André solves the following system of equations by elimination:

[tex]\[
\begin{cases}
3a + 6b = 12 \\
-3a + 6b = -12
\end{cases}
\][/tex]

Which could be the resulting equations when André eliminates one of the variables?



Answer :

To solve the given system of equations using the elimination method, we follow these steps:

1. Write down the system of equations:
[tex]\[ \begin{cases} 3a + 6b = 12 \\ -3a + 6b = -12 \end{cases} \][/tex]

2. Add the two equations together to eliminate variable [tex]\(a\)[/tex]:
[tex]\[ (3a + 6b) + (-3a + 6b) = 12 + (-12) \][/tex]
Simplify the left-hand side and the right-hand side:
[tex]\[ 3a + 6b - 3a + 6b = 0 \][/tex]

3. Combine like terms:
[tex]\[ 0a + 12b = 0 \][/tex]
Which simplifies to:
[tex]\[ 12b = 0 \][/tex]

4. Solve for [tex]\(b\)[/tex]:
[tex]\[ 12b = 0 \implies b = 0 \][/tex]

5. Substitute [tex]\(b = 0\)[/tex] back into one of the original equations to find [tex]\(a\)[/tex]:
Using the first equation:
[tex]\[ 3a + 6(0) = 12 \][/tex]
Simplify:
[tex]\[ 3a = 12 \][/tex]
Now, solve for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{12}{3} = 4 \][/tex]

Therefore, the resulting equation when one of the variables is eliminated is:
[tex]\[ 12b = 0 \][/tex]
And the solutions for the variables are [tex]\(a = 4\)[/tex] and [tex]\(b = 0\)[/tex].