Two positive integers differ by 7. The sum of their squares is 169. Fi

Question

Two positive integers differ by 7. The sum of their squares is 169. Find the larger integer.
a. 4
b. 5
c. 9
d. 12
e. 14

Sanyam_Maheshwari · Answer

Let x and y be the two positive integers.
We know that x - y = 7 ...(1)
And, x^2 + y^2 = 169 ...(2)

From equation (1), express x in terms of y:
x = y + 7

Substitute this expression for x into equation (2):
(y + 7)^2 + y^2 = 169

Expand and simplify:
2y^2 + 14y + 49 = 169

Simplify further:
2y^2 + 14y - 120 = 0

Factor the quadratic:
(y - 5)(y + 12) = 0

Discard the negative solution (y = -12).
Choose the positive solution, y = 5.

Now, find x:
x = y + 7
x = 5 + 7
x = 12

Therefore, the larger integer is 12.

ruis · Answer

I do not understand how you get to (y-5)(y+12). Shouldn´t be -120 the product of the results? and 14 the difference?

Bunuel · Answer

Reduce 2y^2 + 14y - 120 = 0 by 2 to get y^2 + 7y - 60 = 0 and then factor.

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