Which statement about the ordered pairs (3, −8) and (4, 4) is true for the equation 3x−y4=11?

Responses

(4,  4) is a solution to the equation but not (3, −8).
Begin ordered pair 4 comma 4 end orderd pair is a solution to the equation but not Begin ordered pair 3 comma negative 8 end ordered pair .

(3, −8) is a solution to the equation but not (4, 4).
Begin ordered pair 3 comma negative 8 end ordered pair is a solution to the equation but not Begin ordered pair 4 comma 4 end ordered pair .

Both ordered pairs are solutions.
Both ordered pairs are solutions.

Neither ordered pair is a solution.

We need to check whether each ordered pair satisfies the given equation:

\[
3x - \frac{y}{4} = 11
\]

### For the first ordered pair \((3, -8)\):

Substitute \(x = 3\) and \(y = -8\) into the equation:

\[
3(3) - \frac{-8}{4} = 11
\]
\[
9 + 2 = 11
\]
\[
11 = 11
\]

Since the equation holds true, \((3, -8)\) is a solution.

### For the second ordered pair \((4, 4)\):

Substitute \(x = 4\) and \(y = 4\) into the equation:

\[
3(4) - \frac{4}{4} = 11
\]
\[
12 - 1 = 11
\]
\[
11 = 11
\]

Since this equation also holds true, \((4, 4)\) is a solution.

### Conclusion:

Both ordered pairs are solutions. Thus, the correct response is:

**Both ordered pairs are solutions.**

Would you like more details or have any questions? Here are some related questions:

1. How do you solve linear equations with fractions?
2. What is the general method for verifying if an ordered pair is a solution to an equation?
3. How can you rewrite equations to eliminate fractions?
4. What is the difference between an ordered pair and a solution set?
5. How do systems of equations relate to solving for multiple ordered pairs?

**Tip:** Always simplify equations where possible to make the calculation easier, especially when fractions are involved.

Which statement about the ordered pairs (3, −8) and (4, 4) is true for the equation 3x−y/4=11?

Explore how to verify if the ordered pairs (3, -8) and (4, 4) are solutions to the linear equation 3x − y/4 = 11. This problem walks through substitution and solving, and is ideal for students in Grades 8-10.

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