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Help.

 Jan 4, 2018
 #1
avatar+232 
+1

#17 {(4,0), (0,4)}

 

#18 3,-2,2

 

 

those are the answers

 Jan 4, 2018
 #2
avatar+101406 
+1

17

 

x  +  y   =  4  ⇒   y  =  4 - x     (1)

x^2  + y^2  =  16       (2)

 

Sub   (1)  into (2)  and we have that

 

x^2  + (4 - x)^2   =  16

x^2  +  (x - 4)^2  = 16

x^2  + x^2  - 8x  +  16  =  16      subtract 16 from both sides

 

2x^2  - 8x  = 0     factor

2x ( x   - 4)  = 0

 

Setting each factor to 0  we have that x  = 0   and x  = 4

And when x  = 0   , y  = 4 - 0  = 4

And when x = 4, y  = 4 - 4  = 0

 

So.....the solutions are    (0,4)  and (4,0)

 

 

18.....easy,  NSS

 

Just change the signs  on the constants in the parentheses and we get

 

3, -2 , 2

 

 

cool cool cool

 Jan 4, 2018

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