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**Hint:**To find the value of $a$ from $3a+4b=13\text{ and }a+3b=1$ , we will be using the formula

$\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$ for the equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\text{ and }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ . We can compare this cross multiplication formula with the given two equations. When substituting these values, we will be getting an equation of the form $\dfrac{a}{35}=\dfrac{b}{-10}=\dfrac{1}{5}$ . From this, the value of $a$ can be obtained.

**Complete step by step answer:**

We need to find the value of $a$ from $3a+4b=13\text{ and }a+3b=1$ using the cross multiplication method.

Let us first recollect what cross multiplication method is.

Let us consider the equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\text{ and }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$

We can find the values of x and y are found using the cross multiplication formula shown below.

$\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}...(i)$

Now, let us compare this with the given set of equations.

We have $3a+4b=13\text{ and }a+3b=1$

This can be written as $3a+4-13=0\text{ and }a+3b-1=0$

$\begin{align}

& x=a,y=b \\

& {{a}_{1}}=3,{{a}_{2}}=1 \\

& {{b}_{1}}=4,{{b}_{2}}=3 \\

& {{c}_{1}}=-13,{{c}_{2}}=-1 \\

\end{align}$

Now, let us substitute the above values in equation (i). We will get

$\dfrac{a}{4\times -1-3\times \left( -13 \right)}=\dfrac{b}{-13\times 1-\left( -1 \right)\times 3}=\dfrac{1}{3\times 3-4\times 1}$

Let us simplify the denominators. We will get

$\dfrac{a}{-4+39}=\dfrac{b}{-13+3}=\dfrac{1}{9-4}$

$\Rightarrow \dfrac{a}{35}=\dfrac{b}{-10}=\dfrac{1}{5}$

Now let us consider the first and last terms, that is,

$\dfrac{a}{35}=\dfrac{1}{5}$

This can be written as

$a=\dfrac{1\times 35}{5}$

When we solve this, we get

$a=7$

**So, the correct answer is “Option C”.**

**Note:**Be careful when substituting the equations in the standard formula. The given equations must be converted to the standard form ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\text{ and }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ . You must be cautious with the denominators when writing the formula $\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$ .

We can see that the denominator of x does not contain its coefficient. Also the subscripts of the terms of the denominator are different. For example, ${{b}_{1}}{{c}_{2}}\text{ and }{{b}_{2}}{{c}_{1}}$ .

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