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

### 4 Answers

- billrussell42Lv 78 months agoFavourite answer
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

quadratic equation:

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

- ComoLv 78 months ago
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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- KrishnamurthyLv 78 months ago
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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- rotchmLv 78 months ago
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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your opinion. To me it is simple.