Explain or show how you know [tex]\((2,3)\)[/tex] is not a solution to this system of equations:
[tex]\[
\begin{cases}
2x + y = -1 \\
y = 8x - 1
\end{cases}
\][/tex]

Question

14-06-2024
Mathematics

Answered

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Explain or show how you know [tex]\((2,3)\)[/tex] is not a solution to this system of equations:
[tex]\[
\begin{cases}
2x + y = -1 \\
y = 8x - 1
\end{cases}
\][/tex]

Sagot :

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GinnyAnswer GinnyAnswer · Answer 1 · 2024-06-14T15:37:43-04:00

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]

Sagot :

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