Question

if x-3 is directly proportional to the square of y and x = 5 when y = 2, find x when y = 6

Answer 1

\[(x-3) \alpha \y^2 \rightarrow (x-3) =k \y^2\]
Where k is a constant
Now when x=5 and y =2, first we ned to find the value of the constant i.e. k
so let us substitute x=5 and y=2 in the above eq
\[(5-3) =k \times 2^2 \rightarrow 2= k \times 4 \rightarrow k= \frac{2}{4} =\frac{1}{2} \]\\[k= \frac{1}{2}\]
Now we substitute the value of k in the eq \[(x-3) =k y^2\] we find
\[(x-3) =\frac{1}{2} \times y^2\rightarrow (x-3) =\frac{y^2}{2}\]which is the required eq of variation.
Now let us put y=6 to find the value of x
i.e.\[(x-3) =\frac{y^2}{2} \rightarrow (x-3) =\frac{6^2}{2}=\frac{36}{2} = 18\]
\[\rightarrow (x-3) = 18 \rightarrow x= 18 +3 \rightarrow x= 21\]
@bquiz