Given 2x + y ≥ 4,

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

y = 4 and x =2

The required points are (0, 4) and (2, 0)

Checking for origin (0, 0)

0 ≥ 4, this is not true

Hence the origin doesn’t lies in the solution area of the lines graph. The solution area would be given by the right side of the lines graph.

x + y ≤ 3,

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

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true

Hence the solution area would include the origin and hence would be on the left side of the lines graph.

2x – 3y ≤ 6

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

y = -2 and x = 3

The required points are (0, -2), (3, 0)

Checking for the origin (0, 0)

0 ≤ 6 this is true

So the origin lies in the solution area and the area would be on the left of the lines graph.

Hence the shaded area in the graph is the required 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.