Answer :
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.
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.
