+0  
 
+1
247
2
avatar+4690 

Help.

 Jan 4, 2018
 #1
avatar+166 
+1

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

 

#18 3,-2,2

 

 

those are the answers

 Jan 4, 2018
 #2
avatar+94545 
+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

11 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.