Question

1. what is the solution to this system of equations?

$-6x + 10y = 42$
$2x - \frac{1}{3}y = -5$
options: (2, 27), (-2, 3), (2, 0), (-3, -3)

Explanation:

Step1: Simplify the second equation

Multiply the second equation \(2x - \frac{1}{3}y = -5\) by 3 to eliminate the fraction:
\(3(2x) - 3(\frac{1}{3}y) = 3(-5)\)
\(6x - y = -15\)
Rearrange to solve for \(y\): \(y = 6x + 15\).

Step2: Substitute \(y\) into the first equation

The first equation is \(-6x + 10y = 42\). Substitute \(y = 6x + 15\) into it:
\(-6x + 10(6x + 15) = 42\)
Expand: \(-6x + 60x + 150 = 42\)
Combine like terms: \(54x + 150 = 42\)
Subtract 150: \(54x = 42 - 150 = -108\)
Divide by 54: \(x = \frac{-108}{54} = -2\).

Step3: Find \(y\) using \(x = -2\)

Substitute \(x = -2\) into \(y = 6x + 15\):
\(y = 6(-2) + 15 = -12 + 15 = 3\).

Step4: Verify the solution

Check \((-2, 3)\) in both original equations:

1. First equation: \(-6(-2) + 10(3) = 12 + 30 = 42\) (matches).
2. Second equation: \(2(-2) - \frac{1}{3}(3) = -4 - 1 = -5\) (matches).

Answer:

\(\boldsymbol{(-2, 3)}\) (Option: \(\boldsymbol{(-2, 3)}\))