Question

practice

1. carson’s wheelbarrow can hold 345 pounds.

if he has 121 pounds of rock in the
wheelbarrow, what number of pounds, p,
can he put in the wheelbarrow without going
over the weight limit?

1. solve ( x - 19 < 81 ). then graph the solution.

1. solve ( 4\frac{1}{2} < x + 2\frac{1}{8} ). then graph the solution.

1. solve ( 52.3 < n - 22.8 ). show your work.

1. solve ( 7\frac{2}{3} > c + 4\frac{1}{3} ). show your work.

1. solve ( -11.44 > k + 6.78 ). show your work.

Explanation:

Problem 1

Step1: Set up the inequality

The total weight (current weight + additional weight) must be less than or equal to the limit. So, \(121 + p \leq 345\).

Step2: Solve for \(p\)

Subtract 121 from both sides: \(p \leq 345 - 121\).
Calculate \(345 - 121 = 224\).

Step1: Solve the inequality

Add 19 to both sides of \(x - 19 < 81\): \(x < 81 + 19\).

Step2: Calculate the result

\(81 + 19 = 100\), so \(x < 100\).
(Graphing: Draw a number line, put an open circle at 100, and shade to the left.)

Step1: Solve for \(x\)

Subtract \(2\frac{1}{8}\) from both sides of \(4\frac{1}{2} < x + 2\frac{1}{8}\).
First, convert to improper fractions: \(4\frac{1}{2}=\frac{9}{2}=\frac{36}{8}\), \(2\frac{1}{8}=\frac{17}{8}\).
Then, \(\frac{36}{8}-\frac{17}{8} < x\).

Step2: Calculate the difference

\(\frac{36 - 17}{8}=\frac{19}{8}=2\frac{3}{8}\), so \(x > 2\frac{3}{8}\).
(Graphing: Draw a number line, put an open circle at \(2\frac{3}{8}\), shade to the right.)

Answer:

\(p \leq 224\)

Problem 2