Certainly! Let's verify if the given point [tex]\((2, 3)\)[/tex] satisfies the system of equations step-by-step:
### 1. Checking the First Equation [tex]\(2x + y = -1\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 3\)[/tex] into the first equation:
[tex]\[ 2(2) + 3 = -1 \][/tex]
Calculate the left-hand side (LHS):
[tex]\[ 2 \cdot 2 + 3 = 4 + 3 = 7 \][/tex]
The LHS is 7. Compare it to the right-hand side (RHS):
LHS = 7, RHS = -1
Clearly, [tex]\(7 \neq -1\)[/tex]. Hence, the point [tex]\((2, 3)\)[/tex] does not satisfy the first equation.
### 2. Checking the Second Equation [tex]\(y = 8x - 1\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 3\)[/tex] into the second equation:
[tex]\[ 3 = 8(2) - 1 \][/tex]
Calculate the RHS:
[tex]\[ 8 \cdot 2 - 1 = 16 - 1 = 15 \][/tex]
So, the LHS is 3 and the RHS is 15:
LHS = 3, RHS = 15
Clearly, [tex]\(3 \neq 15\)[/tex]. Hence, the point [tex]\((2, 3)\)[/tex] does not satisfy the second equation either.
### Conclusion
Since the point [tex]\((2, 3)\)[/tex] does not satisfy either of the two equations in the system, [tex]\((2, 3)\)[/tex] is not a solution to the system of equations:
[tex]\[
\begin{array}{l}
2x + y = -1 \\
y = 8x - 1
\end{array}
\][/tex]
