solving equations by the method of substitution $\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3},$
$x+1=y$
We have to solve for $x$ and $y$.I have tried to solve for them by finding value of $x$ or $y$ from the second equation and place them in the second.It is obvious that the answers would be $2$ and $3,$but we need something else. I tried to find the relation between $a$ and $b$ and them place them again in the first equation along with the value of $x$ or $y$ ,  but it yields something bizarre. So how do we solve it? A tiny hint will be appreciated. 
 A: Hint: (1) Multiply each side of the first equation by $(2)(3)(x)(y)$; (2) Then replace $y$ by $x+1$; (3) Rearrange, you get a quadratic equation in $x$. 
You could solve the resulting equation using the Quadratic Formula. Actually, the quadratic factors nicely. 
Alternately, you spotted one of the roots. If you know one root of a quadratic, then by glancing at the coefficients you can find the other. This is because in the quadratic equation $px^2+qx+r=0$, the product of the roots is $r/p$ and the sum of the roots is $-q/p$. You will find the product criterion more pleasant. If you use it you don't even need to calculate the messisest coefficient of the quadratic, the coefficient of $x$.
A: $\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3}$...........$(1)$
$x+1=y$............$(2)$
From (2),$x=y-1$. Substituting it into $(1)$,
$\dfrac{a}{y-1}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3}$
$\implies \dfrac{a}{y-1}-\dfrac{a}{2}=\dfrac{b}{3}-\dfrac{b}{y}$
$\implies \dfrac{2a-ay+a}{2(y-1)}=\dfrac{b(y-3)}{3y}$ 
$\implies \dfrac{3a-ay}{2(y-1)}=\dfrac{b(y-3)}{3y}$ 
$\implies \dfrac{a(3-y)}{2(y-1)}=\dfrac{b(y-3)}{3y}$ 
$\implies \dfrac{a(3-y)}{2(y-1)}=-\dfrac{b(3-y)}{3y}$  [Multiplying the fraction on the right side by $\dfrac{-1}{-1}$]
$\implies \dfrac{a(3-y)}{2(y-1)}+=0$
$\implies (3-y)[\dfrac{a}{2(y-1)}+\dfrac{b}{3y}]=0$
Since the product of two numbers is 0,at least one of them must be 0.........
No quadratic formula needed!
