Solving 'AND' Inequalities
Solve AND compound inequalities in Grade 9 algebra by finding values satisfying both conditions simultaneously, writing solutions as intersections and graphing on a number line.
Key Concepts
Property An 'AND' compound inequality, often written in a compact form like a < x < b, is true only if a value for x satisfies both conditions at the same time.
Explanation Think of your variable as being trapped between two walls! To solve for it, you have to do the same thing to all three parts of the inequality—the left side, the middle, and the right side. It’s a package deal; whatever inverse operation you perform, apply it everywhere to isolate the variable.
Examples To solve $ 5 \le 2x + 1 \le 9$, subtract 1 from all three parts to get $ 6 \le 2x \le 8$. Then, divide all parts by 2 to find that $ 3 \le x \le 4$. To solve $10 < 2(x + 3) < 24$, first distribute to get $10 < 2x + 6 < 24$. Then subtract 6 from all parts: $4 < 2x < 18$. Finally, divide by 2: $2 < x < 9$.
Common Questions
What is an AND inequality and how do you solve it?
An AND inequality requires a variable to satisfy two conditions simultaneously, such as x > 2 AND x < 8. The solution is the overlap (intersection) of both solution sets, written as 2 < x < 8.
How do you graph the solution to an AND inequality?
Graph each inequality on a number line separately, then shade only the region where both are true. For 2 < x < 8, shade the segment between 2 and 8 with open circles at both endpoints if the inequalities are strict.
What is the difference between AND and OR compound inequalities?
AND inequalities require both conditions to be true (intersection of solution sets), giving a bounded interval. OR inequalities require at least one condition to be true (union), often producing two separate rays on the number line.