Question

In the system of equations $(x+y)(y+z)=35,(y+z)(z+x)=56$ and $(z+x)(x+y)=40$, the values of $x,y$ and $z$ are:

• A. $\pm 5,\pm 3$ and $\pm 2$
• B. $\pm 3,\pm 2$ and $\pm 5$
• C. $\pm 2,\pm 5$ and $\pm 3$
• D. $\pm 2,\pm 5$ and $\pm 4$

Answer 1

The correct option is B $\pm 3,\pm 2$ and $\pm 5$

On multiplying the given equations together,

$(x+y{)}^2.(y+z{)}^2.(z+x{)}^2=35\times 56\times 40=78400$

$\therefore (x+y)(y+z)(z+x)=\sqrt{7800}=\pm 280$ ......(i)

On dividing equations (i) by the given equations respectively,

$z+x=\pm \frac{280}{35}=\pm 8$ ......(ii)

$x+y=\pm \frac{20}{56}=\pm 5$ ......(iii)

and $y+z=\pm \frac{280}{40}=\pm 7$ .......(iv)

On adding equation (ii), (iii) and (iv)

$2(x+y+z)=\pm (8+5+7)=\pm 20$

$\therefore x+y+z=\pm \frac{20}{2}=\pm 10$ ......(v)

On subtracting from equation (v), the equation (ii), (iii) and (iv) respectively.

$y=\pm (10-8)=\pm 2;z=\pm (10-5)$ $=\pm 5$ and

$x=\pm (10-7)=\pm 3$.