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jenyasd209 [6]
3 years ago
6

A car needs fixing and Abe can fix it for $70 per hour with a $130 part, but Gabe can fix it for $80 an hour with a $40 part. Ho

w long will it take for both Abe and Gabe to cost the same amount?
Mathematics
2 answers:
Shtirlitz [24]3 years ago
8 0
For the same amount set up two equations set equal to each other.
$70h + 130 = 80h + $40
Subtract 80h from both sides
-10h + 130 = 40
subtract 130 from both sides
-10h = -90 divide both sides by -10
h = 9 hours

CHECK
70(9) + 130 = $760 
80(9) + 40 = $760
Shkiper50 [21]3 years ago
8 0

Answer:

9 hours

Step-by-step explanation:

Let for x hours the amount charged by Abe and Gabe is same.

Given,

In fixing by Abe,

Fixed cost  = $ 130,

Additional cost per hour = $ 70,

So, the total charge for x hours = 130 + 70x

In fixing by Gabe,

Fixed cost = $ 40,

Additional cost per hour = $ 80,

So, the total charge for x hours = 40 + 80x

For same charge,

130 + 70x = 40 + 80x

70x = 40 + 80x - 130

70x - 80x = -90

-10x = -90

⇒ x = 9

Hence, it will take 9 hours for both Abe and Gabe to cost the same amount

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