Alright, let's solve the system of equations step by step using the elimination method.
Given the system of equations:
[tex]\[
\begin{array}{l}
8x + 7y = 39 \quad \text{(1)} \\
4x - 14y = -68 \quad \text{(2)}
\end{array}
\][/tex]
### Step 1: Multiply the first equation to enable the elimination of the [tex]\( y \)[/tex]-term
To eliminate the [tex]\( y \)[/tex]-term, we need to get the coefficients of [tex]\( y \)[/tex] to be the same (or opposites). We can do this by multiplying the first equation by 2:
[tex]\[
2 \times (8x + 7y) = 2 \times 39 \\
16x + 14y = 78 \quad \text{(3)}
\][/tex]
### Step 2: Add the equations to eliminate the [tex]\( y \)[/tex]-terms
Now we add equation (3) to equation (2):
[tex]\[
16x + 14y + 4x - 14y = 78 + (-68)
\][/tex]
The [tex]\( y \)[/tex]-terms cancel out:
[tex]\[
16x + 4x = 10 \implies 20x = 10
\][/tex]
### Step 3: Solve the new equation for the [tex]\( x \)[/tex]-value
We divide both sides of the equation by 20:
[tex]\[
x = \frac{10}{20} = \frac{1}{2}
\][/tex]
### Step 4: Substitute the [tex]\( x \)[/tex]-value back into either original equation to find the [tex]\( y \)[/tex]-value
We substitute [tex]\( x = \frac{1}{2} \)[/tex] into the first equation (1):
[tex]\[
8 \left( \frac{1}{2} \right) + 7y = 39 \\
4 + 7y = 39 \\
7y = 39 - 4 \\
7y = 35 \\
y = \frac{35}{7} = 5
\][/tex]
Thus, the solution to the system of equations is [tex]\(\left( \frac{1}{2}, 5 \right)\)[/tex].
### Step 5: Check the solution
We should check our solution by substituting [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( y = 5 \)[/tex] back into both original equations to ensure they are true.
[tex]\[
8 \left( \frac{1}{2} \right) + 7 \times 5 = 39 \\
4 - 14 \times 5 = -68
\][/tex]
Both equations hold true, so our solution is confirmed.
The solution to the system of equations is [tex]\(\left( \frac{1}{2}, 5 \right)\)[/tex].
