Solve using elimination.

–8x + 3y = 10
–9x + 4y = 5

1 answer

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:

  1. \(-32x + 12y = 40\)
  2. \(-27x + 12y = 15\)

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:

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