+0

+2
221
3
+2448

Feb 28, 2018

#1
+18096
+2

Second one:

x=y

y =x^2+3x-3

Substitute  x = y into the second equation

y = y^2 +3y-3

0 = y^2 +2y -3   factor

0 = (y+3)(y-1)                    so y = -3  or 1      and since x= y

-3,-3   or  1,1   are solutions

Feb 28, 2018
#2
+99617
+2

Note, RP

P  = (12, -4)

P'  =  ( 18, - 6)

We added  6  to the x coordnate of P  to get the x coordinate of P'

And we subtracted 2 from the y coordinate of P  to get the y coordinate of P'

So...we need to do the same to the coordinates of Q...so we have

(-16 + 6, 8 - 2)  =  (-10, 6)  =   Q'

x  = y   (1)

y = x^2 + 3x -3    (2)

Sub  (1)  into (2)

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

0  =  x^2 + 2x - 3      factor

0  =  ( x + 3) ( x - 1)

Setting  both of the factors equal to 0  we have that

x + 3  =  0              x - 1  = 0

x  =  - 3                  x  =  1

Note..if  x  = y   .....then  (1, 1)  is a solution  ⇒  "C"

Feb 28, 2018
#3
+2448
+1

Thank you both for your help ^-^

Feb 28, 2018