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avatar+846 

Solve simultaneously:

\(x+y=2\)

and

\(y+1={-3 \over x-1}\)

 

I did this:

\(x+y=2\)

\(x=2-y.....(1)\)

\(y+1={-3 \over x-1}.....(2)\)

 

(1) into (2)

\(y+1={-3 \over (2-y)-1}\)

\(y+1={-3 \over -2+y}\)

\((y+1)(y-2)=-3\)

\(y^2-y=-1\)

\(y(y-1)=-1\)

\(y=-1\)  or  \(y-1=-1\)

                     \(y=0\)

Then I substituted the y values into (1), getting the x values to be 3 and 2 respectively.

 

Then another question follows:

Hence, or otherwise, determine the value of:

\(({1 \over x}+{1 \over y})\)

 

The teacher says the answer should be \(-{1 \over4}\),  I do not get that.

 

where did I go wrong?...please help..Thank you all very much indeed.

 Jun 2, 2021
 #1
avatar+121056 
+1

You're OK  until  here   

 

y + 1   =       - 3

               _________

                (2 - y) - 1 

 

y + 1  =    -3

             ______

                1 - y

 

(1 + y) (1 - y)  =  - 3

 

-y^2  + 1  =   - 3                         

 

-y^2  + 4  =  0      multiply through by -1

 

y^2  - 4  =  0

 

y^2  =  4

 

y = 2                or  y  = -2

 

So

 

x + 2  = 2                                                  or       x   -  2  =  2

 

x = 0                                                                    x =   4

 

reject since  1/x  isn't defined

 

So     x =  4 , y  =  -2

 

1/4  +   1/-2   = 

 

1/4   - 2/4   =

 

-1/4

 

cool cool cool  

 Jun 2, 2021
edited by CPhill  Jun 2, 2021
 #2
avatar+846 
+1

CPhill,

 

I should have minused, not multiply!!...ooh dear me!!!indecision..thank you CPhill!

juriemagic  Jun 3, 2021

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