{r}-16-4y=5\ \ +16\ \ \ \ \ \ +16\end{array}}{\\-4y=21}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953840/gif.latex)
Then we divide each side by 


Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can subtract our equations to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to add to both sides to isolate the
Then we divide each side by
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can subtract our equations to cancel out the 
![\frac{\begin{array}[b]{r}2x+2y=16\ -\ 6x+2y=20\end{array}}{ -4x=-4}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953814/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to subtract
from both sides to isolate the 
![\frac{\begin{array}[b]{r}2+2y=16\ \ -2\ \ \ \ \ \ \ \ -2\end{array}}{\\2y=14}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953820/gif.latex)
Then divide both sides by
to solve for 


Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can subtract our equations to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to subtract from both sides to isolate the
Then divide both sides by to solve for
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can subtract our equations to cancel out the 
![\frac{\begin{array}[b]{r}9x+3y=27\ -\ 4x+3y=12\end{array}}{ 5x=15}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953776/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to subtract
from both sides to isolate the 
![\frac{\begin{array}[b]{r}12+3y=12\ \ -12\ \ \ \ \ \ \ \ -12\end{array}}{\\3y=0}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/953783/gif.latex)
Then divide both sides by
to solve for 


Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can subtract our equations to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to subtract from both sides to isolate the
Then divide both sides by to solve for
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above


Which of the following expresses the solutions to the above system of equations as an ordered pair in the form
?
Which of the following expresses the solutions to the above system of equations as an ordered pair in the form ?
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can subtract our equations to cancel out the 
![\frac{\begin{array}[b]{r}6x+4y=28\ -\ (2x+4y=16)\end{array}}{ 4x=12}](//vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/1156578/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to subtract
from both sides to isolate the 
![\frac{\begin{array}[b]{r}6+4y=16\ \ -6\ \ \ \ \ \ \ \ -6\end{array}}{\\4y=10}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/954197/gif.latex)
Then divide both sides by
to solve for 


Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can subtract our equations to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to subtract from both sides to isolate the
Then divide both sides by to solve for
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our
variables have the same coefficient. We can subtract our equations to cancel out the 
![\frac{\begin{array}[b]{r}-4x+y=20\ -\ 6x+y=30\end{array}}{ -10x=-10}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/954090/gif.latex)
Next, we can divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 


We want to subtract
from both sides to isolate the 
![\frac{\begin{array}[b]{r}6+y=30\ \ -6\ \ \ \ \ \ \ \ -6\end{array}}{\\y=24}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/954096/gif.latex)
Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, elimination makes the most sense because our variables have the same coefficient. We can subtract our equations to cancel out the
Next, we can divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
We want to subtract from both sides to isolate the
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to
, into the
of our second equation:

Next, we need to distribute and combine like terms:


We are solving for the value of
, which means we need to isolate the
to one side of the equation. We can subtract
from both sides:
![\frac{\begin{array}[b]{r}12y+4=28\ \ -4 \ -4\end{array}}{\\12y=24}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/954553/gif.latex)
Then divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 



Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the
of our second equation:
Next, we need to distribute and combine like terms:
We are solving for the value of , which means we need to isolate the
to one side of the equation. We can subtract
from both sides:
Then divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above
Use algebra to solve the following system of linear equations:


Use algebra to solve the following system of linear equations:
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either
or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to
, into the
of our second equation:

Next, we need to distribute and combine like terms:


We are solving for the value of
, which means we need to isolate the
to one side of the equation. We can subtract
from both sides:
![\frac{\begin{array}[b]{r}12y+14=74\ \ -14 -14\end{array}}{\\12y=60}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/954474/gif.latex)
Then divide both sides by
to solve for 


Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both
and
values.
Now that we have the value of
, we can plug that value into the
variable in one of our given equations and solve for 



Our point of intersection, and the solution to the two system of linear equations is 
There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination.
Substitution can be used by solving one of the equations for either or
, and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the
form, and then set both equations equal to each other.
Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable.
For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to , into the
of our second equation:
Next, we need to distribute and combine like terms:
We are solving for the value of , which means we need to isolate the
to one side of the equation. We can subtract
from both sides:
Then divide both sides by to solve for
Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both and
values.
Now that we have the value of , we can plug that value into the
variable in one of our given equations and solve for
Our point of intersection, and the solution to the two system of linear equations is
Compare your answer with the correct one above