Solve a system of equation using substitution - High school Algebra I Exercises

Solve the following systems using the substitution method.

\begin{cases} 2x-y=4 \cr \cr 4x+2y=3 \end{cases}

\left(\dfrac{-9}{17};\dfrac{16}{7}\right)

\left(2;\dfrac{-3}{7}\right)

\left(\dfrac{9}{7};\dfrac{6}{5}\right)

\left(\dfrac{11}{8};\dfrac{-5}{4}\right)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} 2x-y=4 \cr \cr 4x+2y=3 \end{cases}

Step 1

Solve 2x-y=4 for y

2x-y=4

Subtract 2x from both sides:

-y=-2x+4

Multiply both sides by -1:

y=2x-4

Step 2

Substitute y into the other equation

Substitute y=2x-4 into 4x+2y=3. The equation becomes:

4x+2\times\left(2x-4\right)=3

Step 3

Solve 4x+2\times\left(2x-4\right)=3 for x

4x+2\times\left(2x-4\right)=3

4x+4x-8=3

8x-8=3

Add 8 to both sides:

8x=11

Divide both sides by 8:

\dfrac{8x}{8}=\dfrac{11}{8}

x=\dfrac{11}{8}

Step 4

Substitute x=\dfrac{11}{8} into y=2x-4

y=2\times\dfrac{11}{8}-4

y=\dfrac{11}{4}-4

y=\dfrac{-5}{4}

The solution is \left(x,y\right)=\left(\dfrac{11}{8}.\dfrac{-5}{4}\right).

\begin{cases} x-y=2 \cr \cr x+2y=8 \end{cases}

(2,3)

\left(3{,}1\right)

(5,4)

\left(4{,}2\right)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} x-y=2 \cr \cr x+2y=8 \end{cases}

Step 1

Solve x-y=2 for x

x-y=2

Add y to both sides:

x=2+y

Step 2

Substitute x in the other equation

After substituting, the second equation becomes:

\left(2+y\right)+2y=8

Step 3

Solve \left(2+y\right)+2y=8 for y

\left(2+y\right)+2y=8

2+3y=8

Subtract 2 from both sides:

3y=6

Divide both sides by 3:

y=2

Step 4

Substitute y=2 into x-y=2

Substituting gives:

x-2=2

Add 2 to both sides:

x=4

The solution is (x,y)=(4{,}2).

\begin{cases} x-y=5 \cr \cr 3x+y=7 \end{cases}

(3,−2)

(−1,4)

(1,3)

(2,4)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} x-y=5 \cr \cr 3x+y=7 \end{cases}

Step 1

Solve x-y=5 for x

x-y=5

Add y to both sides:

x=5+y

Step 2

Substitute x in the other equation

After substituting, the second equation becomes:

3\left(5+y\right)+y=7

Step 3

Solve 3\left(5+y\right)+y=7 for y

3\left(5+y\right)+y=7

15+4y=7

Subtract 15 from both sides:

4y=-8

Divide both sides by 4:

y=-2

Step 4

Substitute y=-2 into x-y=5

Substituting gives:

x-\left(-2\right)=5

x+2=5

Subtract 2 from both sides:

x=3

The solution is (x,y)=(3,-2).

\begin{cases} 3x-y=1 \cr \cr x+2y=12 \end{cases}

(4,2)

(5,1)

(1,3)

(2,5)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} 3x-y=1 \cr \cr x+2y=12 \end{cases}

Step 1

Solve x+2y=12 for x

x+2y=12

Subtract 2y from both sides:

x=12-2y

Step 2

Substitute x into the other equation

After substituting the second equation becomes:

3\left(12-2y\right)-y=1

Step 3

Solve \left(2+y\right)+2y=8 for y

3\left(12-2y\right)-y=1

36-6y-y=1

36-7y=1

Subtract 36 from both sides:

-7y=-35

Divide both sides by -7:

y=5

Step 4

Substitute y=5 into x+2y=12

Substituting gives:

x+2\left(5\right)=12

x+10=12

Subtract 10 to both sides:

x=2

The solution is (x,y)=(2{,}5).

\begin{cases} x-y=3 \cr \cr x+y=-1 \end{cases}

(2,−4)

(1,2)

(1,−2)

(2,3)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} x-y=3 \cr \cr x+y=-1 \end{cases}

Step 1

Solve x-y=3 for x

x-y=3

Add y to both sides:

x=3+y

Step 2

Substitute x=3+y into the other equation

After substituting, the second equation becomes:

\left(3+y\right)+y=-1

Step 3

Solve \left(2+y\right)+2y=8 for y

3+2y=-1

Subtract 3 from both sides:

2y=-4

Divide both sides by 2:

y=-2

Step 4

Substitute y=-2 into x-y=3

Substituting gives:

x-\left(-2\right)=3

x+2=3

Subtract 2 from both sides:

x=1

The solution is (x,y)=(1,-2).

\begin{cases} 5x-2y=3 \cr \cr 2x+y=3 \end{cases}

(2,4)

(4,2)

(1,1)

(3,1)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} 5x-2y=3 \cr \cr 2x+y=3 \end{cases}

Step 1

Solve 2x+y=3 for y

2x+y=3

Subtract 2x from both sides:

y=-2x+3

Step 2

Substitute y=-2x+3 into the other equation

After substituting, the second equation becomes:

5x-2\left(-2x+3\right)=3

Step 3

Solve 5x-2\left(-2x+3\right)=3 for x

5x-2\left(-2x+3\right)=3

5x+4x-6=3

9x-6=3

Add 6 to both sides:

9x=9

Divide both sides by 9:

y=1

Step 4

Substitute y=1 into 2x+y=3

Substituting gives:

2x+1=3

Subtract 1 from both sides:

2x=2

Divide by 2:

x=1

The solution is (x,y)=(1{,}1).

\begin{cases} 9x-3y=12 \cr \cr 2x+y=6 \end{cases}

(−2,3)

(1,4)

(3,4)

(2,2)

Solving via substitution requires:

Solve one of the equations for one of the variables.
Substitute the value of the variable into the other equation.
Solve for the remaining variable (one variable was removed via substitution)
Plug the value of one variable into any equation with two variables to solve for the second variable.

Here, we need to solve:

\begin{cases} 9x-3y=12 \cr \cr 2x+y=6 \end{cases}

Step 1

Solve 2x+y=6 for y

2x+y=6

Subtract 2x from both sides:

y=6-2x

Step 2

Substitute y=6-2x into the other equation

After substituting, the second equation becomes:

9x-3\left(6-2x\right)=12

Step 3

Solve 9x-3\left(6-2x\right)=12 for x

9x-3\left(6-2x\right)=12

9x-18+6x=12

15x-18=12

Add 10 to both sides:

15x=30

Divide both sides by 15:

y=2

Step 4

Substitute y=2 into 2x+y=6

Substituting gives:

2x+2=6

Subtract 2 from both sides:

2x=4

Divide both sides by 2:

x=2

The solution is (x,y)=(2{,}2).