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Given the function [tex]f(x) = 4|x - 5| + 3[/tex], for what values of [tex]x[/tex] is [tex]f(x) = 15[/tex]?

A. [tex]x = 2, x = 8[/tex]
B. [tex]x = 1.5, x = 8[/tex]
C. [tex]x = 2, x = 7.5[/tex]
D. [tex]x = 0.5, x = 7.5[/tex]

Answer :

To solve for $x$ in

$$
4|x-5| + 3 = 15,
$$

follow these steps:

1. Subtract 3 from both sides to isolate the absolute value term:

$$
4|x-5| = 15 - 3 \quad \Rightarrow \quad 4|x-5| = 12.
$$

2. Divide both sides by 4:

$$
|x-5| = \frac{12}{4} \quad \Rightarrow \quad |x-5| = 3.
$$

3. Recall that if $|A| = B$, then $A = B$ or $A = -B$. Therefore, set up the two equations:

$$
x - 5 = 3 \quad \text{or} \quad x - 5 = -3.
$$

4. Solve each equation:

- For $x - 5 = 3$:

$$
x = 5 + 3 \quad \Rightarrow \quad x = 8.
$$

- For $x - 5 = -3$:

$$
x = 5 - 3 \quad \Rightarrow \quad x = 2.
$$

Thus, the values of $x$ that satisfy the equation are $x = 2$ and $x = 8$.

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Rewritten by : Jeany