**Solve algebratically the simultaneous equations**

**x^2 + y^2 + 25**

**5x - y = 11**

**Here is what i did: **

**-y = 11 - 5x **

**-y^2 = y^2 **

**(-5x+11) (-5x+11) = 25^2 - 110x +121 = 25 **

**= x^2 -25^2 - 110x + 96 = 0**

**= -24^2 - 110x + 96= 0**

**= -12^2 - 110x + 96= 0**

**then i tried to factorise my answer but i can't.**

edit: sorry my working out was -12x^2 - 55x +48 =0

YEEEEEET
Nov 11, 2017

#1**+1 **

x^2 + y^2 = 25 ⇒ y^2 = 25 - x^2 (1)

5x - y = 11 ⇒ 5x - 11 = y (2)

Square both sides of (2) and we have that

(5x - 11)^2 = y^2

25x^2 - 110x + 121 = y^2 set this equal to (1)

25x^2 - 110x + 121 = 25 - x^2 add x^2 to both sides, subtract 25 from both sides

26x^2 -110x + 96 = 0 divide through by 2

13x^2 - 55x + 48 = 0 factor as

(13x - 16) (x - 3) = 0

Set both factors to 0, solve for x and we have that x = 16/13 and x = 3

So using 5x - 11 = y

5(16/13) - 11 = 80/13 - 11 = [ 80 - 143] / 13 = -63/13

5 (3) - 11 = 15 - 11 = 4

So.......the solutions are (16/13, -63/13) and ( 3, 4)

CPhill
Nov 11, 2017

#2**+1 **

OMG after reading your solution i realised my result after first phrase should've been x^2 + 25x^2 - 110x -25 = 0 but i written -25x^2 for some reason thx

YEEEEEET
Nov 11, 2017

#3**0 **

Carelessly transposing numbers is such a frustrating habit. I have stared at many problems for prolonged periods of time without realizing such error.

TheXSquaredFactor
Nov 12, 2017