1Cademy - Solving $$\left\{7x + 8y = 4,\; 3x - 5y = 27\right\}$$ by Elimination

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Solving a System of Linear Equations by Elimination

Example

Solving $\left\{7x + 8y = 4,\; 3x - 5y = 27\right\}$ by Elimination

To solve the system $\left\{\begin{array}{l} 7x + 8y = 4 \\ 3x - 5y = 27 \end{array}\right.$ using the elimination method, follow these steps:

Step 1 — Write both equations in standard form. Both equations are already in standard form.

Step 2 — Make the coefficients of one variable opposites. To eliminate $y$ , note that the $y$ -coefficients are $8$ and $-5$ . Multiply the first equation by $5$ and the second equation by $8$ so that the $y$ -coefficients become $+40$ and $-40$ :

$5(7x + 8y) = 5(4) \implies 35x + 40y = 20$

$8(3x - 5y) = 8(27) \implies 24x - 40y = 216$

The system is now $\left\{\begin{array}{l} 35x + 40y = 20 \\ 24x - 40y = 216 \end{array}\right.$

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

$35x + 40y + 24x - 40y = 20 + 216$

$59x = 236$

The $y$ -terms cancel out.

Step 4 — Solve for the remaining variable. Divide both sides by $59$ :

$x = 4$

Step 5 — Substitute back into an original equation. Substitute $x = 4$ into the first original equation $7x + 8y = 4$ :

$7(4) + 8y = 4$

$28 + 8y = 4$

$8y = -24$

$y = -3$

Step 6 — Write the solution as an ordered pair: $(4, -3)$ .

Step 7 — Check in both original equations:

First equation: $7(4) + 8(-3) = 28 - 24 = 4$ . Since $4 = 4$ is true ✓
Second equation: $3(4) - 5(-3) = 12 + 15 = 27$ . Since $27 = 27$ is true ✓

Both equations are satisfied, confirming that the solution is $(4, -3)$ .

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Updated 2026-05-07

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References

OpenStax Intermediate Algebra

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Intermediate Algebra @ OpenStax

Ch.4 Systems of Linear Equations - Intermediate Algebra @ OpenStax

Algebra

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