Sep 18, 2012 Thread starter #1 B bkarpuz New member Jan 27, 2012 11 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
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
Sep 18, 2012 #2 chisigma Well-known member Feb 13, 2012 1,704 bkarpuz said: 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 Click to expand... 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 $\chi$ $\sigma$
bkarpuz said: 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 Click to expand... 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 $\chi$ $\sigma$
Sep 18, 2012 Admin #3 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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.
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.
Sep 18, 2012 Admin #4 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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 $\displaystyle x+y$.
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 $\displaystyle x+y$.