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# Simultaneous equations

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Heya

I need help with two questions,i thought i figured this topic out but i guess not ahahah

question 1 :
x^2-y^2=12
y=x-2
i know i have to substitute y into the first equation but from there i kind of got stuck
question 2 :
x^2+y-=25
y=x+3
again this one needs to be substituted but i kind of get confused midway whether i need to factorise it or if that's even needed for this one

May 8, 2019

#1
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OK...first one     x^2 - y^2  = 12      and  y = x - 2

Put the secnd into the first for y  and we have

x^2 - ( x - 2)^2  = 12

x^2 - (x^2 - 4x + 4)  = 12

x^2 - x^2 + 4x - 4  = 12

4x - 4  = 12      add 4 to both sides

4x = 16    divide both sides by 4

x =  4

And y  =   x - 2  =   (4) - 2  =   2

So   (x, y)  = (4, 2)

EDIT TO CORRECT A PREVIOUS ERROR>>>>>thanks, heureka  !!!   May 8, 2019
edited by CPhill  May 9, 2019
#2
+2

Second one      x^2 + y  =  25      y = x+3

Using the same idea.....

x^2 + (x + 3) = 25

x^2 + x + 3  = 25       subtract 25 from both sides

x^2 + x - 22 = 0         add 22 to both sides

x^2 + x  = 22          complete the square on x

x^2 + x + 1/4  = 22 + 1/4

(x + 1/2)^2  =  89/4   take both roots

x + 1/2  = ±√89 / 2      subtract 1/2 from both sides

x = ±√89 / 2 - 1/2

And y  =   x + 3  =   ±√89 / 2 - 1/2 + 3   =   ±√89 / 2 - 1/2 + 6/2  =   ±√89 / 2 + 5/2

So  (x , y)  =   (  ±√89 / 2 - 1/2 , ±√89 / 2 + 5/2 )   May 8, 2019