Question

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 two equations:

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

We can eliminate \( y \) by adding the two equations after adjusting them, or we can multiply the first equation by a suitable number to make the coefficients of \( y \) in both equations add to zero.

To eliminate \( y \), we can multiply the first equation by 1 and the second equation by 1 (since the coefficient of \( y \) in the second equation is already suitable):

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

Now add the two equations together:

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

This simplifies to:

\[
6x = 24
\]

Now, solve for \( x \):

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

Next, substitute \( x = 4 \) back into one of the original equations to solve for \( y \). We'll use the first equation:

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

This simplifies to:

\[
8 - 3y = 14
\]

Now, subtract 8 from both sides:

\[
-3y = 14 - 8
\]
\[
-3y = 6
\]

Now, divide by -3:

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

Thus, the solution to the system of equations is:

\[
(x, y) = (4, -2)
\]

In conclusion, the answer is:

\[
\boxed{(4, -2)}
\]