Question

Solve using elimination.

–8x + 3y = 10
–9x + 4y = 5

Answer 1

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

1) \(-8x + 3y = 10\)
2) \(-9x + 4y = 5\)

We want to eliminate one of the variables. Let's manipulate these equations to eliminate \(y\).

First, we can make the coefficients of \(y\) in both equations the same by finding a common multiple. The coefficients of \(y\) are 3 and 4, so the least common multiple is 12.

We can multiply the first equation by 4 and the second equation by 3:

1) \(4(-8x + 3y) = 4(10)\)
\(-32x + 12y = 40\) (Equation 3)

2) \(3(-9x + 4y) = 3(5)\)
\(-27x + 12y = 15\) (Equation 4)

Now we have:

3) \(-32x + 12y = 40\)
4) \(-27x + 12y = 15\)

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

\[
(-32x + 12y) - (-27x + 12y) = 40 - 15
\]

This simplifies to:

\[
-32x + 27x = 25
\]
\[
-5x = 25
\]

Now, divide both sides by -5:

\[
x = -5
\]

Now that we have \(x\), we can substitute it back into one of the original equations to find \(y\). Let's use Equation 1:

\[
-8(-5) + 3y = 10
\]
\[
40 + 3y = 10
\]
\[
3y = 10 - 40
\]
\[
3y = -30
\]
\[
y = -10
\]

So the solution to the system of equations is:

\[
\boxed{(-5, -10)}
\]