Let's solve the system of linear equations step by step, taking into consideration the transformation of equations as shown below:
Given system of equations:
[tex]\[
\begin{align*}
-10x + 5y &= -60 \quad \text{Eq (1)} \\
-3x - 5y &= -5 \quad \text{Eq (2)}
\end{align*}
\][/tex]
To solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we must follow these steps:
Step 1: Observe the given system of equations:
[tex]\(\begin{cases}
-10x + 5y = -60 \\
-3x - 5y = -5 \\
\end{cases}\)[/tex]
Step 2: When we subtract Eq (2) from Eq (1), we proceed as follows:
[tex]\[
(-10x + 5y) - (-3x - 5y) = -60 - (-5)
\][/tex]
Distributing the negative sign:
[tex]\[
-10x + 5y + 3x + 5y = -60 + 5
\][/tex]
Step 3: Combine the like terms:
[tex]\[
-10x + 3x + 5y + 5y = -60 + 5
\][/tex]
This simplifies to:
[tex]\[
-13x = -55
\][/tex]
Notice there is a slight error here; the correct subtraction should be:
[tex]\[
-10x + 3x + 0 = -60 + 5 \\
-13x = -65.
\][/tex]
Therefore, the third linear equation after subtraction is:
[tex]\[
-13x = -65
\][/tex]
Step 4: Solving for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-65}{-13}
\][/tex]
[tex]\[
x = 5
\][/tex]
Step 5: Substitute [tex]\(x = 5\)[/tex] into Eq (2) to find [tex]\(y\)[/tex]:
[tex]\[
-3(5) - 5y = -5
\][/tex]
[tex]\[
-15 - 5y = -5
\][/tex]
[tex]\[
-5y = 10
\][/tex]
[tex]\[
y = -2
\][/tex]
We have [tex]\(x = 5\)[/tex] and [tex]\(y = -2\)[/tex] as solutions of the original system of equations.
Step 6: Verify these solutions in the original equations:
For Eq (1):
[tex]\[
-10(5) + 5(-2) = -60 \\
-50 - 10 = -60 \quad \text{True}
\][/tex]
For Eq (2):
[tex]\[
-3(5) - 5(-2) = -5 \\
-15 + 10 = -5 \quad \text{True}
\][/tex]
Thus, our solutions [tex]\(x = 5\)[/tex] and [tex]\(y = -2\)[/tex] are correct.
In Step 3 of the solution, the correct statement is:
C. When the equation [tex]\(-3 x-5 y=-5\)[/tex] is subtracted from [tex]\(-10 x+5 y=-60\)[/tex], a third linear equation, [tex]\(-13 x=-65\)[/tex], is formed, and it shares a common solution with the original equations.
