Question

What is the solution to the system of equations?

`x+2y=10`

`3x+4y=8`

Answer 1

To solve the system of equations given by:

1. \( x + 2y = 10 \) (Equation 1)
2. \( 3x + 4y = 8 \) (Equation 2)

we can use the substitution or elimination method. Here, we'll use the substitution method:

First, solve Equation 1 for \( x \):

\[
x = 10 - 2y
\]

Now, substitute \( x \) in Equation 2:

\[
3(10 - 2y) + 4y = 8
\]

Distribute \( 3 \):

\[
30 - 6y + 4y = 8
\]

Combine the \( y \) terms:

\[
30 - 2y = 8
\]

Now isolate \( y \):

\[
-2y = 8 - 30
\]
\[
-2y = -22
\]
\[
y = \frac{-22}{-2} = 11
\]

Now that we have \( y = 11 \), substitute it back into the expression for \( x \):

\[
x = 10 - 2(11)
\]
\[
x = 10 - 22
\]
\[
x = -12
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (-12, 11)
\]

To verify, plug \( x = -12 \) and \( y = 11 \) back into the original equations:

1. For \( x + 2y = 10 \):

\[
-12 + 2(11) = -12 + 22 = 10 \quad \text{(True)}
\]

2. For \( 3x + 4y = 8 \):

\[
3(-12) + 4(11) = -36 + 44 = 8 \quad \text{(True)}
\]

Both equations are satisfied, so the final solution is:

\[
\boxed{(-12, 11)}
\]