Question

solve the system of equations using elimination: $8x + 3y = 1$ and $6x + 4y = 6$. answer attempt 1 out of 2

Explanation:

Step1: Multiply equations to eliminate x

Multiply the first equation \(8x + 3y = 1\) by 3: \(3\times(8x + 3y)=3\times1\) gives \(24x + 9y = 3\).
Multiply the second equation \(6x + 4y = 6\) by 4: \(4\times(6x + 4y)=4\times6\) gives \(24x + 16y = 24\).

Step2: Subtract equations to eliminate x

Subtract the first new equation from the second: \((24x + 16y)-(24x + 9y)=24 - 3\).
Simplify: \(24x - 24x + 16y - 9y = 21\) → \(7y = 21\).

Step3: Solve for y

Divide both sides by 7: \(y=\frac{21}{7}=3\).

Step4: Substitute y into original equation

Substitute \(y = 3\) into \(8x + 3y = 1\): \(8x + 3\times3 = 1\) → \(8x + 9 = 1\).

Step5: Solve for x

Subtract 9: \(8x = 1 - 9=-8\).
Divide by 8: \(x=\frac{-8}{8}=-1\).

Answer:

The solution is \(x = -1\), \(y = 3\) (or as an ordered pair \((-1, 3)\)).