Big Ideas Math: Modeling Real Life, Grade 6
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8. Solving Inequalities
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Exercise 58 Page 398

What can we do to isolate a variable in an inequality? Graph both inequalities.

Practice makes perfect
We want to graph the numbers that are solutions of both given inequalities. p+3/4 < 3 1/4 p > 3/8 We can do this by solving both inequalities and graphing them on a number line. Inequalities can be solved in the same way as equations — by performing inverse operations on both sides until the variable is isolated. The only difference is that when we divide or multiply by a negative number, we must reverse the inequality sign. Let's start with the first inequality!
p+3/4 < 3
p + 3/4 - 3/4 < 3 - 3/4
p < 3 - 3/4
p < 4 * 3/4 - 3/4
p < 12/4-3/4
p < 12-3/4
p < 9/4
We can also write 94 as a mixed number. Let's do it!
9/4
8+1/4
8/4 + 1/4
2 + 1/4
2 14
We found that all values of p less than 2 14 will satisfy the first inequality. Now let’s graph the inequality on a number line. Since p cannot equal 2 14, we draw an open circle at this point. We know that p is all values less than 2 14, so we will shade the part of the number line that represents numbers less than 2 14. This means that we shade to the left of our point at 2 14.
Now we can solve the second inequality and graph its solution. Let's do it!
1/4 p > 3/8
4 * 1/4 p > 4 * 3/8
p > 4 * 3/8
p > 4 * 3/4 * 2
p > 4 * 3/4 * 2
p > 1.5
p > 1 12
We found that all values of p greater than 1 12 will satisfy the inequality. Now let’s graph the inequality on a number line. Since p cannot 1 12, we draw an open circle at this point. We know that p is all values greater than 1 12, so we will shade the part of the number line that represents numbers greater than 1 12. This means that we shade to the right of our point at 1 12.

We have solved both inequalities. Let's graph both solutions on the same number line. Any number in the region that is shaded by both inequalities is a solution to both inequalities. Let's see the graph!

As we can see, the numbers greater 1 12 and less than 2 14 are included in the solution set of both inequalities. Let's shade just this region to show the numbers that solve both inequalities. Remember to use open circles at both 1 12 and 2 14! This means that we have to shade starting at 1 12 with an open circle and ending at 2 14 with an open circle too.