Question

Solve the following system of equations with the elimination method. Write your answers as an Ordered Pair. If infinitely many solutions exist, enter
(oo). If no solution exists, enter DNE
7y - 2x =20
4y - 5x= -4

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

1. \( 7y - 2x = 20 \)
2. \( 4y - 5x = -4 \)

First, we can rearrange both equations into a standard form (Ax + By = C) if necessary. However, we’ll go ahead with elimination as they are.

Next, we can eliminate one of the variables by making the coefficients of either \(x\) or \(y\) equal in both equations. Let's eliminate \(x\):

To do this, we can multiply both equations by suitable constants to equalize the coefficients of \(x\):

- For the first equation \( 7y - 2x = 20 \), we multiply by 5:
\[
5(7y - 2x) = 5(20) \implies 35y - 10x = 100 \quad \text{(Equation 3)}
\]

- For the second equation \( 4y - 5x = -4 \), we multiply by 2:
\[
2(4y - 5x) = 2(-4) \implies 8y - 10x = -8 \quad \text{(Equation 4)}
\]

Now we have the following two new equations:
1. \( 35y - 10x = 100 \) (Equation 3)
2. \( 8y - 10x = -8 \) (Equation 4)

Next, we will subtract Equation 4 from Equation 3 to eliminate \(x\):
\[
(35y - 10x) - (8y - 10x) = 100 - (-8)
\]
This simplifies to:
\[
35y - 8y = 100 + 8
\]
\[
27y = 108
\]
Now, solve for \(y\):
\[
y = \frac{108}{27} = 4
\]

Now that we have the value of \(y\), we can substitute it back into one of the original equations to find \(x\). We'll use the first equation:

\[
7y - 2x = 20
\]
Substituting \(y = 4\):
\[
7(4) - 2x = 20
\]
\[
28 - 2x = 20
\]
Now, solve for \(x\):
\[
-2x = 20 - 28
\]
\[
-2x = -8
\]
\[
x = \frac{-8}{-2} = 4
\]

So, the solution to the system of equations is:
\[
(x, y) = (4, 4)
\]

Final answer:
\[
\boxed{(4, 4)}
\]