Systems of Inequalities Worksheet Answer Key – Algebra 1

Problem 1 Test two points

(3, 9): 9 ≥ 2(3) + 1 = 7 is true. (1, 2): 2 ≥ 2(1) + 1 = 3 is false.

(3, 9) is a solution; (1, 2) is not.

Problem 2 Solid or dashed

(a) Strict inequality: dashed. (b) ≥ includes the line: solid. (c) ≤ includes the line: solid.

(a) dashed (b) solid (c) solid

Problem 3 Which side to shade

0 > −6(0) + 14 = 14 is false, so the origin is not in the solution. Shade the side away from the origin, above the line.

Shade above the dashed line.

Problem 4 Write from a graph

The dashed line passes through (0, 2) with slope 1: y = x + 2. The shading is below the line and the line is dashed.

y < x + 2

Problem 5 Check a point in a system

First: 3 < −2 + 6 = 4 is true. Second: 3 ≥ −(−2) = 2 is true.

Yes, (−2, 3) is a solution.

Problem 6 Graph a solid half-plane

The boundary has y-intercept 3 and slope −1/2, through (0, 3) and (6, 0). It is solid. Test (0, 0): 0 ≤ 3 is true, so shade below.

Solid line, shade below

Problem 7 Graph from standard form

The boundary 3x + 2y = 12 crosses the axes at (4, 0) and (0, 6), and is dashed. Test (0, 0): 0 > 12 is false, so shade the side away from the origin, above the line.

Dashed line, shade above

Problem 8 Vertical and horizontal boundaries

x = −1 is a dashed vertical line with shading to the left. y = 2 is a solid horizontal line with shading above. The overlap is the upper-left region.

Test (−3, 4): −3 < −1 and 4 ≥ 2 are both true.

Upper-left region

Problem 9 Solve for y first

−2y ≤ −4x + 10. Divide by −2 and reverse the sign: y ≥ 2x − 5.

The boundary y = 2x − 5 is solid, with shading above.

y ≥ 2x − 5; solid line, shade above

Problem 10 Write a system from a graph

The dashed line y = x has the shading above it: y > x. The solid horizontal line y = 4 has the shading below it: y ≤ 4.

Test (0, 2): 2 > 0 and 2 ≤ 4 are both true.

y > x and y ≤ 4

Problem 11 Graph a system

The boundary y = x − 2 is solid with shading above; y = −x + 4 is dashed with shading below. They meet at x − 2 = −x + 4, so x = 3 and y = 1; this corner is not a solution because one line is dashed.

Test (0, 0): 0 ≥ −2 and 0 < 4 are true.

Yes, (0, 0) is a solution.

Problem 12 Test several points

(4, 1): 9 ≤ 10 and 3 > 1: yes. (5, 3): 13 ≤ 10 is false. (3, −2): 4 ≤ 10 and 5 > 1: yes. (6, 0): 12 ≤ 10 is false.

(4, 1) and (3, −2)

Problem 13 Triangle region

The first two bound the first quadrant; x + y = 9 crosses the axes at (9, 0) and (0, 9). The solution is a triangle.

Corners (0, 0), (9, 0) and (0, 9)

Problem 14 No overlap

Both boundaries have slope 8 and are parallel. The first needs y above the line y = 8x + 3; the second needs y below the lower line y = 8x − 7. No point is both above the upper line and below the lower one.

Algebra: 8x + 3 ≤ y ≤ 8x − 7 would require 3 ≤ −7, which is false.

No solution

Problem 15 Snack cart budget

2.5p + 2j ≤ 30.

(6, 7): 15 + 14 = 29 ≤ 30, allowed. (8, 5): 20 + 10 = 30 ≤ 30, allowed (on the line). (9, 6): 22.5 + 12 = 34.5, not allowed.

The boundary line passes through (12, 0) and (0, 15).

2.5p + 2j ≤ 30; (6, 7) and (8, 5) work, (9, 6) does not

Problem 16 Bake sale constraints

c + b ≤ 40, c ≥ 12, b ≥ 0.

Corners: c = 12 with b = 0 gives (12, 0); b = 0 with c + b = 40 gives (40, 0); c = 12 with c + b = 40 gives (12, 28).

(15, 20): 35 ≤ 40 and 15 ≥ 12, feasible. (10, 25): 10 < 12, not feasible.

Corners (12, 0), (40, 0), (12, 28); (15, 20) yes, (10, 25) no

Problem 17 Shipping limits

With k kits and s sensors: 3k + 0.5s ≤ 30, k ≥ 2, s ≥ 0.

(5, 20): 15 + 10 = 25 ≤ 30 and 5 ≥ 2, allowed. (9, 10): 27 + 5 = 32, too heavy.

With 2 kits: 6 + 0.5s ≤ 30, so s ≤ 48.

(5, 20) yes; (9, 10) no; 48 sensors

Problem 18 Hiking food plan

250b + 400p ≥ 2,000 (which is 5b + 8p ≥ 40) and 2b + 5p ≤ 40.

(4, 3): 1,000 + 1,200 = 2,200 ≥ 2,000 and 8 + 15 = 23 ≤ 40, feasible. (2, 2): 500 + 800 = 1,300 < 2,000, not feasible. (10, 4): 2,500 + 1,600 ≥ 2,000 and 20 + 20 = 40 ≤ 40, feasible (on the boundary).

(4, 3) and (10, 4) are feasible; (2, 2) is not

Problem 19 Weekly earnings plan

12b + 15t ≥ 240 and b + t ≤ 20 (with b, t ≥ 0).

(10, 8): 120 + 120 = 240 ≥ 240 and 18 ≤ 20, feasible. (15, 3): 180 + 45 = 225 < 240, no. (4, 12): 48 + 180 = 228, no. (5, 15): 60 + 225 = 285 ≥ 240 and 20 ≤ 20, feasible.

(10, 8) and (5, 15)

Problem 20 Shirts and hats

15s + 9h ≥ 360 and s + h ≤ 30.

Meeting point: h = 30 − s gives 15s + 9(30 − s) = 360, so 6s = 90, s = 15 and h = 15.

The feasible region is the triangle with corners (24, 0), (30, 0) and (15, 15).

For (20, 8): 300 + 72 = 372 ≥ 360 and 28 ≤ 30, so it works.

Boundaries meet at (15, 15); 20 shirts and 8 hats work