1Cademy - Solving $$\{x + \frac{1}{2}y = 6,\; \frac{3}{2}x + \frac{2}{3}y = \frac{17}{2}\}$$ by Elimination

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Solving a System of Linear Equations by Elimination

Example

Solving $\{x + \frac{1}{2}y = 6,\; \frac{3}{2}x + \frac{2}{3}y = \frac{17}{2}\}$ by Elimination

Solve the system $\left\{\begin{array}{l} x + \frac{1}{2}y = 6 \\ \frac{3}{2}x + \frac{2}{3}y = \frac{17}{2} \end{array}\right.$ using the elimination method.

Because both equations contain fractions, the first step is to clear the fractions from each equation by multiplying it by its own LCD. This converts the system into one with integer coefficients, making the subsequent elimination steps simpler.

Step 1 — Clear fractions from each equation.

First equation: The fractions have denominator $2$ , so the LCD is $2$ . Multiply every term by $2$ :

$2\left(x + \frac{1}{2}y\right) = 2(6) \implies 2x + y = 12$

Second equation: The fractions have denominators $2$ , $3$ , and $2$ . The LCD is $6$ . Multiply every term by $6$ :

$6\left(\frac{3}{2}x + \frac{2}{3}y\right) = 6\left(\frac{17}{2}\right) \implies 9x + 4y = 51$

The system is now $\left\{\begin{array}{l} 2x + y = 12 \\ 9x + 4y = 51 \end{array}\right.$

Step 2 — Make the coefficients of one variable opposites. Both equations are in standard form. To eliminate $y$ , multiply the first equation by $-4$ so that the $y$ -coefficients become $-4$ and $+4$ :

$-4(2x + y) = -4(12) \implies -8x - 4y = -48$

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

$-8x - 4y + 9x + 4y = -48 + 51$

$x = 3$

The $y$ -terms cancel because $-4y + 4y = 0$ .

Step 4 — Substitute back into an original equation. Substitute $x = 3$ into the first original equation $x + \frac{1}{2}y = 6$ :

$3 + \frac{1}{2}y = 6$

$\frac{1}{2}y = 3$

$y = 6$

Step 5 — Write the solution as an ordered pair: $(3, 6)$ .

Step 6 — Check in both original equations:

First equation: $3 + \frac{1}{2}(6) = 3 + 3 = 6$ . Since $6 = 6$ is true ✓
Second equation: $\frac{3}{2}(3) + \frac{2}{3}(6) = \frac{9}{2} + 4 = \frac{9}{2} + \frac{8}{2} = \frac{17}{2}$ . Since $\frac{17}{2} = \frac{17}{2}$ is true ✓

Both equations are satisfied, confirming that $(3, 6)$ is the solution. This example demonstrates the strategy of clearing fractions first when a system contains fractional coefficients: multiply each equation by its own LCD to convert the system into one with integer coefficients, then proceed with standard elimination. Because the two equations may have different denominators, each equation may require a different LCD — here the first equation used $2$ and the second used $6$ .

Updated 2026-05-07

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