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

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Solving a System of Linear Equations by Substitution

Example

Solving $\{y = -2x + 5,\; y = \frac{1}{2}x\}$ by Substitution

Solve the system $\left\{\begin{array}{l} y = -2x + 5 \\ y = \frac{1}{2}x \end{array}\right.$ using the substitution method.

Because both equations are already solved for $y$ , the two right-hand-side expressions can be set equal to each other directly — no algebraic rearrangement is needed for Step 1.

Step 2 — Substitute into the other equation. Replace $y$ in the first equation with $\frac{1}{2}x$ :

$\frac{1}{2}x = -2x + 5$

Step 3 — Solve the resulting one-variable equation. The equation contains a fraction, so begin by clearing it. Multiply both sides by $2$ :

$2 \cdot \frac{1}{2}x = 2(-2x + 5)$

$x = -4x + 10$

Add $4x$ to both sides: $5x = 10$ . Divide both sides by $5$ : $x = 2$ .

Step 4 — Find the other variable. Substitute $x = 2$ into $y = \frac{1}{2}x$ :

$y = \frac{1}{2}(2) = 1$

Step 5 — Write the solution as an ordered pair: $(2, 1)$ .

Step 6 — Check in both original equations:

Second equation: $1 \stackrel{?}{=} \frac{1}{2} \cdot 2 = 1$ . Since $1 = 1$ is true ✓
First equation: $1 \stackrel{?}{=} -2(2) + 5 = -4 + 5 = 1$ . Since $1 = 1$ is true ✓

Both equations are satisfied, confirming that $(2, 1)$ is the solution of the system. This example demonstrates the special case where both equations are already solved for $y$ , so Step 1 is automatically complete and the two expressions for $y$ can be equated directly. It also illustrates using the clearing fractions technique — multiplying both sides by $2$ — to eliminate the fractional coefficient before solving the resulting one-variable equation.

Updated 2026-04-21

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OpenStax Elementary Algebra

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