SOLUTION: I need help with an algebra equation. Here is the set up. A father is looking for the best option for daycare for his 5 year old son. A home-based option (Option A) charges a fl

Question 855281: I need help with an algebra equation. Here is the set up.
A father is looking for the best option for daycare for his 5 year old son. A home-based option (Option A) charges a flat rate per hour of $15. A center-based option (option B) charges a fixed fee per week of $150 for 40 hours and $10 an hour for any hour over the 40. What is the best option for the father? The father works 50 hours a week from his office which is located in his home.
x = number of hours
y = total child care cost
Option A: (A)y = 15x
Option B1: (B1)y= 150 + 10 (x-40)
My paper got sent back because it said I need an additional equation for Option B that represents less than 40 hours of daycare is needed to complete the assessment.
This is what I came up with.
Option B2: (B2)y=150+10x, x≥40 hours � day care with less than 40
Is this right?????? But I am not sure how to solve this.
If that is right I need to show when option A and option B2 will be equal. How do I do that? I also need to be able to graph it so I will need an ordered pair.

Update: I already submitted my paper fixing option A instead of B showing that equation with less than 50 hours a week (less than 40 y=15*30=450) and it got send back for the same reason I will need an additional equation for Option B that represents less than 40 hours of daycare needed to complete the assessment.
Thank you for any help anyone can give and PLEASE don't answer if you really aren't going to help me. I am stuck and need help.

Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
No, Option B for 40 hours or less is a fixed price of $150.
Option B : 1. for
for
for
2. for
Option A is just
.
.
.

.
.
.
Option A : Red Line
Option B : Green Line
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