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avatar+154 

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
edited by YEEEEEET  Nov 11, 2017
 #1
avatar+86919 
+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)

 

 

cool cool cool

CPhill  Nov 11, 2017
 #2
avatar+154 
0

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 cheeky thx 

YEEEEEET  Nov 11, 2017
 #3
avatar+2075 
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

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