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Consider the following system of equations:

[tex]\[
\begin{array}{r}
2x - y = 12 \\
-3x - 5y = -5
\end{array}
\][/tex]

The steps for solving the given system of equations are shown below:

[tex]\[
\begin{array}{rlrl}
\text{Step 1:} & -5(2x - y) & = -60 \\
& -3x - 5y & = -5 \\
\text{Step 2:} & -10x + 5y & = -60 \\
& -3x - 5y & = -5 \\
\text{Step 3:} & -13x & = -65 \\
\text{Step 4:} & x & = 5 \\
\text{Step 5:} & 2(5) - y & = 12 \\
\text{Step 6:} & y & = -2 \\
\end{array}
\][/tex]

Solution: [tex]\((5, -2)\)[/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 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 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 shares a common solution with the original equations.

D. 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.



Answer :

GinnyAnswer
Let's solve the given system of equations step by step to determine the correct statement about step 3.

The given system of equations is:
[tex]\[ \begin{array}{rcl} 2x - y & = & 12 \quad \text{(Equation 1)} \\ -3x - 5y & = & -5 \quad \text{(Equation 2)} \end{array} \][/tex]

Step 1: Multiply Equation 1 by -5 to align the coefficients of y.

[tex]\[ -5 \times (2x - y) = -5 \times 12 \][/tex]
[tex]\[ -10x + 5y = -60 \quad \text{(Equation 3)} \][/tex]

Step 2: Add Equation 3 to Equation 2 to eliminate y.

[tex]\[ \begin{array}{rcl} -10x + 5y & = & -60 \\ -3x - 5y & = & -5 \\ \hline -13x & = & -65 \end{array} \][/tex]

Step 3: Solve for x.

[tex]\[ -13x = -65 \][/tex]
[tex]\[ x = \frac{-65}{-13} = 5 \][/tex]

Step 4: Substitute [tex]\( x = 5 \)[/tex] back into Equation 1 to solve for y.

[tex]\[ 2x - y = 12 \][/tex]
[tex]\[ 2(5) - y = 12 \][/tex]
[tex]\[ 10 - y = 12 \][/tex]
[tex]\[ -y = 12 - 10 \][/tex]
[tex]\[ y = -2 \][/tex]

The solution to the system of equations is [tex]\( (5, -2) \)[/tex].

Now, to verify which statement is correct about Step 3:

Step 3: When the equations [tex]\(-10x + 5y = -60\)[/tex] and [tex]\(-3x - 5y = -5\)[/tex] are added together, they form the equation [tex]\( -13x = -65 \)[/tex].

We found that solving this equation gave us [tex]\( x = 5 \)[/tex], which, when substituted back into the original system, yields [tex]\( y = -2 \)[/tex]. This means the equation [tex]\(-13x = -65\)[/tex] shares a common solution with the original system of equations.

Thus, the correct statement is:
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 shares a common solution with the original equations.

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