Question

algebra 1
chapter 2 progress monitoring tool

1. solve and graph the solution of $-\frac{2}{3}x + 5 < -\frac{1}{3}x + 3$

number line from -2 to 14

1. solve $24 \leq 3x + 27 \leq 30$. graph the solution. choose the correct graph of the solution.

a. number line with dots at -1 and 1
b. number line with open circles at -1 and 1
c. number line with open circles at -1 and 1, line between
d. number line with dots at -1 and 1, line between
solve $x - 7 \leq -6$ or $3x + 3 \geq 15$. graph the solution. choose the correct graph of the solution.
a. number line with dots at 1 and 4
b. number line with dots at 6 and 13
c. number line with dots at 1 and 4, arrows
d. number line with dots at 6 and 13, arrows

Explanation:

Sub - question 1: Solve \(-\frac{2}{3}x + 5<-\frac{1}{3}x + 3\)

Step 1: Add \(\frac{2}{3}x\) to both sides

To eliminate the \(x\) term with the coefficient \(-\frac{2}{3}\) on the left, we add \(\frac{2}{3}x\) to both sides of the inequality.
\(-\frac{2}{3}x+5+\frac{2}{3}x<-\frac{1}{3}x + 3+\frac{2}{3}x\)
Simplifying both sides, we get \(5<\frac{1}{3}x + 3\)

Step 2: Subtract 3 from both sides

To isolate the term with \(x\), we subtract 3 from both sides.
\(5 - 3<\frac{1}{3}x+3 - 3\)
Simplifying, we have \(2<\frac{1}{3}x\)

Step 3: Multiply both sides by 3

To solve for \(x\), we multiply both sides of the inequality by 3 (since 3 is positive, the direction of the inequality sign remains the same).
\(2\times3<\frac{1}{3}x\times3\)
Simplifying gives \(x > 6\)

Step 1: Subtract 27 from all parts

We subtract 27 from the left - hand side, the middle, and the right - hand side of the compound inequality to isolate the term with \(x\).
\(24-27\leq3x + 27-27\leq30 - 27\)
Simplifying each part, we get \(- 3\leq3x\leq3\)

Step 2: Divide all parts by 3

We divide each part of the compound inequality by 3 (since 3 is positive, the direction of the inequality signs remains the same).
\(\frac{-3}{3}\leq\frac{3x}{3}\leq\frac{3}{3}\)
Simplifying gives \(-1\leq x\leq1\)

For the graph, we draw closed circles at \(x=-1\) and \(x = 1\) and shade the region between \(-1\) and \(1\) on the number line. Among the given options, option D has closed circles at \(-1\) and \(1\) and shades the region between them.

Solve \(x - 7\leq-6\)

Step 1: Add 7 to both sides

To solve for \(x\) in the inequality \(x - 7\leq-6\), we add 7 to both sides.
\(x-7 + 7\leq-6 + 7\)
Simplifying gives \(x\leq1\)

Solve \(3x + 3\geq15\)

Step 1: Subtract 3 from both sides

To isolate the term with \(x\), we subtract 3 from both sides of the inequality \(3x + 3\geq15\).
\(3x+3 - 3\geq15 - 3\)
Simplifying gives \(3x\geq12\)

Step 2: Divide both sides by 3

To solve for \(x\), we divide both sides of the inequality \(3x\geq12\) by 3 (since 3 is positive, the direction of the inequality sign remains the same).
\(\frac{3x}{3}\geq\frac{12}{3}\)
Simplifying gives \(x\geq4\)

The solution to the compound inequality \(x - 7\leq-6\) or \(3x + 3\geq15\) is \(x\leq1\) or \(x\geq4\). Among the given options, option C has closed circles at \(x = 1\) and \(x = 4\) and shades the regions to the left of \(1\) and to the right of \(4\).

Answer:

\(x>6\) (For graphing, we draw an open circle at \(x = 6\) and shade the region to the right of \(6\) on the number line)

Sub - question 2: Solve \(24\leq3x + 27\leq30\)