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Question: How do you find the slope and intercepts to graph \(2x-3y=3\)?...

How do you find the slope and intercepts to graph 2x3y=32x-3y=3?

Explanation

Solution

In this problem we need to calculate the slope and the intercepts of the given line. We can solve this problem in many methods. We will follow the easiest one for this time. The solution contains two parts, one part is finding the slope and the second part is finding the intercepts. For slope of the given line, we will convert the given equation in slope intercept form which is y=mx+cy=mx+c. So, we will convert the given equation in the form of y=mx+cy=mx+c where mm is our required slope of the line. Now the first part is over. In the second part we will calculate the intercepts of the line. For this also we will convert the given equation into intercept form which is xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1 where aa is the value of xx intercept and bb is the value of yy intercept.

Complete answer:
The equation of the line is 2x3y=32x-3y=3.
Calculating the slope of the given line.
To calculate the slope of the line we will convert the given equation in slope intercept form which is y=mx+cy=mx+c.
Shifting the term 2x2x which is in right side to left side, then we will get
2x3y=3 3y=32x \begin{aligned} & 2x-3y=3 \\\ & \Rightarrow -3y=3-2x \\\ \end{aligned}
Dividing the above equation with 3-3 on both sides of the above equation, then we will get
3y3=32x3 y=33+2x3 y=(23)x1 \begin{aligned} & \Rightarrow \dfrac{-3y}{-3}=\dfrac{3-2x}{-3} \\\ & \Rightarrow y=\dfrac{3}{-3}+\dfrac{2x}{3} \\\ & \Rightarrow y=\left( \dfrac{2}{3} \right)x-1 \\\ \end{aligned}
Comparing the above equation with y=mx+cy=mx+c, then we will get the slope of the equation as m=23m=\dfrac{2}{3}.
Now we will calculate the intercepts of the given equation by converting the equation into intercept form which is xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1.
Dividing the given equation 2x3y=32x-3y=3 with 33 on both sides, then we will get
2x3y3=33 (23)x+(33)y=1 x(32)+y(1)=1 \begin{aligned} & \Rightarrow \dfrac{2x-3y}{3}=\dfrac{3}{3} \\\ & \Rightarrow \left( \dfrac{2}{3} \right)x+\left( \dfrac{-3}{3} \right)y=1 \\\ & \Rightarrow \dfrac{x}{\left( \dfrac{3}{2} \right)}+\dfrac{y}{\left( -1 \right)}=1 \\\ \end{aligned}
From the above equation we can write xx intercept of the given equation as 32\dfrac{3}{2}, yy intercept of the given equation as 1-1.

Note: In this problem we have calculated the slope and intercept by converting the given equation into various forms. We know that the slope of the equation is the variation in the direction of yy for a unit variation in xx. So, we will calculate any two points on the given line and calculate the value of y2y1x2x1\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} and consider it as required slope. To calculate the intercepts we will substitute x=0x=0 and calculate the yy for yy intercept and for xx intercept we will substitute y=0y=0 and calculate the xx value.