Question

Solve the following system of equations using the elimination method.

2x−3y=14
, 4x+3y=10


(1 point)

Answer 1

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

1. \( 2x - 3y = 14 \) (Equation 1)
2. \( 4x + 3y = 10 \) (Equation 2)

First, we can eliminate \(y\) by adding the two equations together. To do this, we can multiply Equation 1 by 1 (no change necessary) and Equation 2 by 1 (again, no change):

1. \( 2x - 3y = 14 \)
2. \( 4x + 3y = 10 \)

Now, we add the two equations:

\[
(2x - 3y) + (4x + 3y) = 14 + 10
\]

This simplifies to:

\[
6x = 24
\]

Next, we solve for \(x\):

\[
x = \frac{24}{6} = 4
\]

Now that we have \(x\), we will substitute this value back into one of the original equations to solve for \(y\). We'll use Equation 1:

\[
2(4) - 3y = 14
\]

This simplifies to:

\[
8 - 3y = 14
\]

Now, we can isolate \(y\):

\[
-3y = 14 - 8
\]
\[
-3y = 6
\]
\[
y = \frac{6}{-3} = -2
\]

Thus, the solution to the system of equations is:

\[
x = 4, \quad y = -2
\]

Final answer:

\((4, -2)\)