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Hint: Here, we have to square the first equation $2x+y=14$ and then apply the second equation $xy=6$ into the first one.

Complete step-by-step answer:

Here, we are given with two equations:

$2x+y=14\text{ }.....\text{ (1)}$

$xy=6\text{ }.....\text{ (2)}$

With the help of these two equations we have to find the value of $4{{x}^{2}}+{{y}^{2}}$.

First, let us consider the equation (1).

By squaring equation (1) on both the sides we get,

${{(2x+y)}^{2}}={{14}^{2}}\text{ }....\text{ (3)}$

From equation (3) we can say that its LHS is of the form ${{(a+b)}^{2}}$whose expansion we are familiar with. i.e. the expansion for ${{(a+b)}^{2}}$is given as:

${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

In the LHS of equation (3), has the form ${{(a+b)}^{2}}$ where $a=2x$, and $b=y$.

Now by applying the above formula in equation (3) we get:

$\begin{align}

& {{(2x)}^{2}}+2\times 2x\times y+{{y}^{2}}=196 \\

& 4{{x}^{2}}+4xy+{{y}^{2}}=196 \\

\end{align}$

By rearranging the equation we obtain:

$4{{x}^{2}}+{{y}^{2}}+4xy=196\text{ }...\text{ (4)}$

In the LHS of equation (4) we have $4xy$, but we know that $xy=6$. i.e.

In the next step we have to apply equation (2) in equation (4). Hence, we get the equation:

$4{{x}^{2}}+{{y}^{2}}+4\times 6=196$ i.e. by putting $xy=6\text{ }$

$4{{x}^{2}}+{{y}^{2}}+24=196$

In the next step, take 24 to the right side then 24 becomes -24, i.e. take variables to one side and constants to the other side. When the side changes, the sign also changes. i.e. we get the equation:

$\begin{align}

& 4{{x}^{2}}+{{y}^{2}}=196-24 \\

& 4{{x}^{2}}+{{y}^{2}}=172 \\

\end{align}$

Hence, the value of $4{{x}^{2}}+{{y}^{2}}=172$

Note: Here don’t try to solve the equation (1) by putting $y=\dfrac{6}{x}$, it will become more complicated and also there is a large probability that the answer would be wrong. It is better to solve this by squaring the first equation $2x+y=14$.

Complete step-by-step answer:

Here, we are given with two equations:

$2x+y=14\text{ }.....\text{ (1)}$

$xy=6\text{ }.....\text{ (2)}$

With the help of these two equations we have to find the value of $4{{x}^{2}}+{{y}^{2}}$.

First, let us consider the equation (1).

By squaring equation (1) on both the sides we get,

${{(2x+y)}^{2}}={{14}^{2}}\text{ }....\text{ (3)}$

From equation (3) we can say that its LHS is of the form ${{(a+b)}^{2}}$whose expansion we are familiar with. i.e. the expansion for ${{(a+b)}^{2}}$is given as:

${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

In the LHS of equation (3), has the form ${{(a+b)}^{2}}$ where $a=2x$, and $b=y$.

Now by applying the above formula in equation (3) we get:

$\begin{align}

& {{(2x)}^{2}}+2\times 2x\times y+{{y}^{2}}=196 \\

& 4{{x}^{2}}+4xy+{{y}^{2}}=196 \\

\end{align}$

By rearranging the equation we obtain:

$4{{x}^{2}}+{{y}^{2}}+4xy=196\text{ }...\text{ (4)}$

In the LHS of equation (4) we have $4xy$, but we know that $xy=6$. i.e.

In the next step we have to apply equation (2) in equation (4). Hence, we get the equation:

$4{{x}^{2}}+{{y}^{2}}+4\times 6=196$ i.e. by putting $xy=6\text{ }$

$4{{x}^{2}}+{{y}^{2}}+24=196$

In the next step, take 24 to the right side then 24 becomes -24, i.e. take variables to one side and constants to the other side. When the side changes, the sign also changes. i.e. we get the equation:

$\begin{align}

& 4{{x}^{2}}+{{y}^{2}}=196-24 \\

& 4{{x}^{2}}+{{y}^{2}}=172 \\

\end{align}$

Hence, the value of $4{{x}^{2}}+{{y}^{2}}=172$

Note: Here don’t try to solve the equation (1) by putting $y=\dfrac{6}{x}$, it will become more complicated and also there is a large probability that the answer would be wrong. It is better to solve this by squaring the first equation $2x+y=14$.