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Guest Feb 22, 2018
 #1
avatar+87333 
+2

1.

Since y equals y...we can set the frst two equations equal

 

x^2  - 2x + 4   =  -2x + 4       subtract  -2x + 4 from both sides

 

x^2  =  0    ⇒    x  =  0

 

So...only one solution  at  (0, 4)

 

Here's a graph   : https://www.desmos.com/calculator/pmhadfbhr1

 

 

cool cool cool

CPhill  Feb 22, 2018
 #2
avatar+87333 
+2

Again....we can set the equations  equal

 

x^2  - 5x + 6  =  -4x + 6       add 4x, subtract 6   to / from each side

 

x^2 - x = 0     factor

 

x ( x - 1)  = 0

 

Setting each  factor to 0   and solving for x produces x = 0  or x  = 1

 

And using    -4x  +  6

 

When x  = 0 , y  = 6

When x  = 1, y = 2

 

So...the solutions are  (0,6)  and (1,2)

 

 

cool cool cool

CPhill  Feb 22, 2018
 #3
avatar+87333 
+2

x^2 + 4x  - 10  =  8x - 14

 

Subtract  8x , add 14  from / to   both sides

 

x^2 - 4x + 4  =  0       factor

 

(x - 2) ( x - 2)  =  0

 

Setting either factor to 0 and solving for x   prduces  x  = 2

 

And  using  8x  - 14.....  y  =  8(2) - 14  =  16   -  14   = 2

 

So the solution is (2,2)

 

 

cool cool cool

CPhill  Feb 22, 2018

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