Let's solve each equation step-by-step.
### Equation 1: [tex]\( x - 1.79 = -5 \)[/tex]
1. To isolate [tex]\( x \)[/tex], we need to add 1.79 to both sides of the equation:
[tex]\[
x - 1.79 + 1.79 = -5 + 1.79
\][/tex]
2. Simplifying the right side of the equation:
[tex]\[
x = -5 + 1.79
\][/tex]
3. Adding the numbers on the right side, we get:
[tex]\[
x = -3.21
\][/tex]
So, the solution to the first equation is:
[tex]\[
x = -3.21
\][/tex]
### Equation 2: [tex]\( (x + 6)^2 = 31 \)[/tex]
1. To solve this equation, we first take the square root of both sides:
[tex]\[
\sqrt{(x + 6)^2} = \sqrt{31}
\][/tex]
2. Since the square root of a square function [tex]\( (x+6) \)[/tex] results in both a positive and negative solution for [tex]\( x+6 \)[/tex], we have:
[tex]\[
x + 6 = \sqrt{31} \quad \text{or} \quad x + 6 = -\sqrt{31}
\][/tex]
3. Solving for [tex]\( x \)[/tex] in the first case:
[tex]\[
x + 6 = \sqrt{31}
\][/tex]
Subtract 6 from both sides:
[tex]\[
x = \sqrt{31} - 6
\][/tex]
4. Solving for [tex]\( x \)[/tex] in the second case:
[tex]\[
x + 6 = -\sqrt{31}
\][/tex]
Subtract 6 from both sides:
[tex]\[
x = -\sqrt{31} - 6
\][/tex]
Approximations using the value of [tex]\( \sqrt{31} \approx 5.57 \)[/tex]:
[tex]\[
x \approx 5.57 - 6 = -0.4322356371699785
\][/tex]
[tex]\[
x \approx -5.57 - 6 = -11.567764362830022
\][/tex]
Thus, the solutions to the second equation are approximately:
[tex]\[
x \approx -0.43 \quad \text{or} \quad x \approx -11.57
\][/tex]
So, in summary, the solutions to the given equations are:
1. [tex]\( x = -3.21 \)[/tex]
2. [tex]\( x \approx -0.432 \)[/tex] or [tex]\( x \approx -11.568 \)[/tex]
