Answer
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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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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