To determine the value of [tex]\(a\)[/tex] for which the system of equations has infinitely many solutions, we need to establish conditions for which the equations are linearly dependent. This means one equation is a scalar multiple of the other. Given the system:
[tex]\[
\begin{aligned}
5s - 2t - 1 &= -a \quad \text{(Equation 1)} \\
-8s + bt - 2 &= 2 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
First, let's rewrite the equations to align the constants on the right side:
[tex]\[
\begin{aligned}
5s - 2t &= -a + 1 \quad \text{(Equation 1)} \\
-8s + bt &= 4 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
For the system to have infinitely many solutions, the coefficients of [tex]\(s\)[/tex] and [tex]\(t\)[/tex] in the first equation must be proportional to the coefficients of [tex]\(s\)[/tex] and [tex]\(t\)[/tex] in the second equation, and the constants on the right side should have the same proportionality.
Let's find the ratios of the coefficients:
1. The ratio of the coefficients of [tex]\(s\)[/tex]:
[tex]\[
\frac{5}{-8}
\][/tex]
2. The ratio of the coefficients of [tex]\(t\)[/tex]:
[tex]\[
\frac{-2}{b}
\][/tex]
3. The ratio of the constants:
[tex]\[
\frac{(-a + 1)}{4}
\][/tex]
For the equations to be proportional, these ratios must be equal. We set up the equations:
[tex]\[
\frac{5}{-8} = \frac{-2}{b} \quad \text{and} \quad \frac{5}{-8} = \frac{(-a + 1)}{4}
\][/tex]
First, solve [tex]\( \frac{5}{-8} = \frac{-2}{b} \)[/tex]:
[tex]\[
5b = 16
\][/tex]
[tex]\[
b = \frac{16}{5}
\][/tex]
Now solve [tex]\( \frac{5}{-8} = \frac{(-a + 1)}{4} \)[/tex]:
Cross-multiply to solve for [tex]\(a\)[/tex]:
[tex]\[
5 \cdot 4 = -8 \cdot (-a + 1)
\][/tex]
[tex]\[
20 = 8a - 8
\][/tex]
[tex]\[
20 + 8 = 8a
\][/tex]
[tex]\[
28 = 8a
\][/tex]
[tex]\[
a = \frac{28}{8} = \frac{7}{2}
\][/tex]
Thus, the value of [tex]\(a\)[/tex] that ensures the system of equations has infinitely many solutions is:
[tex]\[
\boxed{\frac{7}{2}}
\][/tex]
