Question

Find the number of integer values of m for which the x coordinate of the point of intersection of the line $3x+4y=9$ and $y=mx+1$ is an integer.

Answer 1

Hint: Using the equation $y=mx+1$ , put the value of y in the equation $3x+4y=9$ . Now, solve and get the value of x in terms of m. It is given that the value of m and x should be integers. Since x is an integer so, for integral values of x the possible values of the equation $\left( 3+4m \right)$ can be any of 5, -5, 1 or 1. Now, solve one by one and get the values of m. Ignore those values of m which are not integers.

Complete step-by-step solution -
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According to the question, we have the equation of the two lines intersecting at a point. We have to find the number of integral values of m for which the x coordinate of the point of intersection of these two lines is also an integer.
$3x+4y=9$ ………………….(1)
$y=mx+1$ ……………………(2)
First of all, we have to find the coordinates of the point of intersection of these two lines. To get the x coordinate of the point of intersection, we have to get the value of y from equation (2) and put it in equation (1).
Now, putting the value of y from equation (2) in equation (1), we get
$3x+4\left( mx+1 \right)=9$
On solving the above equation, we get
$3x+4\left( mx+1 \right)=9$

& \Rightarrow 3x+4\left( mx+1 \right)=9 \\\ & \Rightarrow 3x+4mx+4=9 \\\ & \Rightarrow \left( 3+4m \right)x=9-4 \\\ & \Rightarrow \left( 3+4m \right)x=5 \\\ \end{aligned}$$ $$\Rightarrow x=\dfrac{5}{\left( 3+4m \right)}$$ …………………(3) In the question, it is given that the x coordinate of the point of intersection is an integer. In equation (3), we have the value of x. Since it is given that the x coordinate of the point of intersection is an integer. $$x=\dfrac{5}{\left( 3+4m \right)}$$ Since x is an integer so, for integral values of x the possible values of the equation $$\left( 3+4m \right)$$ can be any of 5, -5, 1 or 1. In case $${{1}^{st}}$$ , let us take $$\left( 3+4m \right)$$ equal to 5. $$\begin{aligned} & \left( 3+4m \right)=5 \\\ & \Rightarrow 4m=5-3 \\\ & \Rightarrow 4m=2 \\\ & \Rightarrow m=\dfrac{2}{4}=\dfrac{1}{2} \\\ \end{aligned}$$ Here, we got $$m=\dfrac{1}{2}$$ and $$\dfrac{1}{2}$$ is not an integer. Since it is given that the value of m should be an integer so $$m=\dfrac{1}{2}$$ is not the solution of the equation $$\left( 3+4m \right)$$ . Now, in case $${{2}^{nd}}$$ , let us take $$\left( 3+4m \right)$$ equal to -5. $$\begin{aligned} & \left( 3+4m \right)=-5 \\\ & \Rightarrow 4m=-5-3 \\\ & \Rightarrow 4m=-8 \\\ & \Rightarrow m=\dfrac{-8}{4}=-2 \\\ \end{aligned}$$ Here, we got $$m=-2$$ and -2 is an integer. Since it is given that the value of m should be an integer so $$m=-2$$ is the solution of the equation $$\left( 3+4m \right)$$ . Now, in case $${{3}^{rd}}$$ , let us take $$\left( 3+4m \right)$$ equal to 1. $$\begin{aligned} & \left( 3+4m \right)=1 \\\ & \Rightarrow 4m=1-3 \\\ & \Rightarrow 4m=-2 \\\ & \Rightarrow m=\dfrac{-2}{4}=\dfrac{-1}{2} \\\ \end{aligned}$$ Here, we got $$m=\dfrac{-1}{2}$$ and $$\dfrac{-1}{2}$$ is not an integer. Since it is given that the value of m should be an integer so $$m=\dfrac{-1}{2}$$ is not the solution of the equation $$\left( 3+4m \right)$$ . Now, in case $${{4}^{th}}$$ , let us take $$\left( 3+4m \right)$$ equal to -1. $$\begin{aligned} & \left( 3+4m \right)=-1 \\\ & \Rightarrow 4m=-1-3 \\\ & \Rightarrow 4m=-4 \\\ & \Rightarrow m=\dfrac{-4}{4}=-1 \\\ \end{aligned}$$ Here, we got $$m=-1$$ and -1 is an integer. Since it is given that the value of m should be an integer so $$m=-1$$ is the solution of the equation $$\left( 3+4m \right)$$ . Therefore, the integral values of are -2 and -1. Hence, the number of integer values of m for which the x coordinate of the point of intersection of the line $$3x+4y=9$$ and $$y=mx+1$$ is an integer is 2. Note: In this question, since $$m=\dfrac{1}{2}$$ and $$m=\dfrac{-1}{2}$$ is satisfying the equation $$\left( 3+4m \right)$$ .Therefore, one might include $$m=\dfrac{1}{2}$$ and $$m=\dfrac{-1}{2}$$ as the solution of the equation $$\left( 3+4m \right)$$ . This is contradiction because $$m=\dfrac{1}{2}$$ and $$m=\dfrac{-1}{2}$$ are not integers, and in the question it is given that m should be an integer.