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André solves the following system of equations by elimination:
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].
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