Question

lution of the system of equations.
$4x + 3y = -14$
$8x + 10y = -36$

Explanation:

Step1: Eliminate x by multiplying the first equation by 2

Multiply \(4x + 3y = -14\) by 2: \(2\times(4x + 3y)=2\times(-14)\) gives \(8x + 6y = -28\)

Step2: Subtract the new first equation from the second equation

Subtract \(8x + 6y = -28\) from \(8x + 10y = -36\): \((8x + 10y)-(8x + 6y)=-36-(-28)\)
Simplify: \(8x + 10y - 8x - 6y=-36 + 28\) → \(4y=-8\)

Step3: Solve for y

Divide both sides by 4: \(y=\frac{-8}{4}=-2\)

Step4: Substitute y = -2 into the first original equation to solve for x

Substitute into \(4x + 3y = -14\): \(4x + 3\times(-2)=-14\) → \(4x - 6=-14\)
Add 6 to both sides: \(4x=-14 + 6=-8\)
Divide by 4: \(x=\frac{-8}{4}=-2\)

Answer:

The solution to the system of equations is \(x = -2\) and \(y = -2\) (or the ordered pair \((-2, -2)\))