To solve the system of equations using elimination, we have the two equations:
- \(-8x + 3y = 10\)
- \(-9x + 4y = 5\)
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:
-
\(4(-8x + 3y) = 4(10)\)
\(-32x + 12y = 40\) (Equation 3) -
\(3(-9x + 4y) = 3(5)\)
\(-27x + 12y = 15\) (Equation 4)
Now we have:
- \(-32x + 12y = 40\)
- \(-27x + 12y = 15\)
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:
\[ \boxed{(-5, -10)} \]