Answer :
To solve the system of equations:
[tex]\[ \begin{array}{c} x - 2y = 15 \\ 2x + 4y = -18 \end{array} \][/tex]
First, let's use the substitution or elimination method to find the solution:
1. Start with the first equation:
[tex]\[ x - 2y = 15 \quad \text{(Equation 1)} \][/tex]
2. Solve Equation 1 for [tex]\( x \)[/tex]:
[tex]\[ x = 15 + 2y \][/tex]
3. Substitute [tex]\( x \)[/tex] into the second equation:
[tex]\[ 2(15 + 2y) + 4y = -18 \quad \text{(Equation 2)} \][/tex]
4. Expand and simplify:
[tex]\[ 30 + 4y + 4y = -18 \][/tex]
[tex]\[ 30 + 8y = -18 \][/tex]
5. Isolate [tex]\( y \)[/tex]:
[tex]\[ 8y = -18 - 30 \][/tex]
[tex]\[ 8y = -48 \][/tex]
[tex]\[ y = -6 \][/tex]
6. Now substitute [tex]\( y = -6 \)[/tex] back into the first equation to find [tex]\( x \)[/tex]:
[tex]\[ x - 2(-6) = 15 \][/tex]
[tex]\[ x + 12 = 15 \][/tex]
[tex]\[ x = 3 \][/tex]
Hence, the solution to the system of equations is:
[tex]\[ x = 3, \, y = -6 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{C: x=3, y=-6} \][/tex]
[tex]\[ \begin{array}{c} x - 2y = 15 \\ 2x + 4y = -18 \end{array} \][/tex]
First, let's use the substitution or elimination method to find the solution:
1. Start with the first equation:
[tex]\[ x - 2y = 15 \quad \text{(Equation 1)} \][/tex]
2. Solve Equation 1 for [tex]\( x \)[/tex]:
[tex]\[ x = 15 + 2y \][/tex]
3. Substitute [tex]\( x \)[/tex] into the second equation:
[tex]\[ 2(15 + 2y) + 4y = -18 \quad \text{(Equation 2)} \][/tex]
4. Expand and simplify:
[tex]\[ 30 + 4y + 4y = -18 \][/tex]
[tex]\[ 30 + 8y = -18 \][/tex]
5. Isolate [tex]\( y \)[/tex]:
[tex]\[ 8y = -18 - 30 \][/tex]
[tex]\[ 8y = -48 \][/tex]
[tex]\[ y = -6 \][/tex]
6. Now substitute [tex]\( y = -6 \)[/tex] back into the first equation to find [tex]\( x \)[/tex]:
[tex]\[ x - 2(-6) = 15 \][/tex]
[tex]\[ x + 12 = 15 \][/tex]
[tex]\[ x = 3 \][/tex]
Hence, the solution to the system of equations is:
[tex]\[ x = 3, \, y = -6 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{C: x=3, y=-6} \][/tex]