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One integer is 9 less than 5 times another. Their product is 18. Find the integers.

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Answer:

x = 3 and x=-6/5

Explanation:

x = one integer

y = other integer

One integer is 9 less than 5 times another

x= 5y-9

product is 18

xy = 18

Substitute in for x

(5y-9) *y = 18

Distribute

5y*y -9y = 18

5y^2 - 9y = 18

Subtract 18 from each side.

5y^2 - 9y -18= 18-18

5y^2 - 9y -18 = 0

Using the quadratic formula

-b ± sqrt(b^2 -4ac)

-----------------------

2a

-(-9) ±sqrt(9^2 -4*5*(-18))

--------------------------------------

2(5)

9 ±sqrt(81 +360))

--------------------------------------

10

9 ±sqrt(441)

--------------------------------------

10

9±21

----------

10

x = (9+21)/10 and x = (9-21)/10

x = 30/10 and x = (-12)/10

x = 3 and x=-6/5

User Steve Lage
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