Question

If $p,q,r,b,c$ are non zero real numbers such that $p+q+r=16$, $b+c=4$, then find the value of $x+y+z$

Where $x,y,z$ satisfy the system of equations:
$\frac{x}{p}+\frac{y}{p-b}+\frac{z}{p-c}=1$

$\frac{x}{q}+\frac{y}{q-b}+\frac{z}{q-c}=1$

$\frac{x}{r}+\frac{y}{r-b}+\frac{z}{r-c}=1$

Answer 1

Write given equations in matrix form and apply crammer's rule

$x=\frac{{\Delta }_1}{\Delta };y=\frac{{\Delta }_2}{\Delta };z=\frac{{\Delta }_3}{\Delta }$

where

${\Delta }_1=\left|\begin{array}{ccc}1& \frac{1}{(p-b)}& \frac{1}{(p-c)}\\ 1& \frac{1}{(q-b)}& \frac{1}{(q-c)}\\ 1& \frac{1}{(r-b)}& \frac{1}{(r-c)}\end{array}\right|{\Delta }_2=\left|\begin{array}{ccc}\frac{1}{p}& 1& \frac{1}{(p-c)}\\ \frac{1}{q}& 1& \frac{1}{(q-c)}\\ \frac{1}{r}& 1& \frac{1}{(r-c)}\end{array}\right|$

${\Delta }_3=\left|\begin{array}{ccc}\frac{1}{p}& \frac{1}{(p-b)}& 1\\ \frac{1}{q}& \frac{1}{(q-b)}& 1\\ \frac{1}{r}& \frac{1}{(r-b)}& 1\end{array}\right|\Delta =\left|\begin{array}{ccc}\frac{1}{p}& \frac{1}{(p-b)}& \frac{1}{(p-c)}\\ \frac{1}{q}& \frac{1}{(q-b)}& \frac{1}{(q-c)}\\ \frac{1}{r}& \frac{1}{(r-b)}& \frac{1}{(r-c)}\end{array}\right|$

Simplyfing the determinants gives

$x=\frac{pqr}{bc};y=\frac{(p-b)(q-b)(r-b)}{b(b-c)};z=\frac{(p-c)(q-c)(r-c)}{c(c-b)}$

Expanding the above terms and simplyfing

$x+y+z=\frac{(p+q+r)bc(b-c)+bc(c-b)(b+c)}{bc(b-c)}$

$=(p+q+r)-(b+c)$

Hence $x+y+z=16-4=12$

Ans: 12