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 $x = 8$.
$$
4|x-5| + 3 = 15,
$$
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 $x = 8$.
