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Question

Question: How do you graph \(3x-y<0\) ?...

How do you graph 3xy<03x-y<0 ?

Explanation

Solution

We first assume the inequality to be an equation and draw the line accordingly. Then, we need to take any arbitrary point not lying on the line and check if it satisfies the inequality or not and then decide accordingly which side is the required region.

Complete step by step solution:
The given inequality that we have at our disposal is,
3xy<03x-y<0
By the graph of an inequality, we mean a semi-infinite region with a line as its boundary on one side. The general procedure of plotting a graph of an inequality is to at first, not identify it as an inequality. By this statement, we mean to say that, we first need to draw a line by assuming an “equal to” sign in place of the inequality sign. After that, we then take a random arbitrary point on the x-y plane such that it does not line on the line formed by the equation. We substitute this point in the given inequality and see if it satisfies the given inequality or not. If it satisfies, then the side of the line corresponding to the point will be the required graphical region of the inequality. If it does not satisfy, then the other side will be a solution.
We first assume 3xy=03x-y=0 . This will be a line passing through the origin and the point (1,3)\left( 1,3 \right) . Then, let us take another point, say (1,2)\left( 1,2 \right) and see if it lies on the line or not. We can see that it does not lie on the line. We now put this value of x and y in the expression 3xy3x-y to get 3(1)2=13\left( 1 \right)-2=1 . 11 being greater than 00 , does not satisfy the inequality and so the other side of the line will be the required region.

Note: We should choose the point wisely and then judge the respective region correctly as the entire problem depends on the region. There is no other way than this to solve the problem, which forces us to be more cautious in this problem.