Home > AC > Chapter 10 > Lesson 10.2.5 > Problem10-95

10-95.

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
\begin{aligned} 6.6 \left|x\right| &> 18 \\ 6 \left|x\right| &= 18 \\ \left|x\right| &=3 \end{aligned}

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. $\left|3x-2\right|\le2$

See the help for part (b).

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

Use a Fraction Buster to eliminate the fractions.

\begin{aligned} \left(\frac{4}{5}\right)30 - \left(\frac{2x}{3} \right)30 &= \left(\frac{3}{10}\right)30 \\ 25 - 20x &= 9 \end{aligned}

1. $(4x−2)^2\le100$

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

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

Change the inequality into an equation. Then square root both sides.
Make two equations, one with the positive root and one with the negative.
$\left(4x-2\right)^2\le100$
$\left(4x-2\right)^2=100$

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

1. $\left(x-1\right)^3=8$

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