Question

One positive integer is 7 less than twice another. The sum of their squares is 346. Find the integers. should be 2 answers

Answer 1

Answer:

the two positive integers are x= 15, and y = 11

Step-by-step explanation:

Let the first integer be x

Let the second integer be y

from the problem we can decode the following equations

$x=2y-7$ ---------------------------- equation 1

$x^{2} + y^{2}= 346$ -------------------------equation 2

substituting the value of x into equation 2, we have

$(2y-7)^{2} + y^{2}= 346 --------------equation 3$

expanding, we have

$4y^{2}-28y+49+y^{2}=346$

$5y^{2}-28y-297 = 0$

from this, y = 11 or y = -5.4

since our answer is a positive integer, we will have to pick the first value of y which is y = 11

substituting the value of y into equation 1, we have

$x= 2(11)-7=15$

hence x = 15

Therefore, we have x= 15, and y = 11

these are the two positive integers