Given, 4x + 3y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 20 and x = 15

The required points are (0, 20) and (15, 0)

Checking for the origin (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area. The required area would include be on the left of the lines graph.

We have y ≥ 2x,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence the line would pass through origin.

To check which side would be included in the lines graph solution area, we would check for point (15, 0)

⇒ 0 ≥ 15, this is not true so the required solution area would be to the left of the line’s graph.

Consider, x ≥ 3,

For any value of y, the value of x would be same.

Also the origin (0, 0) doesn’t satisfies the inequality as 0 ≥ 3

So the origin doesn’t lies in the solution area, hence the required solution area would be the right of the lines graph.

We have x, y ≥ 0

Since given both x and y are greater than 0 the solution area would be in the first I^{st} quadrant only.

The shaded area in the graph shows the solution area for the given inequalities

Answered by Pragya Singh | 1 year ago

Solve each of the following in equations and represent the solution set on the number line \( \dfrac{5x}{4}-\dfrac{4x-1}{3}>1,\) where x ϵ R.

Solve each of the following in equations and represent the solution set on the number line.\( \dfrac{5x-8}{3}\geq \dfrac{4x-7}{2}\), where x ϵ R.

Solve each of the following in equations and represent the solution set on the number line. 3 – 2x ≥ 4x – 9, where x ϵ R.

Solve each of the following in equations and represent the solution set on the number line. 3x – 4 > x + 6, where x ϵ R.

Solve each of the following in equations and represent the solution set on the number line. 5x + 2 < 17, where

**(i) **x ϵ Z,

**(ii)** x ϵ R.