Consider the following system of equations: [tex]
\begin{aligned}
2x - y &= 12 \\
-3x - 5y &= -5
\end{aligned}
[/tex]
The steps for solving the given system of equations are shown below: [tex]
\begin{array}{l}
\text{Step 1:} \quad -5(2x - y) = -5(12) \\
-3x - 5y = -5 \\
\text{Step 2:} \quad -10x + 5y = -60 \\
-3x - 5y = -5 \\
\text{Step 3:} \quad -13x = -65 \\
\text{Step 4:} \quad x = 5 \\
\text{Step 5:} \quad 2(5) - y = 12 \\
\text{Step 6:} \quad y = -2 \\
\text{Solution:} \quad (5, -2) \\
\end{array}
[/tex]
Select the correct statement about Step 3:
A. When the equation [tex]-3x - 5y = -5[/tex] is subtracted from [tex]-10x + 5y = -60[/tex], a third linear equation, [tex]-13x = -65[/tex], is formed, and it shares a common solution with the original equations.
B. When the equation [tex]-3x - 5y = -5[/tex] is subtracted from [tex]-10x + 5y = -60[/tex], a third linear equation, [tex]-13x = -65[/tex], is formed, and it has a different solution from the original equations.
C. When the equations [tex]-10x + 5y = -60[/tex] and [tex]-3x - 5y = -5[/tex] are added together, a third linear equation, [tex]-13x = -65[/tex], is formed, and it has a different solution from the original equations.
D. When the equations [tex]-10x + 5y = -60[/tex] and [tex]-3x - 5y = -5[/tex] are added together, a third linear equation, [tex]-13x = -65[/tex], is formed, and it shares a common solution with the original equations.
Given the system of equations: [tex]\[
\begin{aligned}
2x - y &= 12 \\
-3x - 5y &= -5
\end{aligned}
\][/tex]
Follow the steps to solve the system:
1. Step 1: Multiply the first equation by -5: [tex]\[
-5(2x - y) = -5(12)
\][/tex] which results in: [tex]\[
-10x + 5y = -60
\][/tex]
Now our system is: [tex]\[
\begin{aligned}
-10x + 5y &= -60 \\
-3x - 5y &= -5
\end{aligned}
\][/tex]
2. Step 2: Rewrite the system with the new equation from Step 1: [tex]\[
\begin{aligned}
-10x + 5y &= -60 \\
-3x - 5y &= -5
\end{aligned}
\][/tex]
3. Step 3: Subtract the second equation from the first: [tex]\[
(-10x + 5y) - (-3x - 5y) = -60 - (-5)
\][/tex] Simplifying this, you get: [tex]\[
-10x + 5y + 3x + 5y = -60 + 5
\][/tex] [tex]\[
-7x + 10y = -55
\][/tex] Since the first variable terms combine to [tex]\(-7x\)[/tex] and the constants on the right-hand side combine to [tex]\(-55\)[/tex], you end up with: [tex]\[
-7x = -55
\][/tex] This simplifies to: [tex]\[
x = 5
\][/tex]
4. Now, substitute [tex]\(x = 5\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. Choose [tex]\(2x - y = 12\)[/tex]: [tex]\[
2(5) - y = 12
\][/tex] [tex]\[
10 - y = 12
\][/tex] [tex]\[
-y = 2
\][/tex] [tex]\[
y = -2
\][/tex]
Thus, the solution to the original system of equations is [tex]\((x, y) = (5, -2)\)[/tex].
Given the steps and the solution, the correct statement about Step 3 is: A. When the equation [tex]\( -3x - 5y = -5 \)[/tex] is subtracted from [tex]\( -10x + 5y = -60 \)[/tex], a third linear equation, [tex]\( -13x = -65 \)[/tex], is formed, and it shares a common solution with the original equations.
