We've already been introduced to inequalities which are expressions that explain a relationship between two quantities that aren't equal. We can also solve inequalities and graph these solutions on a number plane.
Let's look at this process using an example: $y\ge2x+4$y≥2x+4
1. Graph the line as if it was an equation. If your inequality is "$\ge$≥" or "$\le$≤", use a solid line. If your inequality is "$>$>" or "$<$<", use a dashed line. Since our inequality is $y\ge2x+4$y≥2x+4, we're going to use a solid line to draw the line $y=2x+4$y=2x+4.
2. Work out which side of the line you should shade by seeing whether a point on the number plane (that doesn't lie on the line you've drawn) satisfies the inequality. To do this, you need to see whether the substituted $x$x and $y$y values satisfy the inequality. I've picked one point on either side of the line (note: not on the line): $\left(0,0\right)$(0,0) marked in blue and $\left(-5,5\right)$(−5,5) marked in green.
Let's test he origin $\left(0,0\right)$(0,0) first:
The origin does not satisfy our equation, so we will not shade this side of the line.
Let's check another point above the line, say $\left(-5,5\right)$(−5,5).
Now that's looking good! Let's shade this side on our graph.
And that's how we do it.
Is $\left(3,2\right)$(3,2) a solution of $3x+2y$3x+2y $\ge$≥ $12$12?
Write the inequality that describes the points in the shaded region.
Consider the line $y=-2x+2$y=−2x+2.
Find the intercepts of the line.
Which of the following points satisfies the inequality $y$y $\le$≤ $-2x+2$−2x+2?
Sketch a graph of $y$y$\le$≤$-2x+2$−2x+2.
Do the points on the line satisfy the inequality $y$y $\le$≤ $-2x+2$−2x+2?