Question

1. andrew needs at least 60 pencils (x) and markers (y) for his class to play the review game he created. his supply closet has boxes of pencils with 12 in each and boxes of markers with 6 in each.

part a: write the inequality for the situation above.
part b: graph the solution set to the two - variable inequality.

1. graph the inequalities below to show their solutions.

a. $7x + 6y > 30$
b. $y - 5 \leq \frac{1}{2}(x + 3)$

Explanation:

Problem 31 - Part A

Step1: Define total pencils and markers

Total pencils: \( 12x \), total markers: \( 6y \). Total needed: at least 60.

Step2: Write the inequality

Combine: \( 12x + 6y \geq 60 \). Simplify (divide by 6): \( 2x + y \geq 10 \).

Step1: Rewrite inequality as line

From \( 12x + 6y = 60 \), simplify to \( y = -2x + 10 \).

Step2: Determine line type and shading

Inequality is \( \geq \), so solid line. Test (0,0): \( 0 + 0 \geq 60 \)? No. Shade above the line.

Step3: Graph the line

Plot intercepts: when \( x = 0 \), \( y = 10 \); when \( y = 0 \), \( x = 5 \). Draw solid line through (0,10) and (5,0), shade above.

Step1: Rewrite as line equation

\( 7x + 6y = 30 \) → \( y = -\frac{7}{6}x + 5 \).

Step2: Determine line type and shading

Inequality is \( > \), so dashed line. Test (0,0): \( 0 + 0 > 30 \)? No. Shade above the line.

Step3: Graph the line

Find intercepts: \( x = 0 \) → \( y = 5 \); \( y = 0 \) → \( x = \frac{30}{7} \approx 4.29 \). Draw dashed line, shade above.

Answer:

\( 12x + 6y \geq 60 \) (or simplified \( 2x + y \geq 10 \))

Problem 31 - Part B