To solve the system of linear equations:
[tex]\[
\begin{aligned}
A - 2B &= -6 \quad \text{(Equation 1)} \\
2A &= 5 + B \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
we will find the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] step-by-step.
### Step 1: Solve for one variable in terms of the other
From Equation 2:
[tex]\[
2A = 5 + B
\][/tex]
Solve for [tex]\(B\)[/tex] in terms of [tex]\(A\)[/tex]:
[tex]\[
B = 2A - 5 \quad \text{(Rearrange the equation to isolate \(B\))}
\][/tex]
### Step 2: Substitute this expression into Equation 1
Substitute [tex]\(B = 2A - 5\)[/tex] into Equation 1:
[tex]\[
A - 2(2A - 5) = -6
\][/tex]
Expand and simplify:
[tex]\[
A - 4A + 10 = -6
\][/tex]
Combine like terms:
[tex]\[
-3A + 10 = -6
\][/tex]
Subtract 10 from both sides:
[tex]\[
-3A = -16
\][/tex]
Divide by [tex]\(-3\)[/tex]:
[tex]\[
A = \frac{16}{3}
\][/tex]
### Step 3: Substitute [tex]\(A\)[/tex] back into the expression for [tex]\(B\)[/tex]
Substitute [tex]\(A = \frac{16}{3}\)[/tex] back into [tex]\(B = 2A - 5\)[/tex]:
[tex]\[
B = 2 \left( \frac{16}{3} \right) - 5
\][/tex]
Convert 5 to a fraction with a common denominator:
[tex]\[
B = \frac{32}{3} - \frac{15}{3}
\][/tex]
Combine the fractions:
[tex]\[
B = \frac{32 - 15}{3} = \frac{17}{3}
\][/tex]
### Solution
The solution to the system of equations is:
[tex]\[
A = \frac{16}{3}, \quad B = \frac{17}{3}
\][/tex]
Thus, [tex]\( A \)[/tex] and [tex]\( B \)[/tex] satisfy both equations with the values:
[tex]\[
\left( \frac{16}{3}, \frac{17}{3} \right)
\][/tex]
