Answer:
Seth:
Equation 7.50x + 15y = 750, where x is the number of hours Seth works and y is the number of lawns he mows.
Karen:
Equation 6x + 4y = 700, where x is the number of hours Karen works and y is the number of dogs she walks.
Part A: Converting it in function for by solving for y first.
Solving 7.50x + 15y = 750, equation we get
7.50x + 15y = 750
15y = -7.50x + 750
y = -1/2x + 50
Solving 6x + 4y = 700 for y, we get
6x + 4y = 700
4y = -6x + 700
y= -3/2x + 175.
On comaring with slope-intercept form y=mx+b, we got slope for Seth equation is -1/2 and slope for Karen is -3/2.
If we take absolute of those -1/2 and -3/2 we get 1/2 and 3/2.
Therefore, Karen has the higher slope. We can see the graph that blue line has greater slope as it's increasing/decreasing with greater rate.
In function form we could write equation as
f(x) = -1/2x + 50 and
f(x) = -3/2x + 175.
Part B : y-intercept of Karen equation is 175 and y-intercept of Seth equation is 50.
Therefore, Karen has greater y-intercept.
This means Karen earns $175 by just walking dogs.
Part C: If both students work 80 hours for the month.
Let us plug x =80 in each of the equations.
y = -1/2x + 50 => y = -1/2(80) + 50 = -40 +50 = $10.
Seth would earn $10 from mowing lawn.
Therefore, the number of lawns Seth has to mow = 10/15 = 0.67 that is approximately 1 lawn.
y = -3/2x + 175 => y = -3/2(80) + 175 = -120 +175 = $55.
Karen would earn $50 from walking dog.
So, the number of dogs Karne has to walk = 55/4 = 13.75 that ia approximately 14 dogs.
Therefore, the number of lawns Seth has to mow is less than the number of dogs Karen has to walk.
Part D: If Karen begins charging $8 per hour for babysitting and Seth begins earning $8 per hour.
The equations would become : 8x + 15y = 750, 8x + 4y = 700.
It would not effect the graph much. Even the y-intercepts would remain same.
There would be a slightly change in slopes.
Step-by-step explanation:
