It is quite simple you first write x in terms of y from first equation and put x in second equation and solve for y . Then , you will get y in terms of a and b , then put y in any of the two equation and get x .
Question : If $a^{2}=by+cz$, $b^{2}=cz+ax$, $c^{2}=ax+by$, then the value of $\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$ is:
Option 1: $1$
Option 2: $a+b+c$
Option 3: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Option 4: $0$
Question : The third proportional of the following numbers $(x-y)^2, (x^2-y^2)^2$ is:
Option 1: $(x+y)^3(x-y)^2$
Option 2: $(x+y)^4(x-y)^2$
Option 3: $(x+y)^2(x-y)^2$
Option 4: $(x+y)^2(x-y)^3$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile