Math 1 · Lesson 7 · part 7 of 7
Practice Problems
Try them, get help one step at a time, then print how it went.
Warm-up: distance, then the word
Four short ones before the real solving. Say the word out loud: and, or or.
Practice: try it, then get help one step at a time
Try each problem yourself first. Hint gives a nudge, Show next step shows one worked step at a time, Reveal answer checks, and Hide help / start over clears the card. Every problem asks two things: the word (and, or or) and the solution. Once the absolute value is alone on its side, a Show me the picture button appears: it draws the number line and sets up the two problems for you.
My results
Finished the practice? Print a page that shows each question, what you typed or chose, the correct answer, and how it went. Good for showing a teacher or keeping in a folder.
Exit ticket
On paper or an index card. Use the math word and the plain-English word.
- In your own words: why does |x| = 4 have two answers, and what are they?
- Solve |x - 6| < 2. Say the word, show both problems, and draw the answer.
- Solve |3x + 3| ≥ 6. Say the word, show both problems, and check a number from the gap.
- A friend wrote the answer to |x + 1| > 3 as −4 < x < 2. What went wrong, and what is the right answer?
Check your exit ticket
- Absolute value is distance from zero, and there are two numbers exactly four steps from zero: 4 to the right and −4 to the left. So x = 4 or x = −4.
- Step 1: |x + (−6)| < 2. Less than, so between, the word is and: −2 < x + (−6) and x + (−6) < 2. Add 6 to both sides of both (addition property of inequality): 4 < x and x < 8, so 4 < x < 8. Open circles at 4 and 8, shade between.
- Greater than or equal to, so outside, the word is or: 3x + 3 ≤ −6 or 3x + 3 ≥ 6. Add −3 to both sides of both: 3x ≤ −9 or 3x ≥ 3. Multiply both by 13, positive, no flip: x ≤ −3 or x ≥ 1. Gap check: x = 0 gives |3| = 3 ≥ 6, false, so the gap is correctly left out.
- Greater than means outside, so the word is or, not and, and the answer is never a sandwich. The friend solved the right two problems (x + 1 < −3 gives x < −4, and x + 1 > 3 gives x > 2) but joined them the wrong way. Right answer: x < −4 or x > 2. Test x = 0: |1| = 1 > 3 is false, so 0 is not a solution, but the friend's sandwich says it is.