1-02 Solve Linear Systems Algebraically (2025)

Objectives:

• Solve a system of equations by substitution.
• Solve a system of inequalities by elimination.

SDA NAD Content Standards (2018): AII.4.1, AII.6.1

Solve a System of Equations by Substitution

To solve a system by substitution,

1. Solve one of the equations for one of its variables.
2. Substitute the expression from Step 1 into the other equation and solve for the other variable.
3. Substitute the value from Step 2 into the revised equation from Step 1 and solve.

Example 1

Solve \(\left\{ \begin{align} 2x+5y=-5 \\ x=3-3y \end{align} \right.\).

Solution

Equation 2 is solved for x already.

x = 3 − 3y

Substitute this into the first equation.

2x + 5y = −5

2(3 − 3y) + 5y = −5

Distribute.

6 - 6y + 5y = −5

Combine like terms and solve for y.

6 − y = −5

−y = −11

y = 11

Substitute that back into 1st equation.

x = 3 − 3y

x = 3 − 3(11) = −30

The solution is (−30, 11).

Try It 1

Solve \(\left\{ \begin{align} 3x+2y=1 \\ y=2x+4 \end{align} \right.\).

Answer

The solution is (−1, 2).

Example 2

Solve \(\left\{ \begin{align} x-2y & =4 \\ 3x-6y & = 9 \end{align} \right.\).

Solution

Solve the 1st equation for x.

x − 2y = 4

x = 2y + 4

Substitute this into the other equation.

3x − 6y = 8

3(2y + 4) − 6y = 8

Distribute the 3 and solve for y.

6y + 12 − 6y = 8

12 = 8

Notice that all the y's disappeared. Look at the equation 12 = 8. This is never true. That means there is no solution.

No Solution/Many Solutions

A system of equations has no solution or infinitely many solutions if all the variables disappear while solving and what remains is a

• True statement, then there are infinitely many solutions.
• False statement, then there are no solutions.

Try It 2

Solve \(\left\{ \begin{align} x-y & =4 \\ -6x+6y & =-24 \end{align} \right.\).

Answer

All the variables disappeared and −24 = −24, so infinitely many solutions.

Solve a System of Linear Equations by Elimination

To solve by elimination,

1. Line up the equations into columns of like terms.
2. Multiply one or both of the equations by a constant to obtain coefficients that differ only in sign for one of the variables.
3. Add the revised equations from Step 1. Combining like terms will eliminate one of the variables. Solve for the remaining variable.
4. Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

Example 3

Solve \(\left\{ \begin{align} 3x-7y & =10 \\ 6x-8y & =8 \end{align} \right.\).

Solution

Notice that multiplying the 3x by −2 will produce −6x in the top equation and 6x in the bottom equation.

$$ \left\{ \begin{align} \color{red}{-2}\left(3x-7y\right) & = \color{red}{-2}\left(10\right) \\ 6x-8y & =8 \end{align} \right. $$

$$ \left\{ \begin{align} -6x+14y & =-20 \\ 6x-8y & =8 \end{align} \right. $$

Add the equations.

6y = −12

y = −2

Substitute this into one of the original equations.

3x − 7y = 10

3x − 7(−2) = 10

3x + 14 = 10

3x = −4

$$ x = -\frac{4}{3} $$

The solution is \(\left(-\frac{4}{3}, -2\right)\).

Try It 3

Solve \(\left\{ \begin{align} 8x+2y & =4 \\ -2x+3y & =13 \end{align} \right. \).

Answer

The solution is \(\left(-\frac{1}{2}, 4\right)\).

Example 4

Solve \(\left\{ \begin{align} 2x-3y & =8 \\ 3x+4y & =-5 \end{align} \right.\).

Solution

Neither x or y are convenient, so choose one. This example will use x.

Multiply the 1st equation by 3 and the 2nd equation by −2 to get 6x and −6x.

$$ \left\{ \begin{align} \color{red}{3}\left(2x-3y\right) & =\color{red}{3}\left(8\right) \\ \color{red}{-2}\left(3x+4y\right) & =\color{red}{-2}\left(-5\right) \end{align} \right. $$

$$ \left\{ \begin{align} 6x-9y & =24 \\ -6x-8y & =10 \end{align} \right. $$

Add the equation together.

−17y = 34

y = −2

Substitute this into one of the original equations.

2x − 3y = 8

2x − 3(−2) = 8

2x = 2

x = 1

The solution is (1, −2).

Try It 4

Solve \(\left\{ \begin{align} 4x-2y & =8 \\ 3x-7y & =-5 \end{align} \right.\).

Answer

The solution is (3, 2).

Example 5

Solve \(\left\{ \begin{align} 4x-6y & =10 \\ -6x+9y & =-9 \end{align} \right.\).

Solution

Multiply the 1st equation by 3 and the 2nd equation by 2 to get 12x and −12x.

$$ \left\{ \begin{align} \color{red}{3}\left(4x-6y\right) & =\color{red}{3}\left(10\right) \\ \color{red}{2}\left(-6x+9y\right) & =\color{red}{2}\left(-9\right) \end{align} \right. $$

$$ \left\{ \begin{align} 12x-18y & =30 \\ -12x+18y & =-18 \end{align} \right. $$

Add the equations together.

0 = 12

All the variables disappeared and false statement, so there is no solution.