Let's solve the system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
2x - 3y = -15 \\
8x - 4y = -1
\end{array}
\right.
\][/tex]
First, we label the equations for reference:
[tex]\[
\begin{array}{c}
\text{(1)} \quad 2x - 3y = -15 \\
\text{(2)} \quad 8x - 4y = -1
\end{array}
\][/tex]
### Step 1: Normalize equation (2)
To make equation [tex]\((2)\)[/tex] simpler, let's divide it by [tex]\(4\)[/tex], simplifying it:
[tex]\[
\frac{8x - 4y = -1}{4} \quad \Rightarrow \quad 2x - y = -\frac{1}{4}
\][/tex]
Now we have the system:
[tex]\[
\left\{
\begin{array}{l}
2x - 3y = -15 \\
2x - y = -\frac{1}{4}
\end{array}
\right.
\][/tex]
### Step 2: Eliminate [tex]\(x\)[/tex]
Next, we eliminate [tex]\(x\)[/tex] by subtracting the second equation from the first. Let's rewrite the equations:
[tex]\[
\text{(1)} \quad 2x - 3y = -15
\][/tex]
[tex]\[
\text{(3)} \quad 2x - y = -\frac{1}{4}
\][/tex]
Now subtract equation (3) from equation (1):
[tex]\[
(2x - 3y) - (2x - y) = -15 - \left(-\frac{1}{4}\right)
\][/tex]
Simplify:
[tex]\[
2x - 3y - 2x + y = -15 + \frac{1}{4}
\][/tex]
[tex]\[
-2y = -15 + \frac{1}{4}
\][/tex]
Convert [tex]\(-15\)[/tex] to a fraction with the same denominator:
[tex]\[
-15 = -\frac{60}{4}
\][/tex]
Now:
[tex]\[
-2y = -\frac{60}{4} + \frac{1}{4}
\][/tex]
Combine the fractions:
[tex]\[
-2y = -\frac{59}{4}
\][/tex]
Divide both sides by [tex]\(-2\)[/tex]:
[tex]\[
y = \frac{\frac{59}{4}}{2} = \frac{59}{8}
\][/tex]
### Step 3: Solve for [tex]\(x\)[/tex]
Now that we have [tex]\(y = \frac{59}{8}\)[/tex], substitute this value back into equation (3):
[tex]\[
2x - y = -\frac{1}{4}
\][/tex]
Substitute [tex]\(y\)[/tex]:
[tex]\[
2x - \frac{59}{8} = -\frac{1}{4}
\][/tex]
Convert [tex]\(-\frac{1}{4}\)[/tex] to a fraction with the same denominator:
[tex]\[
-\frac{1}{4} = -\frac{2}{8}
\][/tex]
So the equation becomes:
[tex]\[
2x - \frac{59}{8} = -\frac{2}{8}
\][/tex]
Add [tex]\(\frac{59}{8}\)[/tex] to both sides:
[tex]\[
2x = \frac{59}{8} - \frac{2}{8} = \frac{57}{8}
\][/tex]
Divide both sides by 2:
[tex]\[
x = \frac{\frac{57}{8}}{2} = \frac{57}{16}
\][/tex]
### Final Answer
Thus, the solution to the system of equations is:
[tex]\[
x = \frac{57}{16}, \quad y = \frac{59}{8}
\][/tex]
