The given inequalities are
2x + y < 1
y ≥ x/2 + 2
The first step is to write both inequalities as equations. We have
2x + y = 1
y = x/2 + 2
The next step is to plot the straight line graphs for each equation. Since we are given options, we would find the x and y intercepts of each equation and select the graph that matches them.
Recall, x intercept is the value of x when y is zero
y intercept is the value of y when x is zero
For the first equation,
when x = 0, we have
2(0) + y = 1
0 + y = 1
y = 0. This means that y intercept = 0
when y = 0,
2x + 0 = 1
2x = 1
x = 1/2 = 0.5
This means that x intercept = 0.5
The line representing this equation should pass through the y axis at y = 0 and the x axis at x = 0.5
Also, the line would be dashed(not solid) because the solutions to the inequality do not include the values on the line.
For the second equation,
when x = 0, y = 0/2 + 2 = 0 + 2 = 2
when y = 0, we have
0 = x/2 + 2
x/2 = 0 - 2 = - 2
x = - 2 * 2 = - 4
The line representing this equation should pass through the y axis at y = 2 and the x axis at x = - 4
Also, the line would be a solid line because the solutions to the inequality include the values on the line.
The next step is to determine the shaded region for each inequality. Where the shaded regions overlap is the solution. The shaded regions in each option indicate the solution. We would pick points which we would test in each of the shaded regions
For the first option, let us pick a point in the shaded region. Let us pick x = - 2 and y = 2
We would plug these values in both inequalities and see if they satisfy it.
For the first,
2x + y < 1
y < 1 - 2x
Plugging in x = - 2 and y = 2, we have
2< 1 - 2(-2)
2 < 1 + 4
2 < 5
This is true
For the second,
y≥ x/2 + 2
Plugging in x = - 2 and y = 2, we have
2 ≥ -2/2 + 2
2 ≥ - 1 + 2
2 ≥ 1
This is also true
Thus, the first option is correct