Two Equations: Find x+y

[bkarpuz]

Dear MHB members,

here is my problem.
If $\frac{y^{2}-4}{y+3}=\frac{x^{2}-2x-3}{x+2}=\frac{2}{3}$, then $x+y=?$

Thank you.
bkarpuz

 

[chisigma]

Dear MHB members,

here is my problem.
If $\frac{y^{2}-4}{y+3}=\frac{x^{2}-2x-3}{x+2}=\frac{2}{3}$, then $x+y=?$

Thank you.
bkarpuz

Solve in x and y the two equations...

$\displaystyle \frac{y^{2}-4}{y+3}= \frac{2}{3}$


$\displaystyle \frac{x^{2}-2x-3}{x+2}\ =\frac{2}{3}$

... and then compute x+y...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

You might also write:

$\displaystyle \frac{y^2-4}{y+3}=\frac{x^2-2x-3}{x+2}$

$\displaystyle \frac{(y+2)(y-2)}{y+3}=\frac{(x+1)(x-3)}{x+2}$

$\displaystyle \frac{((y+3)-1)((y+3)-5)}{y+3}=\frac{((x+2)-1)((x+2)-5)}{x+2}$

This implies:

$\displaystyle y+3=x+2\:\therefore\:x-y=1\:\therefore\:x+y=2x-1$

Now solve:

$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$

To find x.

 

[MarkFL]

Here's another method:

$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$

implies:

(1) $\displaystyle 3x^2-8x-13=0$

$\displaystyle\frac{y^2-4}{y+2}=\frac{2}{3}$

implies:

(2) $\displaystyle 3x^2-2y-18=0$

Adding (1) and (2) we get:

(3) $\displaystyle 3x^2+3y^2-8x-2y-31=0$

If we multiply (1) by 2 we have:

$\displaystyle 6x^2-16x-26=0$

which we may write as:

$\displaystyle -2x-8(x-1)+6x(x-1)-34=0$

Using $\displaystyle y=x-1$ this becomes:

(4) $\displaystyle-2x-8y+6xy-34=0$

Adding (3) and (4) we obtain:

$\displaystyle 3x^2+6xy+3y^2-10x-10y-65=0$

(5) $\displaystyle 3(x+y)^2-10(x+y)-65=0$

Now we have a quadratic in $\displaystyle x+y$.