Question

If conditions are given by $2x+3y=13$, $xy=6$ then find the value of $4{{x}^{2}}+9{{y}^{2}}$.

Answer 1

To solve this question, we have to find the values of x and y which satisfy the equations $2x+3y=13\text{ and } xy=6$. We have to substitute $y=\dfrac{6}{x}$ in the equation $2x+3y=13$ and solve for x to get the values of x and y and then substituting in the term $4{{x}^{2}}+9{{y}^{2}}$ gives the required answer.

Complete step-by-step solution:
In the question, given constraints are
$2x+3y=13\to \left( 1 \right)$
$xy=6\to \left( 2 \right)$
From equation -$2$, we should substitute $y=\dfrac{6}{x}$ in the equation - $(1)$. We get
$2x+3\times \dfrac{6}{x}=13$.
Simplifying, we get
$2{{x}^{2}}-13x+18=0$
By doing factorisation, we get,
$\begin{aligned} & 2{{x}^{2}}-4x-9x+18=0 \\\ & 2x\left( x-2 \right)-9\left( x-2 \right)=0 \\\ & \left( x-2 \right)\left( 2x-9 \right)=0 \\\ \end{aligned}$
When $AB = 0$, either A is 0 or B is 0. So, we can get conditions as
$\begin{aligned} & x-2=0\Rightarrow x=2 \\\ & 2x-9=0\Rightarrow x=\dfrac{9}{2} \\\ \end{aligned}$
$\therefore x = 2$ or $\dfrac{9}{2}$
We have the relation that y = $\dfrac{6}{x}$
For $x = 2, y = \dfrac{6}{2}=3$
For $x =\dfrac{9}{2}$, $y = \dfrac{6}{\dfrac{9}{2}}=\dfrac{6\times 2}{9}=\dfrac{4}{3}$
The required expression is $4{{x}^{2}}+9{{y}^{2}}$.
Case-1 let us consider $x = 2, y = 3$.
Substituting in the expression $4{{x}^{2}}+9{{y}^{2}}$ gives,
$4\times \left( {{2}^{2}} \right)+9\left( {{3}^{2}} \right)=4\times 4+9\times 9=16+81=97$
Case-1 gives the required value as $97$.
Case-2 let us take $x =\dfrac{9}{2}$, $y = \dfrac{4}{3}$
Substituting in the expression $4{{x}^{2}}+9{{y}^{2}}$ gives,
$4\times {{\left( \dfrac{9}{2} \right)}^{2}}+9{{\left( \dfrac{4}{3} \right)}^{2}}=4\times \dfrac{81}{4}+9\times \dfrac{16}{9}=81+16=97$
Case-2 also gives the required value as $97$.
$\therefore$ The required value of the expression $4{{x}^{2}}+9{{y}^{2}}$, given that $2x+3y=13$ and $xy=6$, is $97$.

Note: The alternate way to do the problem is to look at the required expression $4{{x}^{2}}+9{{y}^{2}}$ carefully. It is of the form${{\left( 2x \right)}^{2}}+{{\left( 3y \right)}^{2}}$. The terms $2x$ and $3y$ are present in the constraint given by$2x+3y=13$. There is a relation that states that
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$. Here in the question, $a = 2x$ and $b = 3y$. Comparing them with the formula, we get
$4{{x}^{2}}+9{{y}^{2}}={{\left( 2x \right)}^{2}}+{{\left( 3y \right)}^{2}}={{\left( 2x+3y \right)}^{2}}-2\times 2x\times 3y={{\left( 2x+3y \right)}^{2}}-12xy$
Substituting the values of $2x+3y=13$ and $xy=6$, we get
${{\left( 2x+3y \right)}^{2}}-12xy={{13}^{2}}-12\times 6=169-72=97$.
This tally with the answer we got in the process of the solution. The key to this method is to find that the constraints and the required expression are directly related to each other.