If x^2-y^2 = 55 and x-y=11, then what is y?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that x^2 - y^2 = 55 and x - y = 11.

x^2 - y^2 = (x - y)(x + y)

=> 55 = 11*(x + y)

=> x + y = 5 ...(1)

x - y = 11 ...(2)

(1) - (2)

=> x + y - x + y = 5 - 11

=> 2y = -6

=> y = -3

The value of y = -3

cathyle | Student, Undergraduate | (Level 1) eNoter

Posted on

x-y = 11 => x= 11+y

Replace x=11+y in x^2 - y^2 = 55

And then, apply a^2 - b^2 = (a-b)(a+b)

(11+y)^2 - y^2 = 55 <=> (11+y-y)(11+y+y) = 55

<=>   11(11+2y)            = 55

<=>   11+2y                   = 5

<=>   2y                         = -6

<=>     y                         = -3.

Hope easier to understand.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll recall the fact that the difference of two squares returns the product (x-y)(x+y) = x^2 - y^2

Therefore, we can re-write the first equation:

(x-y)(x+y) = 55

But (x-y) = 11 => 11*(x+y) = 55

We'll divide by 11 both sides:

x + y = 5 (3)

We'll subtract (2) from (3):

x + y - x + y = 5 - 11

2y = -6

y = -3

The requested value of y is: y = -3.