# find three consecutive multiples of five such that the product of the first and second numbers minus seven times the 3rd one is 125?

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find three consecutive multiples of five such that the product of the first and second numbers minus seven times the 3rd one is 125?

call the numbers x, x+5, x+10

x(x+5) – 7(x+10) = 125

x² + 5x – 7x – 70 – 125 = 0

x² – 2x – 195 = 0

to solve ax² + bx + c = 0

x = [–b ± √(b²–4ac)] / 2a

x = [2 ± √(4+4•196)] / 2

x = [2 ± √(784)] / 2

x = [2 ± 28] / 2

x = 1 ± 14

x = 15, –13

so answer is 15, 20, 25

check

15(20) – 7(25) = 125

300 – 175 = 125

ok

• your opinion. To me it is simple.

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• 5x ( 5x. + 5 ) - 7 ( 5x + 10 ) = 125

25x² + 25x - 35 x - 70 = 125

25x² - 10 x - 195 = 0

5x² - 2x - 39 = 0

( 5x + 13 ) ( x - 3 ) = 0

x = 3 is acceptable integer

Three consecutive integers are 15 , 20 and 25

Check

(15 x 20) - 175 = 300 - 175 = 125 ____as required

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• Three consecutive multiples of five such that

the product of the first and second numbers

minus seven times the 3rd one is 125:

5(x - 1)5x - 7(5(x + 1)) = 125

x = 4

15, 20 and 25

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• let the three multiples be 5(n-1), 5n, 5(n+1).

You Are told that 5(n-1) * 5n -7*5(n+1) = 125

Solve for n.

Done.

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