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# solve simultaneously

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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.

Jun 2, 2021

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

CPhill,

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

juriemagic  Jun 3, 2021