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Solve for [tex]\( x \)[/tex]:
[tex]\[ 3x = 6x - 2 \][/tex]

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Steps for solving the given system of equations are shown below:
[tex]\[ -3x - 5y = -b \][/tex]

\begin{tabular}{|c|c|c|c|}
\hline
Step 1: & [tex]$-5(2x - y$[/tex] & [tex]$= 12$[/tex] & 2) \\
\hline
& [tex]$-3x - 5y$[/tex] & [tex]$= -5$[/tex] & \\
\hline
Step 2: & [tex]$-10x + 5y$[/tex] & [tex]$= -60$[/tex] & \\
\hline
& [tex]$-3x - 5y$[/tex] & [tex]$= -5$[/tex] & \\
\hline
Step 3: & [tex]$-13x$[/tex] & [tex]$= -65$[/tex] & \\
\hline
Step 4: & [tex]$x$[/tex] & [tex]$= 5$[/tex] & \\
\hline
Step 5: & [tex]$2(5) - y$[/tex] & [tex]$= 12$[/tex] & \\
\hline
Step 6: & [tex]$y$[/tex] & [tex]$= -2$[/tex] & \\
\hline
\end{tabular}

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 has a different solution from the original equations.

B. 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.

C. 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.

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 has a different solution from the original equations.



Answer :

GinnyAnswer
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.

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