Question

Question:
Solve the following system of equations using the elimination method.
2x+y=3.5
−x−2y=2.5

Answer 1

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

1) \( 2x + y = 3.5 \)
2) \( -x - 2y = 2.5 \)

First, let's manipulate the first equation to make the elimination process easier. We can multiply the first equation by 2 so that the coefficients of \( y \) in both equations will be easier to eliminate:

Multiply equation (1) by 2:
\[
2(2x + y) = 2(3.5)
\]
This simplifies to:
\[
4x + 2y = 7
\]

Now we have a new system:

3) \( 4x + 2y = 7 \)
4) \( -x - 2y = 2.5 \)

Next, we'll add equations (3) and (4) together to eliminate \( y \):

\[
(4x + 2y) + (-x - 2y) = 7 + 2.5
\]

Combining like terms gives:

\[
4x - x + 2y - 2y = 9.5
\]
\[
3x = 9.5
\]

Now solving for \( x \):

\[
x = \frac{9.5}{3} = 3.1667 \quad \text{(or } x = \frac{19}{6} \text{)}
\]

Next, we substitute the value of \( x \) back into one of the original equations to find \( y \). We'll use equation (1):

\[
2x + y = 3.5
\]

Substituting \( x = \frac{19}{6} \):

\[
2\left(\frac{19}{6}\right) + y = 3.5
\]
\[
\frac{38}{6} + y = 3.5
\]
\[
\frac{19}{3} + y = 3.5
\]

To express \( 3.5 \) with a denominator of 3, we convert:

\[
3.5 = \frac{10.5}{3}
\]
Now, we have:

\[
\frac{19}{3} + y = \frac{10.5}{3}
\]

Subtract \( \frac{19}{3} \) from both sides:

\[
y = \frac{10.5}{3} - \frac{19}{3} = \frac{10.5 - 19}{3} = \frac{-8.5}{3} = -\frac{17}{6}
\]

Now we have both values:
\[
x = \frac{19}{6}, \quad y = -\frac{17}{6}
\]

Thus, the solution to the system of equations is:
\[
\left( x, y \right) = \left( \frac{19}{6}, -\frac{17}{6} \right)
\]