HELP!! Two mechanics worked on a car. The first mechanic worked for 20 hours, and the second mechanic worked for 5 hours. Together they charged a total of 2375. What was the rate charged per hour by each mechanic if the sum of the two rates was 160 per hour?

Step-by-step explanation:

Let r represent the rate charged by the first mechanic. Then 160-r is the rate charged by the second mechanic. The total of charges is ...

  20r +5(160-r) = 2375

  15r +800 = 2375 . . . . . . eliminate parentheses

  15r = 1575 . . . . . . . . . . . . subtract 800

  r = 105 . . . . . . . . . . . . . . . divide by 15. First mechanic's rate.

  160-r = 55 . . . . . Second mechanic's rate

The first mechanic charged $105 per hour; the second, $55 per hour.

the rate charged per hour by the first mechanic is 105 per hour.

the rate charged per hour by the second mechanic is 55 per hour.

Step-by-step explanation:

Let x represent the rate charged per hour by the first mechanic.

Let y represent the rate charged per hour by the second mechanic.

The first mechanic worked for 20 hours, and the second mechanic worked for 5 hours. Together they charged a total of 2375. This means that

20x + 5y = 2375 - - - - - - - - - - -1

if the sum of the two rates was 160 per hour, it means that

x + y = 160

Substituting x = 160 - y into equation 1, it becomes.

20(160 - y) + 5y = 2375

3200 - 20y + 5y = 2375

- 20y + 5y = 2375 - 3200

- 15y = - 825

y = - 825/ - 15

y = 55

x = 160 - y = 160 - 55

x = 105

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb