1Cademy - Solving $$\left\{\frac{1}{3}x - \frac{1}{2}y = 1,\; \frac{3}{4}x

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Solving a System of Linear Equations by Elimination

Example

Solving $\left\{\frac{1}{3}x - \frac{1}{2}y = 1,; \frac{3}{4}x - y = \frac{5}{2}\right\}$ by Elimination

To solve the system $\left\{\begin{array}{l} \frac{1}{3}x - \frac{1}{2}y = 1 \frac{3}{4}x - y = \frac{5}{2} \end{array}\right.$ using the elimination method, follow these steps:

Step 1 — Clear fractions.

First equation: The fractions have denominators $3$ and $2$ , so the LCD is $6$ . Multiply every term by $6$ : $6\left(\frac{1}{3}x - \frac{1}{2}y\right) = 6(1) \implies 2x - 3y = 6$
Second equation: The fractions have denominators $4$ and $2$ , so the LCD is $4$ . Multiply every term by $4$ : $4\left(\frac{3}{4}x - y\right) = 4\left(\frac{5}{2}\right) \implies 3x - 4y = 10$

The system is now $\left\{\begin{array}{l} 2x - 3y = 6 3x - 4y = 10 \end{array}\right.$

Step 2 — Make the coefficients of one variable opposites. To eliminate $x$ , note that the $x$ -coefficients are $2$ and $3$ . Multiply the first equation by $3$ and the second equation by $-2$ so that the $x$ -coefficients become $6$ and $-6$ :

$3(2x - 3y) = 3(6) \implies 6x - 9y = 18$

$-2(3x - 4y) = -2(10) \implies -6x + 8y = -20$

Step 3 — Add the equations to eliminate $x$ .

$6x - 9y - 6x + 8y = 18 - 20$

$-y = -2$

The $x$ -terms cancel out.

Step 4 — Solve for the remaining variable. Divide both sides by $-1$ :

$y = 2$

Step 5 — Substitute back into an original equation. It is easier to substitute $y = 2$ into the cleared first equation $2x - 3y = 6$ :

$2x - 3(2) = 6$

$2x - 6 = 6$

$2x = 12$

$x = 6$

Step 6 — Write the solution as an ordered pair: (6, 2).

Step 7 — Check in both original equations:

First equation: $\frac{1}{3}(6) - \frac{1}{2}(2) = 2 - 1 = 1$ . Since $1 = 1$ is true ✓
Second equation: $\frac{3}{4}(6) - 2 = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}$ . Since $\frac{5}{2} = \frac{5}{2}$ is true ✓

Both equations are satisfied, confirming that the solution is (6, 2).

Updated 2026-06-17

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