Solving Linear Systems by Substitution

This summary covers how to solve a linear system algebraically by substitution, what to do when the answer is a fraction, and how to check a solution with a left-side / right-side check.

The method of substitution

To solve a linear system by substitution:

1. In one equation, isolate a variable. Choose a variable with a coefficient of $1$ (or $-1$) if you can.
2. Substitute that expression into the other equation, then solve for the one variable that is left.
3. Substitute the value you found into either original equation to find the other variable.
4. State the solution as an ordered pair $(x,y)$: the point of intersection.

Example 1

Find the point of intersection (that is, solve for $x$ and $y$).

$\begin{array}{rlrl}& \text{A}:& 2x+y& =2\\ & \text{B}:& 3x-2y& =10\end{array}$

From A: $y=-2x+2$

Substitute into B:

$$\begin{array}{rl}3x-2y& =10\\ 3x-2(-2x+2)& =10\\ 3x+4x-4& =10\\ 7x-4+4& =10+4\\ 7x& =14\\ \frac{7x}{7}& =\frac{14}{7}\\ x& =2\end{array}$$

This is only half a solution. We need an ordered pair $(x,y)$: the intersection point. We have $x$, so how do we find $y$?

Substitute $x=2$ into A, the other equation:

$$\begin{array}{rl}2x+y& =2\\ 2(2)+y& =2\\ 4+y-4& =2-4\\ y& =-2\end{array}$$

$\therefore$ the solution, or point of intersection, is $(2,-2)$.

Example 2

Find the point of intersection (that is, solve for $x$ and $y$).

$\begin{array}{rlrl}& \text{A}:& 3x+6y& =4\\ & \text{B}:& x-2y& =1\end{array}$

From B (since it has a term with a coefficient of $1$): $x=1+2y$

Substitute into A:

$$\begin{array}{rl}3x+6y& =4\\ 3(1+2y)+6y& =4\\ 3+6y+6y& =4\\ 3+12y-3& =4-3\\ 12y& =1\\ \frac{12y}{12}& =\frac{1}{12}\\ y& =\frac{1}{12}\end{array}$$

Leave your answer as a fraction in lowest terms, unless the decimal equivalent terminates.

Substitute $y=\frac{1}{12}$ into B, the other equation:

$$\begin{array}{rl}x-2y& =1\\ x-2\left(\frac{1}{12}\right)& =1\\ x-\frac{2}{12}& =1\\ x-\frac{1}{6}+\frac{1}{6}& =1+\frac{1}{6}\\ x& =\frac{6}{6}+\frac{1}{6}\\ x& =\frac{7}{6}\end{array}$$

Use improper fractions, instead of mixed fractions.

$\therefore$ the solution, or point of intersection, is $\left(\frac{7}{6},\frac{1}{12}\right)$.

Checking a solution

Example 3

Check the solution to the linear system.

By the method of substitution, you found that the solution to the system

$\begin{array}{rlrl}& \text{A}:& x-y& =7\\ & \text{B}:& 2x+y& =-10\end{array}$

was $(-1,-8)$.

Check the solution. You must do a left-side / right-side check. Form matters!

Equation A

$$\begin{array}{rl}[t]\text{L.S.}& =x-y\\ & =(-1)-(-8)\\ & =-1+8\\ & =7\end{array}\begin{array}{rl}[t]\text{R.S.}& =7\end{array}$$

Equation B

$$\begin{array}{rl}[t]\text{L.S.}& =2x+y\\ & =2(-1)+(-8)\\ & =-2-8\\ & =-10\end{array}\begin{array}{rl}[t]\text{R.S.}& =-10\end{array}$$

$\because$ L.S. $=$ R.S. in both equations, $(-1,-8)$ is the solution.

Curriculum connection

B1.1

solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination (Sample problem: Solve $y=\frac{1}{2}x-5$, $3x+2y=-2$ for $x$ and $y$ algebraically, and verify algebraically and graphically);

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