Certainly! Let's solve the given system of linear equations using the elimination method step-by-step. The system of equations is:
[tex]\[
\begin{cases}
3a + 5b = 12 \\
-3a + 5b = -12
\end{cases}
\][/tex]
Step 1: Adding the Equations
First, let's add the two equations together. Adding the left-hand sides and the right-hand sides, we get:
[tex]\[
(3a + 5b) + (-3a + 5b) = 12 + (-12)
\][/tex]
Simplifying both sides, we get:
[tex]\[
3a - 3a + 5b + 5b = 0
\][/tex]
[tex]\[
0 + 10b = 0
\][/tex]
This simplifies to:
[tex]\[
10b = 0
\][/tex]
Dividing both sides by 10, we get:
[tex]\[
b = 0
\][/tex]
Step 2: Substitute [tex]\( b = 0 \)[/tex] into one of the original equations
Now that we have [tex]\( b = 0 \)[/tex], let's substitute [tex]\( b \)[/tex] back into one of the original equations to solve for [tex]\( a \)[/tex]. We can use the first equation:
[tex]\[
3a + 5b = 12
\][/tex]
Substitute [tex]\( b = 0 \)[/tex]:
[tex]\[
3a + 5(0) = 12
\][/tex]
Simplify:
[tex]\[
3a = 12
\][/tex]
Dividing both sides by 3, we get:
[tex]\[
a = 4
\][/tex]
Resulting System
The solutions [tex]\( a = 4 \)[/tex] and [tex]\( b = 0 \)[/tex] satisfy the system. If we focus on the resulting equations from the elimination method, the key resulting equation when eliminating [tex]\( a \)[/tex] was:
[tex]\[
10b = 0
\][/tex]
Additionally, after isolating [tex]\( b \)[/tex], we further simplified to find:
[tex]\[
a = 4
\][/tex]
Summary of Resulting Equations
- From eliminating [tex]\( a \)[/tex]: [tex]\( 10b = 0 \)[/tex] (or simplified as [tex]\( b = 0 \)[/tex])
- From solving for [tex]\( a \)[/tex]: [tex]\( 3a = 12 \)[/tex]
Thus, the correct answer from the given options would be:
1. [tex]\( 12b = 0 \)[/tex] (derived from [tex]\( b = 0 \)[/tex] after elimination and simplification)
There are no other options needed because we processed through correct elimination and found the unique results appropriate for given coefficients and constants.
