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To solve the question we will develop a quadratic equation with the help of a linear equation and another relation between x and y. Before starting this question we must first know what is a linear and quadratic equation .A linear equation in two variables is a type of equation which contains two independent variables x and y. A quadratic equation is a form of equation which has maximum power of x=2. Now, we consider that the number at unit place be x and the number at tens place be y. It is also given that y>x. It is also given that the difference between y and x is 3. Thus, we get the following equation:

\[\text{y - x = 3}~\ldots \ldots \ldots \ldots \ldots ..\left( \text{i} \right)\]

Another information given in the question is that the product of x and y is 40. Thus, we will get the following equation:

\[\begin{align}

& \Rightarrow \text{x}\times \text{y = 40}\, \\

& \Rightarrow \text{xy = 40}\ldots \ldots \ldots \ldots \ldots ..\left( \text{ii} \right) \\

\end{align}\]

Now, we will substitute the value of y from equation (i) to equation (ii). From equation (i) we will have

\[\text{x = y}\,+\,\text{3}\,~\ldots \ldots \ldots \ldots \ldots ..\left( \text{iii} \right)\]

We will put this value of y from equation (iii) into equation (ii). After doing this, we will get\[\begin{align}

& \Rightarrow \text{x}\left( x+3 \right)\text{ = 40} \\

& \Rightarrow \,{{\text{x}}^{2}}\text{+}\,\text{3x = 40} \\

& \Rightarrow {{\text{x}}^{2}}\text{+}\,\text{3x-40}\,\,\text{=}\,\text{0}\ldots \ldots \ldots \ldots \ldots ..\left( \text{iv} \right) \\

\end{align}\]

Now we will apply the quadratic formula to calculate the value of x. The quadratic formula for calculating the roots of equation \[\text{a}{{\text{x}}^{2}}+bx\,+\,c\,=0\] is,

\[\begin{align}

& \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\

& \Rightarrow x\,=\,\,\,\,\dfrac{-3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( -40 \right)}}{2\left( 1 \right)} \\

& \Rightarrow \,x\,=\,\,\,\,\dfrac{-3\pm \sqrt{169}}{2} \\

& \Rightarrow \,x\,=\,\,\,\,\dfrac{-3\pm 13}{2} \\

& \Rightarrow \,x\,=\,5\,,\,-8 \\

\end{align}\]

Here, we will consider the positive value of x because the digit of the number cannot be negative. Therefore x=5. Now, the value of y becomes

\[\begin{align}

& \Rightarrow y\,=\,\,x\,+\,3 \\

& \Rightarrow \,y\,=\,\,5\,+\,3 \\

& \Rightarrow y\,=\,\,8 \\

\end{align}\]

Thus, the two digit number =85