Question

# A two digit number contains the smaller of the two digits in the unit place. The product of the digits is 40 and the difference between the digits is 3. Find the numbers.

Hint: To solve the question, we will first derive a linear equation. For this, we will assume that the unit place digit is x and tens place digit is y. Now the difference between them is 3. So, from here we will get the deserved linear equation. Now, we will write y in terms of x. Another equation can be obtained by multiplying x and y and equating to 40. We will substitute the value of y in this equation. On doing this, we will get a quadratic equation, which we will solve to get the value of x and substitute in y to get the number.

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

Note: We can also calculate the value of x by plotting the graphs of line y=x+3 and the hyperbola xy=40. The value of x will be the intersection point of these graphs.