In order to at least earn $90, Katie can work exercising horses for 6 hours and cleaning stalls for 6 hours.
Step-by-step explanation:
Let hours of exercising horses be x --> $5 per hour = 5x
Let hours of cleaning stalls be y --> $10 per hour = 10y
Total earning = 
Total hours = 12
<u>Equation 1:</u>
5x + 10y = 
<u>Equation 2:</u>
x + y 
<em>1. Multiply equation 2 by -5</em>
x + y
(*-5)
5x + 10y = 
-5x - 5y
5x + 10y = 
<em>2. Solve</em>
-5x + 5x -5y + 10y = -60 + 90
5y = 30
y = 6
x + y = 12
x + 6 = 12
x = 6
Therefore, in order to at least earn $90, Katie can work exercising horses for 6 hours and cleaning stalls for 6 hours.
Keywords: Simultaneous, hours, equations
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Adding the equations to eliminate p would be the correct step because when you add the equations together, the variable p would cancel out and be eliminated leaving you with 2w=35 you can then solve for the value of w which you can then input to find the value of p, allowing you to solve for the values of both p and w.
First car is x
second car is y
25x+30y=1550
x+y=55
Multiply the second equation by -25 for elimination method to cancel out x
-25x-25y=-1375
Add the two equations
5y=175
y=35
Plug this value into the second equation
x+35=55
x=20
Final answers:
First car=20 gallons
Second car=35 gallons
It's multily
f(x) time g(x) is basically what they wnat
so
(x^2+8x+15) times (5/(x^2-9))
hold a sec, first factor each
(x+3)(x+5) times 5/((x-3)(x+3))
we get
((x+3)(x+5)(5))/((x-3)(x+3))
(x+30 cancels out
((x+5)(5))/(x-3)
(5x+25)/(x-3)
3rd option