Question

Find the solution of the system of equations.
minus, 6, x, minus, 9, y, equals, minus, 42
−6x−9y=
−42
6, x, plus, 5, y, equals, 10
6x+5y=
10
Answer
Attempt 1 out of 2

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,
right parenthesis) Submit Answer

Answer 1

To solve the system of equations:

1. \(-6x - 9y = -42\) (Equation 1)
2. \(6x + 5y = 10\) (Equation 2)

we can use the method of substitution or elimination. Here, we'll use elimination.

First, let's rewrite the first equation in a more manageable form:

From Equation 1, we can divide everything by -3 to simplify it:

\[
2x + 3y = 14 \quad \text{(Equation 3)}
\]

Now we have the new system of equations:

1. \(2x + 3y = 14\) (Equation 3)
2. \(6x + 5y = 10\) (Equation 2)

Next, we can multiply Equation 3 by 3 to make the coefficients of \(x\) in both equations comparable:

\[
6x + 9y = 42 \quad \text{(Equation 4)}
\]

Now we have:

1. \(6x + 9y = 42\) (Equation 4)
2. \(6x + 5y = 10\) (Equation 2)

Next, we can subtract Equation 2 from Equation 4:

\[
(6x + 9y) - (6x + 5y) = 42 - 10
\]

This simplifies to:

\[
4y = 32
\]

Now, divide both sides by 4:

\[
y = 8
\]

Now that we have the value of \(y\), we can substitute it back into either Equation 1 or 2 to find \(x\). Let's use Equation 2:

\[
6x + 5(8) = 10
\]

This simplifies to:

\[
6x + 40 = 10
\]

Subtract 40 from both sides:

\[
6x = 10 - 40
\]
\[
6x = -30
\]

Now, divide both sides by 6:

\[
x = -5
\]

Now we have the values for \(x\) and \(y\):

\[
x = -5, \quad y = 8
\]

The solution to the system of equations is:

\[
\boxed{(-5, 8)}
\]