# Find x and y is 2x+ y= 11  and x^2 - y^2 = 24

hala718 | Certified Educator

2x+ y = 11.............(1)

x^2 - y^2 = 24............(2)

First we will rewrite (1):

2x + y = 11

==> y= 11- 2x

Now we will substitute in (2):

==> x^2 - y^2 = 24

==> x^2 - ( 11-2x)^2 = 24

==> x^2 - ( 121 - 44x + 4x^2 ) = 24

==? x^2 - 121 + 44x - 4x^2 = 24

==> -3x^2 + 44x - 121 - 24 = 0

==> -3x^2 + 44x - 145 = 0

Now we will calculate the roots:

==> x1= ( -44 + sqrt( 44^2 -4*-3*-145) / 2*-3

= ( -44 + 14) / -6 = -30/-6 = 5

==> x1= 5  ==> y1= 11-2*5= 1==> y1= 1

==> x2= ( -44 - 14) /-6 = -58/ -6 = 29/3

==> x2= 29/3  ==> y1= 11-2*29/3 = -25/3

neela | Student

2x+y = 11.............(1)

x^2-y^2 = 24.........(2)

To solve the system.

From (1) we get y =11-2x.

We put y = 11-2x in (2):

x^2- (11-2x)^2 = 24.

x^2 - (121-44x +4x^2) = 24.

We get together the like terms:

x^2-4x^2  +44x -121-24 = 0.

-3x^2 +44x -145 = 0.

We multiply by -1:

3x^2 -44x+145 = 0.

3x^2 - 29x -15x +145 = 0

x(3x-29) - 5(3x-29) = 0.

(3x-29)(x-5).

x = 5 or x= 29/3.

For x= 5, y = 11-2x = 11-10 =1.

For x= 29/3, y = 11 - 2*29/3 = -25/3.

Therefore  (x,y) = (5,1) , Or (x,y) = (29/3 , -25/3) are the solutions.