Question

x = a + b + 4 and a $\alpha$ y² ; b $\alpha$$\frac{1}{4}$, When y = 2, x = 18 and when y = 1, x =$-$3. Find x when y = 4.

Answer 1

$$\begin{gathered} x=a+b+4\left(\mathrm{Given}\right) \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ a\alpha y^2 \\ \therefore a=k{y}^2\left(k\mathrm{is}\mathrm{constant}.\right)...\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ b\alpha \frac{1}{y} \\ \therefore b=\frac{m}{y}\left(m\mathrm{is}\mathrm{another}\mathrm{constant}.\right)...\left(\mathrm{ii}\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}\hspace{0.17em}\left(\mathrm{i}\right)\hspace{0.17em}\&\left(\mathrm{ii}\right)\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ x=k{y}^2+\frac{m}{y}+4 \\ \mathrm{For}y=2,x=18. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ 18=k{\left(2\right)}^2+\frac{m}{2}+4 \\ \Rightarrow 18=4k+\frac{m}{2}+4 \\ \Rightarrow 2\left(14-4k\right)=m...\left(\mathrm{iii}\right) \\ \mathrm{For}y=1,x=-3. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ -3=k{\left(1\right)}^2+\frac{m}{1}+4 \\ \therefore -7-k=m...\left(\mathrm{iv}\right) \\ \mathrm{From}\left(\mathrm{iii}\right)\&\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}: \\ 2\left(14-4k\right)=-7-k \\ 28-8k=-7-k \\ 35=7k \\ \therefore k=5 \\ \mathrm{From}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}: \\ -7-5=m \\ \therefore m=-12 \\ \mathrm{Now}, \\ x=5y^2-\frac{12}{y}+4\left(\mathrm{Equation}\mathrm{of}\mathrm{variation}\right) \\ \Rightarrow x=5{\left(4\right)}^2-\frac{12}{4}+4\left(\because y=4\right) \\ \Rightarrow x=80-3+4 \\ \therefore x=81 \end{gathered}$$