Answer :
To solve the given system of equations by eliminating one of the variables, we can use the method of elimination. The given system is:
[tex]\[ \begin{array}{l} 4x + y = 4 \quad \text{(Equation 1)} \\ 2x + 6y = 24 \quad \text{(Equation 2)} \end{array} \][/tex]
Step-by-step, let's analyze each option provided:
1. Multiply the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex].
- Multiplying the entire first equation [tex]\(4x + y = 4\)[/tex] by -4:
[tex]\[ -4 \times (4x + y) = -4 \times 4 \][/tex]
This results in:
[tex]\[ -16x - 4y = -16 \][/tex]
Let's align this with the original second equation [tex]\(2x + 6y = 24\)[/tex]:
[tex]\[ \begin{array}{l} -16x - 4y = -16 \\ 2x + 6y = 24 \end{array} \][/tex]
By adding these equations together, the [tex]\(y\)[/tex] terms will cancel each other out if the second equation is appropriately manipulated.
2. Multiply the second equation by -4 to get [tex]$-8x - 24y = -96$[/tex].
- Multiplying the entire second equation [tex]\(2x + 6y = 24\)[/tex] by -4:
[tex]\[ -4 \times (2x + 6y) = -4 \times 24 \][/tex]
This results in:
[tex]\[ -8x - 24y = -96 \][/tex]
Using this transformation to eliminate [tex]\(x\)[/tex] or [tex]\(y\)[/tex] with the modified first equation presents a challenge because their coefficients do not align in a way that simple addition would cancel out one variable.
3. Multiply the first equation by -2 to get [tex]$-8x - 2y = -8$[/tex].
- Multiplying the entire first equation [tex]\(4x + y = 4\)[/tex] by -2:
[tex]\[ -2 \times (4x + y) = -2 \times 4 \][/tex]
This results in:
[tex]\[ -8x - 2y = -8 \][/tex]
With this modified first equation and the original second equation:
[tex]\[ \begin{array}{l} -8x - 2y = -8 \\ 2x + 6y = 24 \end{array} \][/tex]
While this may not be straightforward to use for elimination.
4. Multiply the second equation by -2 to get [tex]$-4x - 12y = -48$[/tex].
- Multiplying the entire second equation [tex]\(2x + 6y = 24\)[/tex] by -2:
[tex]\[ -2 \times (2x + 6y) = -2 \times 24 \][/tex]
This results in:
[tex]\[ -4x - 12y = -48 \][/tex]
Aligning this with the original first equation:
[tex]\[ \begin{array}{l} 4x + y = 4 \\ -4x - 12y = -48 \end{array} \][/tex]
It is not evident how this modifies the system for easy elimination without further simplification.
Correct Transformation:
The goal is to manipulate the first equation in a way that, when combined with the second equation, allows one variable to cancel out. The best candidate here is:
1. Multiplying the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex].
Therefore, the correct action to create an equivalent system that helps in variable elimination is to multiply the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex]. This transformation aligns the system for easier elimination of [tex]\(y\)[/tex] when added to a corresponding transformation of the second equation.
[tex]\[ \begin{array}{l} 4x + y = 4 \quad \text{(Equation 1)} \\ 2x + 6y = 24 \quad \text{(Equation 2)} \end{array} \][/tex]
Step-by-step, let's analyze each option provided:
1. Multiply the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex].
- Multiplying the entire first equation [tex]\(4x + y = 4\)[/tex] by -4:
[tex]\[ -4 \times (4x + y) = -4 \times 4 \][/tex]
This results in:
[tex]\[ -16x - 4y = -16 \][/tex]
Let's align this with the original second equation [tex]\(2x + 6y = 24\)[/tex]:
[tex]\[ \begin{array}{l} -16x - 4y = -16 \\ 2x + 6y = 24 \end{array} \][/tex]
By adding these equations together, the [tex]\(y\)[/tex] terms will cancel each other out if the second equation is appropriately manipulated.
2. Multiply the second equation by -4 to get [tex]$-8x - 24y = -96$[/tex].
- Multiplying the entire second equation [tex]\(2x + 6y = 24\)[/tex] by -4:
[tex]\[ -4 \times (2x + 6y) = -4 \times 24 \][/tex]
This results in:
[tex]\[ -8x - 24y = -96 \][/tex]
Using this transformation to eliminate [tex]\(x\)[/tex] or [tex]\(y\)[/tex] with the modified first equation presents a challenge because their coefficients do not align in a way that simple addition would cancel out one variable.
3. Multiply the first equation by -2 to get [tex]$-8x - 2y = -8$[/tex].
- Multiplying the entire first equation [tex]\(4x + y = 4\)[/tex] by -2:
[tex]\[ -2 \times (4x + y) = -2 \times 4 \][/tex]
This results in:
[tex]\[ -8x - 2y = -8 \][/tex]
With this modified first equation and the original second equation:
[tex]\[ \begin{array}{l} -8x - 2y = -8 \\ 2x + 6y = 24 \end{array} \][/tex]
While this may not be straightforward to use for elimination.
4. Multiply the second equation by -2 to get [tex]$-4x - 12y = -48$[/tex].
- Multiplying the entire second equation [tex]\(2x + 6y = 24\)[/tex] by -2:
[tex]\[ -2 \times (2x + 6y) = -2 \times 24 \][/tex]
This results in:
[tex]\[ -4x - 12y = -48 \][/tex]
Aligning this with the original first equation:
[tex]\[ \begin{array}{l} 4x + y = 4 \\ -4x - 12y = -48 \end{array} \][/tex]
It is not evident how this modifies the system for easy elimination without further simplification.
Correct Transformation:
The goal is to manipulate the first equation in a way that, when combined with the second equation, allows one variable to cancel out. The best candidate here is:
1. Multiplying the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex].
Therefore, the correct action to create an equivalent system that helps in variable elimination is to multiply the first equation by -4 to get [tex]$-16x - 4y = -16$[/tex]. This transformation aligns the system for easier elimination of [tex]\(y\)[/tex] when added to a corresponding transformation of the second equation.
