Certainly! Let's solve the given system of linear equations step-by-step:
[tex]\[
\begin{cases}
-3x - 4y = -2 \quad \text{(Equation 1)} \\
y = 2x - 5 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
### Step 1: Express [tex]\( y \)[/tex] from Equation 2
The second equation is already solved for [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 5 \][/tex]
### Step 2: Substitute [tex]\( y \)[/tex] from Equation 2 into Equation 1
Now we will replace [tex]\( y \)[/tex] in Equation 1 with [tex]\( 2x - 5 \)[/tex]:
[tex]\[ -3x - 4(2x - 5) = -2 \][/tex]
### Step 3: Simplify the Equation
Distribute [tex]\(-4\)[/tex] through [tex]\((2x - 5)\)[/tex]:
[tex]\[ -3x - 4 \cdot 2x + 4 \cdot 5 = -2 \][/tex]
[tex]\[ -3x - 8x + 20 = -2 \][/tex]
Combine like terms:
[tex]\[ -11x + 20 = -2 \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Isolate [tex]\( x \)[/tex] on one side:
[tex]\[ -11x + 20 = -2 \][/tex]
Subtract 20 from both sides:
[tex]\[ -11x = -2 - 20 \][/tex]
[tex]\[ -11x = -22 \][/tex]
Divide both sides by [tex]\(-11\)[/tex]:
[tex]\[ x = \frac{-22}{-11} \][/tex]
[tex]\[ x = 2 \][/tex]
### Step 5: Substitute [tex]\( x \)[/tex] back into Equation 2 to find [tex]\( y \)[/tex]
Using [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2(2) - 5 \][/tex]
[tex]\[ y = 4 - 5 \][/tex]
[tex]\[ y = -1 \][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[ x = 2 \][/tex]
[tex]\[ y = -1 \][/tex]
Thus, the values are:
[tex]\[ (x, y) = (2, -1) \][/tex]
This means the system of equations:
[tex]\[
\begin{cases}
-3x - 4y = -2 \\
y = 2x - 5
\end{cases}
\][/tex]
is satisfied by [tex]\( x = 2 \)[/tex] and [tex]\( y = -1 \)[/tex].
