+0  
 
+2
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3
avatar+2077 

2 q's, please help, ty!

RainbowPanda  Feb 28, 2018
 #1
avatar+13665 
+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 

ElectricPavlov  Feb 28, 2018
 #2
avatar+91255 
+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"

 

cool cool cool

CPhill  Feb 28, 2018
 #3
avatar+2077 
+1

Thank you both for your help ^-^

RainbowPanda  Feb 28, 2018

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