Solving Systems by Substitution

Solving by Substitution

By the end of this lesson, you will be able to solve a system by substitution, choose which variable to isolate so the arithmetic stays clean, and check the pair you get.

Turn Two Unknowns into One

You can solve an equation with one unknown. You cannot solve one with two. So the whole job is getting rid of a variable, and substitution does it by replacing one with what the other equation says it equals.

$$\begin{cases} y = 2x + 1 \ 3x + y = 11 \end{cases}$$

The first equation says $y$ IS $2x + 1$. So in the second, write that instead of $y$:

$$3x + (2x + 1) = 11$$

One unknown. Solve it, then go back for the other.

Four moves, always the same:

1. Get one variable alone in one equation
2. Substitute that expression into the OTHER equation
3. Solve for the one remaining variable
4. Back-substitute to find the first, then check both

Substituting into the Other Equation

Step 2 has a trap worth naming before you meet it.

Substituting $y = 2x + 1$ back into $y = 2x + 1$ gives $2x + 1 = 2x + 1$, which is true and useless. It has to go into the equation you did NOT rearrange, or nothing is eliminated.

And keep the brackets. Substituting $y = 2x + 1$ into $3x - y = 4$ gives:

$$3x - (2x + 1) = 4$$

not $3x - 2x + 1 = 4$. The minus sign applies to the whole expression, and that missing bracket is the single most common error in this method.

Example 1: When a Variable Is Already Alone

Here is the whole method on one system, worked all the way through. $$\begin{cases} y = 2x + 1 \ 3x + y = 11 \end{cases}$$

• Substitute the first into the second, keeping the bracket 3x + (2x + 1) = 11
• Simplify the left side 5x + 1 = 11
• Solve for x 2
• Back-substitute into y = 2x + 1 to find y 5
• Write the solution as a pair (2, 5)

Worked Example 1: Your Turn

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

Choosing What to Isolate

When nothing is alone yet, the choice is yours, and it decides whether the arithmetic is pleasant.

$$\begin{cases} 2x + 3y = 12 \ x - 4y = -5 \end{cases}$$

Isolating $x$ in the second equation costs nothing, because its coefficient is already $1$:

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