Two-Variable Linear Inequalities

Question 1

If Ezra works less than or equal to 50 hours doing lawn care, he must spend at least hours walking dogs.

Revolt · Accepted Answer

35

Question 2

To see how the solution set changes, suppose the coordinate pair (a, b) satisfies the original inequality. If a stays the same, b must be its previous value to satisfy the new inequality.

Revolt · Accepted Answer

greater than

Question 3

Now suppose b stays the same. Then, a must be its previous value to satisfy the new inequality.

Revolt · Accepted Answer

greater than

Question 4

Ezra works two summer jobs to save for a laptop that costs at least $1100. He charges $15/hr to mow lawns and $10/hr to walk dogs. Recall the inequality that represents this situation: .Is (60, 20) a solution? How do you know? What does this point represent?

Revolt · Accepted Answer

Yes, (60, 20) is a solution. When you substitute the values into the inequality, 15(60) + 10(20) ≥ 1100 becomes 900 + 200 ≥ 1100, or 1100 ≥ 1100. Since this is a true statement, the point is a solution. This point represents Ezra working 60 hours mowing lawns and 20 hours walking dogs, which earns him exactly the $1,100 he needs for the laptop.

Question 5

Ezra works two summer jobs to save for a laptop that costs at least $1100. He charges $15/hr to mow lawns and $10/hr to walk dogs. Recall the inequality that represents this situation: Explain how to graph the solution set.

Revolt · Accepted Answer

Graph the solid boundary line 15x + 10y = 1100 through (1100/15, 0) and (0, 110). Shade the region on or above the line in the first quadrant, where 15x + 10y ≥ 1100.

Question 6

Ezra works two summer jobs to save for a laptop that costs at least $1100. He charges $15/hr to mow lawns and $10/hr to walk dogs. He currently charges in half-hour intervals. Recall the inequality that represents this situation: .Ezra decides to charge per dog instead of per hour. How does the solution set change? What are the limitations on x and y? Explain.

Revolt · Accepted Answer

The solution set changes because the allowed values for y are now whole numbers, as Ezra cannot walk a fraction of a dog. Previously, y represented hours and was restricted to non-negative multiples of 0.5. The limitations on x remain the same: x must be a non-negative multiple of 0.5 because Ezra still charges by the half-hour for mowing lawns. Both variables must be non-negative because he cannot work negative hours or walk a negative number of dogs.

Question 7

15x + 10y .

Revolt · Accepted Answer

60

Question 8

Ezra works two summer jobs to save for a laptop that costs at least $1100. He decides to adjust his rates to $250 per lawn (x) for the summer and $300 per dog (y) for the summer, and to limit himself to 2 dogs.Describe how to find a solution graphically.

Revolt · Accepted Answer

First, write the inequality 250x + 300y ≥ 1100 and graph the boundary line in the first quadrant, since the number of lawns and dogs must be non-negative. Because the laptop costs at least $1100, shade the half-plane above the line (the side that does not include the origin). Then, draw a horizontal line at y = 2 to represent the limit of 2 dogs. The solutions are the points with integer coordinates that fall within the intersection of the shaded region and the area on or below the line y = 2.

Question 9

Question 12

Revolt · Accepted Answer

The coordinate pair (60, 20) is a solution because it satisfies the inequality .; The coordinate pair (60, 20) is a solution because Ezra earns 60 x $15 = $900 mowing lawns and 20 x $10 = $200 walking dogs, for a total of $1100.

Question 10

Question 13

Revolt · Accepted Answer

Graph only the first quadrant since x and y must be nonnegative.; Graph the boundary line using the related equation, y = –1.5x + 110.

Two-Variable Linear Inequalities Answers

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