1Cademy - Solving $$\{3x + y = 5,\; 2x + 4y = -10\}$$ by Substitution

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Solving a System of Linear Equations by Substitution

Example

Solving $\{3x + y = 5,\; 2x + 4y = -10\}$ by Substitution

Solve the system $\left\{\begin{array}{l} 3x + y = 5 \\ 2x + 4y = -10 \end{array}\right.$ using the substitution method.

Step 1 — Solve one equation for one variable. In the first equation, $y$ has a coefficient of $1$ , making it the easiest variable to isolate. Subtract $3x$ from both sides:

$y = -3x + 5$

Step 2 — Substitute into the other equation. Replace $y$ in the second equation with $-3x + 5$ :

$2x + 4(-3x + 5) = -10$

Step 3 — Solve the resulting one-variable equation. Distribute $4$ across the parentheses:

$2x - 12x + 20 = -10$

Combine the like terms $2x - 12x = -10x$ :

$-10x + 20 = -10$

Subtract $20$ from both sides: $-10x = -30$ . Divide both sides by $-10$ : $x = 3$ .

Step 4 — Find the other variable. Substitute $x = 3$ into the first original equation $3x + y = 5$ :

$3(3) + y = 5$

$9 + y = 5$

$y = -4$

Step 5 — Write the solution as an ordered pair: $(3, -4)$ .

Step 6 — Check in both original equations:

First equation: $3(3) + (-4) = 9 - 4 = 5$ . Since $5 = 5$ is true ✓
Second equation: $2(3) + 4(-4) = 6 - 16 = -10$ . Since $-10 = -10$ is true ✓

Both equations are satisfied, confirming that $(3, -4)$ is the solution of the system. This example illustrates that when neither equation is already solved for a variable, looking for a variable with a coefficient of $1$ identifies the simplest starting point for Step 1 — here, $y$ in the first equation.

Updated 2026-04-21

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OpenStax Elementary Algebra

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