+134 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
thanks in advance^^
+128408 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 !!!
+128408 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 )
