Question

Solve the system of equations.
−4x+6y=38

x−5y=−20

Answer 1

To solve the system of equations:

\[
-4x + 6y = 38 \quad (1)
\]
\[
x - 5y = -20 \quad (2)
\]

We can use the substitution or elimination method. Here, I'll use the substitution method:

From equation (2), we can express \(x\) in terms of \(y\):

\[
x = 5y - 20
\]

Now we can substitute this expression for \(x\) into equation (1):

\[
-4(5y - 20) + 6y = 38
\]

Expanding this gives:

\[
-20y + 80 + 6y = 38
\]

Combining like terms:

\[
-20y + 6y + 80 = 38
\]
\[
-14y + 80 = 38
\]

Now, isolate \(y\):

\[
-14y = 38 - 80
\]
\[
-14y = -42
\]
\[
y = \frac{-42}{-14} = 3
\]

Now that we have \(y\), we can substitute it back into the equation for \(x\):

\[
x = 5(3) - 20
\]
\[
x = 15 - 20
\]
\[
x = -5
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (-5, 3)
\]

To verify, we can substitute \(x = -5\) and \(y = 3\) back into both original equations:

For equation (1):

\[
-4(-5) + 6(3) = 20 + 18 = 38 \quad (True)
\]

For equation (2):

\[
-5 - 5(3) = -5 - 15 = -20 \quad (True)
\]

Since both equations are satisfied, the solution is indeed:

\[
\boxed{(-5, 3)}
\]