1Cademy - Solving $$\{4x + 2y = 4,\; 6x - y = 8\}$$ by Substitution

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Solving a System of Linear Equations by Substitution

Example

Solving $\{4x + 2y = 4,\; 6x - y = 8\}$ by Substitution

Solve the system $\left\{\begin{array}{l} 4x + 2y = 4 \\ 6x - y = 8 \end{array}\right.$ using the substitution method.

Step 1 — Solve one equation for one variable. We will solve the first equation for $y$ . Subtract $4x$ from both sides of $4x + 2y = 4$ :

$2y = -4x + 4$

Divide both sides by $2$ :

$y = -2x + 2$

Step 2 — Substitute into the other equation. Replace $y$ in the second equation with $-2x + 2$ :

$6x - (-2x + 2) = 8$

Step 3 — Solve the resulting one-variable equation. Distribute the negative sign across the parentheses:

$6x + 2x - 2 = 8$

Combine the like terms $6x + 2x = 8x$ :

$8x - 2 = 8$

Add $2$ to both sides: $8x = 10$ . Divide both sides by $8$ : $x = \frac{10}{8} = \frac{5}{4}$ .

Step 4 — Find the other variable. Substitute $x = \frac{5}{4}$ into the first original equation $4x + 2y = 4$ :

$4\left(\frac{5}{4}\right) + 2y = 4$

$5 + 2y = 4$

$2y = -1$

$y = -\frac{1}{2}$

Step 5 — Write the solution as an ordered pair: $\left(\frac{5}{4},\; -\frac{1}{2}\right)$ .

Step 6 — Check in both original equations:

First equation: $4\left(\frac{5}{4}\right) + 2\left(-\frac{1}{2}\right) = 5 - 1 = 4$ . Since $4 = 4$ is true ✓
Second equation: $6\left(\frac{5}{4}\right) - \left(-\frac{1}{2}\right) = \frac{15}{2} + \frac{1}{2} = \frac{16}{2} = 8$ . Since $8 = 8$ is true ✓

Both equations are satisfied, confirming that $\left(\frac{5}{4},\; -\frac{1}{2}\right)$ is the solution of the system. In this example, the textbook chose to isolate $y$ from the first equation — where its coefficient is $2$ rather than $1$ or $-1$ — requiring an extra division step before substituting. The resulting solution is a pair of fractions, $\frac{5}{4}$ and $-\frac{1}{2}$ , which would be very difficult to read accurately from a graph. This reinforces the practical advantage of the substitution method over graphing when solutions are not integers.

Updated 2026-04-21

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OpenStax Elementary Algebra

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