Question

al from lesson 2.2 and lesson 2.3 absolute - value equations or inequalities in one variable and to solve problems. (lesson 2.2) solve each absolute - value equation algebraically. graph the solutions on a number line. (1 point each) 16. 5+|2x| = 14 how i did (circle one) i got it im still learning it. 17. 6|x - 1| = 30 (lesson 2.3) solve each inequality and graph the solutions. (1 point each) 18. 3|x - 6|≤9 19. the graph shows the solution to which equation? a. |x + 1|≤2 b. |x|+1≥2 c. |x + 1|≥2 d. |x|+1≤2 checkpoint score (\frac{square}{15})

Explanation:

Step1: Solve equation 16

Isolate the absolute - value term.
\[

$$\begin{align*} 5+\vert2x\vert&=14\\ \vert2x\vert&=14 - 5\\ \vert2x\vert&=9 \end{align*}$$

\]
Then, \(2x = 9\) or \(2x=-9\), so \(x=\frac{9}{2}\) or \(x =-\frac{9}{2}\).

Step2: Solve equation 17

Isolate the absolute - value term.
\[

$$\begin{align*} 6\vert x - 1\vert&=30\\ \vert x - 1\vert&=5 \end{align*}$$

\]
Then, \(x - 1=5\) or \(x - 1=-5\). Solving gives \(x = 6\) or \(x=-4\).

Step3: Solve inequality 18

Isolate the absolute - value term.
\[

$$\begin{align*} 3\vert x - 6\vert&\leq9\\ \vert x - 6\vert&\leq3 \end{align*}$$

\]
This implies \(-3\leq x - 6\leq3\). Adding 6 to all parts gives \(3\leq x\leq9\).

Step4: Solve question 19

For \(\vert x + 1\vert\leq2\), we have \(-2\leq x + 1\leq2\), which gives \(-3\leq x\leq1\).
For \(\vert x\vert+1\geq2\), we have \(\vert x\vert\geq1\), which gives \(x\geq1\) or \(x\leq - 1\).
For \(\vert x + 1\vert\geq2\), we have \(x + 1\geq2\) or \(x + 1\leq - 2\), which gives \(x\geq1\) or \(x\leq - 3\).
For \(\vert x\vert+1\leq2\), we have \(\vert x\vert\leq1\), which gives \(-1\leq x\leq1\). The graph corresponds to \(\vert x + 1\vert\leq2\), so the answer is a.

Answer:

1. \(x=\frac{9}{2},x =-\frac{9}{2}\)
2. \(x = 6,x=-4\)
3. \(3\leq x\leq9\)
4. a. \(\vert x + 1\vert\leq2\)