Problem Solving

Running at their respective constant rates, Machine x takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce 5/4w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A)4
B)6
C)8
D)10
E)12

I'm having trouble solving this algebraicly (?). Pls help!

This is what I did:

so given info above : rate for x is w/ [d+2] & rate for y is w/d.

working together they can make 5/[4w] widgets in 3 days so since, rate=work/time ...combined rate= [5/(4w)]/3 = 5/12w

So we can solve for d : w/[d +2] + w/d = 5/12w

5/12w = dw +w[d+2] / d[d +2]
= dw +dw +2w / d^2 +2d
= 2wd + 2w / d^2 +2d
5wd^2 + 10wd = 24dw + 24w
5wd^2 + 10wd - 24dw - 24w = 0
5wd^2 - 14wd - 24w = 0

divide the entire eq by w , we get: 5d^2 - 14d - 24 = 0

Now what do i do??????

it would be very time consuming for me to find two no.s that add upto 14/5 and equal 24/5 when multiplied.

I tried the quadratic eq formula & i was able to get d=4 though.

so if d=4, then d+2=6. so it takes x 6 days to make w widgets & 12 days to make 2w widgets.

BUT STILL i want to know how i can solve the above algebraicly! or is this the only way?!

You reached here fine:

=> 5d^2 - 14d - 24 = 0

Look at it this way : Find 2 numbers a and b such that a + b = - 14 and a * b = -120
a and b can thus be -20 and 6

=> 5d^2 - 20d + 6d - 24 = 0
=> 5d(d - 4) + 6(d - 4) = 0

Therefore d = 4

albatross86 wrote:You reached here fine:

=> 5d^2 - 14d - 24 = 0

Look at it this way : Find 2 numbers a and b such that a + b = - 14 and a * b = -120
a and b can thus be -20 and 6

=> 5d^2 - 20d + 6d - 24 = 0
=> 5d(d - 4) + 6(d - 4) = 0

Therefore d = 4

That's exactly what i was looking for, thank you!!