Question

Solve the following system using any method

3x+7y=-5
8x+9y=6

Answer 1

To solve the system of equations:

1. \( 3x + 7y = -5 \) (Equation 1)
2. \( 8x + 9y = 6 \) (Equation 2)

We can use the elimination method. We'll multiply each equation by a suitable value so that the coefficients of \(x\) or \(y\) are the same. We'll aim to eliminate \(x\).

First, let's make the coefficients of \(x\) the same. We can do this by multiplying Equation 1 by 8 and Equation 2 by 3:

\[
\begin{align*}
8(3x + 7y) & = 8(-5)\\
24x + 56y & = -40 \quad \text{(Equation 3)}
\end{align*}
\]

\[
\begin{align*}
3(8x + 9y) & = 3(6)\\
24x + 27y & = 18 \quad \text{(Equation 4)}
\end{align*}
\]

Now we have the system:

3. \( 24x + 56y = -40 \)
4. \( 24x + 27y = 18 \)

Next, we can subtract Equation 4 from Equation 3 to eliminate \(x\):

\[
(24x + 56y) - (24x + 27y) = -40 - 18
\]

This simplifies to:

\[
56y - 27y = -58
\]

\[
29y = -58
\]

Now, solve for \(y\):

\[
y = \frac{-58}{29} = -2
\]

Now that we have \(y = -2\), we can substitute this value back into one of the original equations to find \(x\). We'll use Equation 1:

\[
3x + 7(-2) = -5
\]

This simplifies to:

\[
3x - 14 = -5
\]

Adding 14 to both sides:

\[
3x = 9
\]

Now solve for \(x\):

\[
x = \frac{9}{3} = 3
\]

So the solution to the system of equations is:

\[
\boxed{(3, -2)}
\]

You can double-check the solution by substituting \(x = 3\) and \(y = -2\) back into both original equations to ensure they hold true.