  ### Home > CCA > Chapter 11 > Lesson 11.2.1 > Problem11-30

11-30.

Solve the equations and inequalities below, if possible.

1.  $\sqrt{x-1}+13=13$

Isolate the square root by subtracting $13$ from both sides.

$\sqrt{x-1}+13-13\ =\ 13-13$

$\sqrt{x-1}\ =\ 0$

Look inside.

If $\sqrt{x-1}\ =\ 0$, then $x-1=0$.

$x=1$

1.  $6\left|x\right|>18$

Change the inequality into an equation.
Then isolate the absolute value by dividing both sides by
$6.6|x|>18\\ \ \ \ 6|x|=18\\ \ \ \ \ \ |x|=3$

Use two equations to find all solutions for $x$.
$x=3$
$x=-3$

Graph both values of $x$ on a number line, then test values from between the points, to the left of them, and to the right of them in the original equation to find the solution region(s).

$x>3$ or $x<-3$

1.  $|3x-2|\le2$

See the help for part (b).

1.  $\frac{4}{5}-\frac{2x}{3}=\frac{3}{10}$

Use a Fraction Buster to eliminate the fractions.

$\left (\frac{4}{5} \right )30-\left (\frac{2x}{3} \right )30\ =\ \left (\frac{3}{10} \right )30\\ \qquad \quad \ \ \ 25 - 20x = 9$

1.  $(4x−2)^2\le100$

$\sqrt{(4x-2)^2}\ =\ \sqrt{100}$

$4x-2=10$
$4x-2=-10$

Change the inequality into an equation. Then square root both sides.
Make two equations, one with the positive root and one with the negative.

$(4x-2)^2\le100$
$(4x-2)^2=100$

Solve both equations for $x$. Then use the method from part (b), Step 4, to find the solution.

1.  $(x-1)^3=8$

The cube root of $8$ is $2$.