To solve the system of equations by the elimination method, follow these steps:
[tex]\[
\begin{array}{l}
2a + 3b = 6 \quad \text{(1)} \\
5a + 2b - 4 = 0 \quad \text{(2)}
\end{array}
\][/tex]
First, rearrange the second equation to match the form of the first equation:
[tex]\[
5a + 2b = 4 \quad \text{(3)}
\][/tex]
Now, we have the system:
[tex]\[
\begin{array}{l}
2a + 3b = 6 \quad \text{(1)} \\
5a + 2b = 4 \quad \text{(3)}
\end{array}
\][/tex]
The goal is to eliminate one of the variables. To do this, we need the coefficients of either [tex]\(a\)[/tex] or [tex]\(b\)[/tex] to be the same in both equations. Let's eliminate [tex]\(b\)[/tex].
We will multiply equation (1) by 2 and equation (3) by 3 to make the coefficients of [tex]\(b\)[/tex] equal:
[tex]\[
4a + 6b = 12 \quad \text{(4)}
\][/tex]
[tex]\[
15a + 6b = 12 \quad \text{(5)}
\][/tex]
Next, we subtract equation (4) from equation (5):
[tex]\[
(15a + 6b) - (4a + 6b) = 12 - 12
\][/tex]
[tex]\[
11a = 0
\][/tex]
[tex]\[
a = 0
\][/tex]
Now we know [tex]\(a = 0\)[/tex]. To find [tex]\(b\)[/tex], substitute [tex]\(a = 0\)[/tex] back into either original equation. Let's use equation (1):
[tex]\[
2(0) + 3b = 6
\][/tex]
[tex]\[
3b = 6
\][/tex]
[tex]\[
b = 2
\][/tex]
So the solution to the system of equations is [tex]\(a = 0\)[/tex] and [tex]\(b = 2\)[/tex].
To check the solution, substitute [tex]\(a = 0\)[/tex] and [tex]\(b = 2\)[/tex] back into the original equations:
For equation (1):
[tex]\[
2(0) + 3(2) = 6
\][/tex]
[tex]\[
6 = 6 \quad \text{(True)}
\][/tex]
For equation (2):
[tex]\[
5(0) + 2(2) - 4 = 0
\][/tex]
[tex]\[
4 - 4 = 0
\][/tex]
[tex]\[
0 = 0 \quad \text{(True)}
\][/tex]
Both equations are satisfied with [tex]\(a = 0\)[/tex] and [tex]\(b = 2\)[/tex].
Thus, the answer is:
[tex]\[
\boxed{0,2}
\][/tex]
