Part a) Use a system of inequalities to model the scenario
Let
x-------> represent babysitting hours
y-------> represent tutoring hours
we know that
[tex]x+y \leq20[/tex] --------> inequality [tex]1[/tex]
[tex]6x+10y \geq 75[/tex] --------> inequality [tex]2[/tex]
Part 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
we know that
[tex]x+y \leq20[/tex] --------> inequality [tex]1[/tex]
This inequality represent Jeanne’s possible number of hours worked this week
Using a graph tool
the solution of this inequality is the shaded area
see the attached figure N [tex]1[/tex]
[tex]6x+10y \geq 75[/tex] --------> inequality [tex]2[/tex]
This inequality represent Jeanne’s probable income earned this week
Using a graph tool
the solution of this inequality is the shaded area
see the attached figure N [tex]2[/tex]
Part 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
we have the system of inequalities
[tex]x+y \leq20[/tex] --------> inequality [tex]1[/tex]
[tex]6x+10y \geq 75[/tex] --------> inequality [tex]2[/tex]
the solution of the system of inequalities is the shaded area
using a graph tool
see the attached figure N [tex]3[/tex]
Let
[tex]A(10,10)[/tex] ------> see the attached figure N [tex]3[/tex]
The point belongs to the shaded area, so it is a system solution.
If the point is a solution, it must satisfy both inequalities.
Prove algebraically
[tex]A(10,10)[/tex]
substitute the value of x and y in the inequalities
inequality [tex]1[/tex]
[tex]10+10 \leq20[/tex]
[tex]20 \leq20[/tex] ------> is ok
inequality [tex]2[/tex]
[tex]6*10+10*10 \geq 75[/tex]
[tex]160 \geq 75[/tex] -------> is ok
the point A satisfy both inequalities
therefore
the point A is a solution of the system
