Respuesta :
Answer:
Let charge per day be $x and charge per mile be $y
then:
as per the given condition we have;
3x + 300y =159 .....[1]
5x + 600y = 289 .....[2]
Multiply equation [1] both sides by 5 we get;
[tex]5(3x+300y) = 5 \cdot 159[/tex]
Using distributive property: [tex]a \cdot(b+c) = a\cdot b+ a\cdot c[/tex]
15x+1500y = 795 .....[3]
Multiply equation [2] by 3 both sides we get;
[tex]3(5x+600y) = 3 \cdot 289[/tex]
Using distributive property: [tex]a \cdot(b+c) = a\cdot b+ a\cdot c[/tex]
15x+1800y = 867 .....[4]
Subtract equation [3] from [4] to eliminate x and solve for y;
15x+1800y-15x-1500y = 867 -795
Combined like terms;
300y = 72
Divide both sides by 300 we get;
[tex]y = \frac{72}{300}=0.24[/tex]
⇒ y =$0.24 per mile.
Now substitute this value y in equation [1] to solve for x;
3x + 300(0.24) =159
3x + 72 =159
Subtract 72 both sides we get;
3x + 72 -72 = 159 -72
Simplify:
3x = 87
Divide both sides by 3 we get;
[tex]x = \frac{87}{3} = 29[/tex]
Therefore, the best rental charges per day is, $ 29 and the charges per mile is, $0.24
