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Question

Question: How do you find the slope of \(5x+2y=-20\)?...

How do you find the slope of 5x+2y=205x+2y=-20?

Explanation

Solution

Change of form of the given equation will give the slope, y intercept, and x-intercept of the line 5x+2y=205x+2y=-20. We change it to the form of y=mx+ky=mx+k to find the slope m. Then, we get into the form of xp+yq=1\dfrac{x}{p}+\dfrac{y}{q}=1 to find the x intercept, and y intercept of the line as p and q respectively.

Complete step by step answer:
The given equation 5x+2y=205x+2y=-20 is of the form ax+by=cax+by=c. Here a, b, c are the constants.
We convert the form to y=mx+ky=mx+k. m is the slope of the line.
So, converting the equation we get
5x+2y=20 2y=5x20 y=5x202=5x210 \begin{aligned} & 5x+2y=-20 \\\ & \Rightarrow 2y=-5x-20 \\\ & \Rightarrow y=\dfrac{-5x-20}{2}=\dfrac{-5x}{2}-10 \\\ \end{aligned}
This gives that the slope of the line 5x+2y=205x+2y=-20 is 52-\dfrac{5}{2}.
Now we have to find the y intercept, and x-intercept of the same line 5x+2y=205x+2y=-20.
For this we convert the given equation into the form of xp+yq=1\dfrac{x}{p}+\dfrac{y}{q}=1. From the form we get that the x intercept, and y intercept of the line will be p and q respectively.
The given equation is 5x+2y=205x+2y=-20. Converting into the form of xp+yq=1\dfrac{x}{p}+\dfrac{y}{q}=1, we get
5x+2y=20 5x20+2y20=1 x4+y10=1 \begin{aligned} & 5x+2y=-20 \\\ & \Rightarrow \dfrac{5x}{-20}+\dfrac{2y}{-20}=1 \\\ & \Rightarrow \dfrac{x}{-4}+\dfrac{y}{-10}=1 \\\ \end{aligned}
Therefore, the x intercept, and y intercept of the line 5x+2y=205x+2y=-20 is 44 and 1010 respectively. The intercepting points for the line with the axes are (4,0)\left( -4,0 \right) and (0,10)\left( 0,-10 \right) respectively.

Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance and hence we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to \infty .