To solve the given system of linear equations using the elimination method, follow these steps:
The given system of linear equations is:
[tex]\[
\begin{array}{l}
-2a + 5b = 124 \quad \text{(Equation 1)} \\
3a + 5b = 139 \quad \text{(Equation 2)}
\end{array}
\][/tex]
1. Subtract Equation 2 from Equation 1 to eliminate [tex]\( b \)[/tex]:
[tex]\[
(-2a + 5b) - (3a + 5b) = 124 - 139
\][/tex]
Simplify this equation:
[tex]\[
-2a + 5b - 3a - 5b = -15
\][/tex]
[tex]\[
-2a - 3a + 5b - 5b = -15
\][/tex]
[tex]\[
-5a = -15
\][/tex]
2. Solve for [tex]\( a \)[/tex]:
[tex]\[
a = \frac{-15}{-5}
\][/tex]
[tex]\[
a = 3
\][/tex]
3. Substitute the value of [tex]\( a \)[/tex] into one of the original equations to find [tex]\( b \)[/tex]. We can use either Equation 1 or Equation 2. Let's use Equation 2:
[tex]\[
3a + 5b = 139
\][/tex]
Substitute [tex]\( a = 3 \)[/tex]:
[tex]\[
3(3) + 5b = 139
\][/tex]
[tex]\[
9 + 5b = 139
\][/tex]
4. Solve for [tex]\( b \)[/tex]:
[tex]\[
5b = 139 - 9
\][/tex]
[tex]\[
5b = 130
\][/tex]
[tex]\[
b = \frac{130}{5}
\][/tex]
[tex]\[
b = 26
\][/tex]
5. Verify the solution by substituting [tex]\( a = 3 \)[/tex] and [tex]\( b = 26 \)[/tex] back into the original equations to ensure they both hold true:
Check Equation 1:
[tex]\[
-2(3) + 5(26) = 124
\][/tex]
[tex]\[
-6 + 130 = 124
\][/tex]
[tex]\[
124 = 124
\][/tex]
This is true.
Check Equation 2:
[tex]\[
3(3) + 5(26) = 139
\][/tex]
[tex]\[
9 + 130 = 139
\][/tex]
[tex]\[
139 = 139
\][/tex]
This is also true.
Therefore, the solution to the system of equations is:
[tex]\[
a = 3 \quad \text{and} \quad b = 26
\][/tex]
