Jeanne babysits for $6 per hour. She also works as a reading tutor for $10 per hour. She is only allowed to work 20 hours per we

Question

7 days ago

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Jeanne babysits for $6 per hour. She also works as a reading tutor for $10 per hour. She is only allowed to work 20 hours per we

ek. This week, her goal is to make at least $75.
A. Use a system of inequalities to model the scenario above. Let x represent babysitting hours and y represent tutoring hours.
B. Use the model created in part A to create a graph representing Jeanne’s probable income earned and possible number of hours worked this week.
C. Analyze the set of coordinate values that represent solutions for the model created in part A. Choose one of the coordinates within the solution and algebraically prove that the coordinate represents a true solution for the model.

Mathematics

2 answers:

zzz [9K] · Answer 1 · 2026-03-20T22:46:59+03:00

For the response to the previous question, I would initiate with a simple expression 6x + 10y is greater than or equal to 75
A.
If we denote x as babysitting hours and y as tutoring hours:
x + y ≤ 20
6x + 10y ≥ 75

B. The relation: x + y ≤ 20 can be represented graphically by plotting the line x + y = 20 and shading the area beneath it.
The relation 6x + 10y ≥ 75 can also be graphed by drawing the line 6x + 10y = 75 and shading the area above this line.

C. The zone where the two shaded areas from the inequalities overlap indicates the feasible hours for tutoring and babysitting. Mathematically:
x ≤ 20 - y
6(20 - y) + 10y ≥ 75
y ≤ 11.25
x ≤ 8.75

zzz [9K] · Answer 2 · 2026-03-20T22:47:49+03:00

Solution:

x + y ≤ 20

6x + 10y ≥ 75

Step-by-step clarification:

Let the hours spent babysitting be denoted as x, while y represents the hours of tutoring

Based on the provided requirements

The total working hours must not exceed 20

Thus

x+y≤20

Additionally, her goal is to earn no less than $75

Thus, the second inequality is

6x+10y≥75

Therefore, the system of inequalities representing the conditions above is

x+y≤20

6x+10y≥75

To graph these, we first plot the lines x+y=20 and 6x+10y=75, shading the regions that satisfy the respective inequalities by testing the coordinate (0,0).

Refer to the attached graph for clarification.

The shaded area illustrates the set of coordinates likely representing solutions to the previous inequalities.

Selecting a coordinate (10,5) from the shaded area to test the solution.

Substituting (10,5) into both inequalities will confirm their validity.

10+5≤20

15≤20 True

6(10)+10(5) ≥75

60+50≥75

110≥75 true

Both inequalities hold true for (10,5)