To determine if the ordered pair $(6, \frac{7}{6})$ is a solution to the system of equations, we substitute $x = 6$ and $y = \frac{7}{6}$ into both equations and check if they satisfy them.

The system of equations is:

1. $-6x + 4y = -8$
2. $4x + 2y = 24$

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Step 1: Substituting into the first equation

$-6(6) + 4\left(\frac{7}{6}\right) = -8$ $-36 + \frac{28}{6} = -8$ Simplify $\frac{28}{6}$ to $\frac{14}{3}$: $-36 + \frac{14}{3} = -8$ Convert $-36$ to have a denominator of 3: $\frac{-108}{3} + \frac{14}{3} = \frac{-94}{3}$ $\frac{-94}{3} \neq -8$ Thus, $(6, \frac{7}{6})$ does not satisfy the first equation.

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Step 2: Substituting into the second equation

$4(6) + 2\left(\frac{7}{6}\right) = 24$ $24 + \frac{14}{6} = 24$ Simplify $\frac{14}{6}$ to $\frac{7}{3}$: $24 + \frac{7}{3} = 24$ Convert 24 to have a denominator of 3: $\frac{72}{3} + \frac{7}{3} = \frac{79}{3}$ $\frac{79}{3} \neq 24$ Thus, $(6, \frac{7}{6})$ does not satisfy the second equation.

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Final Answer:

The ordered pair $(6, \frac{7}{6})$ is not a solution to the system of equations.

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Would you like a detailed explanation of any part? Here are related questions you might find useful:

1. How do you verify ordered pairs as solutions for equations with fractions?
2. What are the steps for solving systems of equations by substitution or elimination?
3. How can you graphically verify solutions for systems of linear equations?
4. What happens if there are no solutions to a system of equations?
5. Can a system of equations have infinitely many solutions?

Tip: Always simplify fractions and convert all terms to the same denominator when verifying equations.