To solve the system of equations:
[tex]\[
\left\{\begin{array}{l}
7x - 3y = 27 \\
5x + y = 35
\end{array}\right.
\][/tex]
we will use the method of elimination. Follow these steps:
1. Rewrite the equations:
[tex]\[
7x - 3y = 27 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
5x + y = 35 \quad \text{(Equation 2)}
\][/tex]
2. Eliminate [tex]\(y\)[/tex] by making the coefficients of [tex]\(y\)[/tex] in both equations opposites.
- To do that, multiply Equation 2 by 3, so the coefficient of [tex]\(y\)[/tex] will be [tex]\(3\)[/tex]:
[tex]\[
3(5x + y) = 3 \cdot 35
\][/tex]
Which simplifies to:
[tex]\[
15x + 3y = 105 \quad \text{(Modified Equation 2)}
\][/tex]
3. Add Equation 1 and the Modified Equation 2:
[tex]\[
(7x - 3y) + (15x + 3y) = 27 + 105
\][/tex]
Simplify this to get:
[tex]\[
7x + 15x = 132
\][/tex]
[tex]\[
22x = 132
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{132}{22}
\][/tex]
[tex]\[
x = 6
\][/tex]
5. Substitute [tex]\(x = 6\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We will use Equation 2:
[tex]\[
5x + y = 35
\][/tex]
Substitute [tex]\(x = 6\)[/tex]:
[tex]\[
5(6) + y = 35
\][/tex]
[tex]\[
30 + y = 35
\][/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[
y = 35 - 30
\][/tex]
[tex]\[
y = 5
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
x = 6, \quad y = 5
\][/tex]
