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The equation provided is \( h(x) = 3x + 4 \). You are asked to solve for \( x \) when \( h(x) = 1 \).

To solve for \( x \), follow these steps:

1. Substitute \( h(x) = 1 \) into the equation:
   \[
   1 = 3x + 4
   \]

2. Subtract 4 from both sides to isolate the term with \( x \):
   \[
   1 - 4 = 3x
   \]
   \[
   -3 = 3x
   \]

3. Divide both sides by 3 to solve for \( x \):
   \[
   x = \frac{-3}{3} = -1
   \]

Thus, the solution is \( x = -1 \).

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:
1. How would the solution change if \( h(x) = 2x + 5 \) instead?
2. What if \( h(x) \) was a quadratic function, such as \( h(x) = x^2 + 3x + 2 \)?
3. How would you approach solving \( h(x) = 3x + 4 \) if \( h(x) = 0 \)?
4. Can you find \( x \) when \( h(x) = 3x + 4 \) equals another constant, like \( h(x) = 7 \)?
5. How would the solution process change if the coefficient of \( x \) was a fraction, such as \( h(x) = \frac{1}{2}x + 4 \)?

**Tip**: Always double-check your algebraic manipulations, especially when isolating variables.

Given h(x) = 3x + 4, solve for x when h(x) = 1.

Learn how to solve the linear equation h(x) = 3x + 4 when h(x) equals 1. This problem involves key algebraic concepts and steps to isolate the variable x. Suitable for students in Grades 7-9.

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