Question

Solving systems by elimination 10x+3y=-30 2x+9y=-6

Answer 1

To solve the system of equations using the elimination method, we have the following two equations:

1. \( 10x + 3y = -30 \)
2. \( 2x + 9y = -6 \)

Our goal is to eliminate one of the variables. Let's eliminate \(y\) by making the coefficients of \(y\) in both equations equal.

First, we can multiply the entire first equation by 3, and the entire second equation by -1:

\[
3(10x + 3y) = 3(-30) \implies 30x + 9y = -90 \quad \text{(Equation 1')}
\]

\[
-1(2x + 9y) = -1(-6) \implies -2x - 9y = 6 \quad \text{(Equation 2')}
\]

Now the system looks like this:

1. \(30x + 9y = -90\) (Equation 1')
2. \(-2x - 9y = 6\) (Equation 2')

Next, we can add these two equations together to eliminate \(y\):

\[
(30x + 9y) + (-2x - 9y) = -90 + 6
\]

This simplifies to:

\[
30x - 2x + 9y - 9y = -90 + 6
\]

\[
28x = -84
\]

Now, solve for \(x\):

\[
x = \frac{-84}{28} = -3
\]

Now that we have the value of \(x\), we can substitute it back into one of the original equations to find \(y\). Let's use the second equation:

\[
2x + 9y = -6
\]

Substituting \(x = -3\):

\[
2(-3) + 9y = -6
\]

\[
-6 + 9y = -6
\]

\[
9y = 0
\]

\[
y = 0
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (-3, 0)
\]

You can verify this solution by substituting \(x = -3\) and \(y = 0\) back into the original equations, and they will both hold true.