Question

consider the system of equations below
{-2x + y = 2
{2x + 5y = 19
what is the solution of the system


A. (1 1/8, 4 1/4)
B. (1 2/3, 5 1/3)
C. (3 1/4, 2 1/2)
D. (3/4, 3 1/2)

Answer 1

To solve the system of equations:

1. \(-2x + y = 2\)
2. \(2x + 5y = 19\)

We can use substitution or elimination. Here, we'll use substitution.

First, solve the first equation for \(y\):

\[
y = 2 + 2x
\]

Now, substitute \(y\) in the second equation:

\[
2x + 5(2 + 2x) = 19
\]

Expanding this:

\[
2x + 10 + 10x = 19
\]

Combine like terms:

\[
12x + 10 = 19
\]

Now, isolate \(x\):

\[
12x = 19 - 10
\]
\[
12x = 9
\]
\[
x = \frac{9}{12} = \frac{3}{4}
\]

Now substitute \(x = \frac{3}{4}\) back into the equation for \(y\):

\[
y = 2 + 2\left(\frac{3}{4}\right)
\]
\[
y = 2 + \frac{6}{4} = 2 + \frac{3}{2} = \frac{4}{2} + \frac{3}{2} = \frac{7}{2}
\]

Converting \(\frac{7}{2}\) to a mixed number gives:

\[
y = 3.5 = 3 \frac{1}{2}
\]

Thus, the solution to the system of equations is:

\[
\left(\frac{3}{4}, \frac{7}{2}\right) = \left(\frac{3}{4}, 3 \frac{1}{2}\right)
\]

So the solution is:

**D. \((\frac{3}{4}, 3 \frac{1}{2})\)**