To solve the given system of equations using the addition method, follow these steps:
1. Write down the given system of equations:
[tex]\[
\begin{cases}
7x + 3y = 10 \quad \text{(Equation 1)} \\
4x + 4y = 8 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
2. Aim for elimination of one variable by making the coefficients of either [tex]\( y \)[/tex] or [tex]\( x \)[/tex] the same in both equations. Here, we will eliminate [tex]\( y \)[/tex]. To do this, we need the coefficients of [tex]\( y \)[/tex] in both equations to be the same.
- Multiply Equation 1 by 4:
[tex]\[
4(7x + 3y) = 4 \cdot 10 \implies 28x + 12y = 40 \quad \text{(Equation 3)}
\][/tex]
- Multiply Equation 2 by 3:
[tex]\[
3(4x + 4y) = 3 \cdot 8 \implies 12x + 12y = 24 \quad \text{(Equation 4)}
\][/tex]
3. Subtract Equation 4 from Equation 3 to eliminate [tex]\( y \)[/tex]:
[tex]\[
(28x + 12y) - (12x + 12y) = 40 - 24
\][/tex]
Simplifying, we get:
[tex]\[
(28x - 12x) + (12y - 12y) = 16
\][/tex]
[tex]\[
16x = 16
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
16x = 16
\][/tex]
[tex]\[
x = \frac{16}{16}
\][/tex]
[tex]\[
x = 1
\][/tex]
5. Substitute [tex]\( x = 1 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Let's use Equation 2:
[tex]\[
4x + 4y = 8
\][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[
4(1) + 4y = 8
\][/tex]
[tex]\[
4 + 4y = 8
\][/tex]
[tex]\[
4y = 8 - 4
\][/tex]
[tex]\[
4y = 4
\][/tex]
[tex]\[
y = \frac{4}{4}
\][/tex]
[tex]\[
y = 1
\][/tex]
Thus, the solution set is [tex]\((x, y) = (1, 1)\)[/tex].
[tex]\[
\boxed{(1, 1)}
\][/tex]
