An algebra problem by Arpit Sah

xy+x+y=23 ,so (x+1)(y+1)=24 similarly (y+1)(z+1)=32,(x+1)(z+1)=48

solving these eq. we get x=5,y=3,z=7

$$ Rewrite $$ $\begin{aligned} xy+x+y=23&\quad\quad\Rightarrow& x+(1+x)y=23&\quad(1)\\ yz+y+z=31&\quad\quad\Rightarrow& y+(1+y)z=31&\quad(2)\\ zx+z+x=47&\quad\quad\Rightarrow& x+(1+x)z=47&\quad(3) \end{aligned}$ $$ First we eliminate $\,z$ by multiplying $(2)$ with $(1+x)$ and multiplying $(3)$ with $(1+y)$ and then subtracting, yield $$ $\frac{ \begin{array}{c}[b]{r} (1+y)x+(1+y)(1+x)z=47(1+y)\\ (1+x)y+(1+x)(1+y)z=31(1+x) \end{array} }{ \quad\quad\quad\quad\quad\;\;(1+x)y-(1+y)x=16+47y-31x }-$ $$ Hence $$ $\begin{aligned} (1+x)y-(1+y)x&=16+47y-31x\\ y+xy-x-xy-47y+31x&=16\\ 32x-48y&=16\\ 2x-3y&=1\\ 3y&=2x-1\\ y&=\frac{2x-1}{3}. \end{aligned}$ $$ Substitute $\,y$ to $(1)$ , yield $$ $\begin{aligned} x+(1+x)\left(\frac{2x-1}{3}\right)&=23\\ 3x+(x+1)(2x-1)&=69\\ 3x+2x^2+x-1-69&=0\\ 2x^2+4x-70&=0\\ x^2+2x-35&=0\\ (x+7)(x-5)&=0\\ x_1=-7\quad&\text{or}\quad x_2=5. \end{aligned}$ $$ Take the positive value of $\,x$ and find $\,y$ and $\,z$ . $$ $\begin{aligned} y&=\frac{2x-1}{3}\\ y&=\frac{2(5)-1}{3}\\ &=3 \end{aligned}$ and $\begin{aligned} y+(1+y)z&=31\\ 3+(1+3)z&=31\\ 3+4z&=31\\ 4z&=28\\ z&=7. \end{aligned}$ $$ Thus, $x+y+z=5+3+7=\boxed{15}$ . $$

we can use substitution method to solve this problem...

lets take 'x' as the subject & represent y,z in terms of x

♦(eqn.1) xy+ x +y = 23 ------> y = (23-x)/(1+x)

♦(eqn.3) zx+ z + x = 47 ------>z = (47-x)/(1+x)

substitute for y,z in (eqn.2) yz+y+z =31

===>{ (23-x)(47-x)/(1+x)² } + { (23-x)/(1+x) } + { (47-x)/(1+x) } = 31 ----- multiply both sides by (1+x)²

==>(23-x)(47-x) + ( (23-x )+(47-x) ) (1+x) = 31 (1+x)²

===> -32x²-64x +1120 =0

===>x²+2x -35 =0

-->(x+7)(x-5) = 0 ie x= 5 or -7

x>0(given) ====> x=5

y=(23-x)/(1+x)-----put x=5 ----> y=3

z=(47-x)/(1+x)-----put x=5-----> z=7

so (x,y,z) = (5,3,7) =====> x+y+z = 15

HENCE , (X+Y+Z =15 ) IS THE ANSWER