Math 1 · Lesson 7 · part 6 of 7
Three worked examples, every line named
One and-example, one or-example, and one where the bars have to be freed first.
Three worked examples, every line named
One and-example, one or-example, and one where the bars have to be freed first. Each ends with its number line.
|x - 3| < 4
The original|x - 3| < 4Read it: the distance of x - 3 from zero is less than 4. The absolute value is already alone on its side, so nothing to peel off.
Step 1 rewrite|x + (−3)| < 4Inside the bars, subtracting 3 is adding −3, its additive inverse (the opposite). Step 1 works inside absolute value bars the same as anywhere else.
Read the signless than, so: betweenLess than 4 means the inside is somewhere between −4 and 4. That is two facts at once, joined by and: the inside is bigger than −4, and the inside is smaller than 4.
Set up two problems−4 < x + (−3) and x + (−3) < 4The absolute value bars are gone. What is left is two Lesson 6 problems with the same inside.
Solve both−4 + 3 < x + (−3) + 3 and x + (−3) + 3 < 4 + 3, so −1 < x and x < 7Add 3, the additive inverse (the opposite) of −3, to both sides of both problems: the addition property of inequality (add the same thing to both sides, and the sign stays the same). (−3) + 3 = 0 and x + 0 = x by the additive identity (the do-nothing number 0).
Answerx > −1 and x < 7, written together as −1 < x < 7Both must be true at once, so the answer is the stretch of number line between −1 and 7. The compact way to write an and-answer puts x in the middle with the small number on the left.
CheckTry x = 3, inside: |3 + (−3)| = |0| = 0, and 0 < 4 is true. Try x = 7, a boundary: |7 + (−3)| = |4| = 4, and 4 < 4 is false, so 7 is not included. Try x = 10, outside: |10 + (−3)| = 7, and 7 < 4 is false. Try x = −5, outside the other way: |−5 + (−3)| = |−8| = 8, and 8 < 4 is false.An and-answer has two outsides, so check a number past each end.
|2x + 1| ≥ 7
The original|2x + 1| ≥ 7Read it: the distance of 2x + 1 from zero is 7 or more. The bar under the sign means the 7 itself counts. Nothing to rewrite, the inside is already all plus and times.
Read the signgreater than, so: outsideA distance of 7 or more from zero means the inside is at or left of −7, or at or right of 7. Only one of those can happen for any single x, so the word is or.
Set up two problems2x + 1 ≤ −7 or 2x + 1 ≥ 7Watch the left problem: the sign faces the other way and the number is negative. The inside is on the far left of the number line, so it is smaller than −7.
Solve both, the plain number2x + 1 + (−1) ≤ −7 + (−1) or 2x + 1 + (−1) ≥ 7 + (−1), so 2x ≤ −8 or 2x ≥ 6Add −1, the additive inverse (the opposite) of 1, to both sides of both: the addition property of inequality (add the same thing to both sides, and the sign stays the same).
Solve both, the reciprocal12 × 2x ≤ 12 × (−8) or 12 × 2x ≥ 12 × 6, so x ≤ −4 or x ≥ 3Multiply both sides of both by 12, the multiplicative inverse (the reciprocal) of the coefficient (the number on the variable) 2. It is positive, so neither sign flips: the multiplication property of inequality for a positive number.
Answerx ≤ −4 or x ≥ 3Two separate pieces of the number line, pointing away from each other. An or-answer is never written as one sandwich: −4 ≥ x ≥ 3 would be nonsense, because nothing is both below −4 and above 3.
CheckTry x = −5: |−10 + 1| = |−9| = 9, and 9 ≥ 7 is true. Try x = 3, a boundary: |6 + 1| = 7, and 7 ≥ 7 is true, so 3 is included. Try x = 0, in the gap: |0 + 1| = 1, and 1 ≥ 7 is false.An or-answer has a gap in the middle, so check a number in the gap as well as one on each piece.
3|x + 2| - 5 < 10
The original3|x + 2| - 5 < 10The absolute value is not alone: it is multiplied by 3 and then 5 is subtracted. Peel those off first, treating |x + 2| like a single block.
Step 1 rewrite3|x + 2| + (−5) < 10Subtracting 5 is adding −5, its additive inverse (the opposite).
Get the bars alone3|x + 2| + (−5) + 5 < 10 + 5, so 3|x + 2| < 15, then 13 × 3|x + 2| < 13 × 15, so |x + 2| < 5Add 5 to both sides (addition property of inequality), then multiply both sides by 13, the multiplicative inverse (the reciprocal) of 3 (multiplication property of inequality, positive, no flip). Now the absolute value is alone. Never split into two problems before this point.
Read the signless than, so: between, joined by andThe inside, x + 2, is somewhere between −5 and 5.
Set up two problems−5 < x + 2 and x + 2 < 5Two Lesson 6 problems with the same inside.
Solve both−5 + (−2) < x + 2 + (−2) and x + 2 + (−2) < 5 + (−2), so −7 < x and x < 3Add −2, the additive inverse (the opposite) of 2, to both sides of both: the addition property of inequality (add the same thing to both sides, and the sign stays the same).
Answer−7 < x < 3Both at once, so one stretch of number line, written as a sandwich with x in the middle.
CheckTry x = 0, inside: 3|0 + 2| + (−5) = 6 + (−5) = 1, and 1 < 10 is true. Try x = 3, a boundary: 3|5| + (−5) = 10, and 10 < 10 is false, so 3 is not included. Try x = −9, outside: 3|−7| + (−5) = 21 + (−5) = 16, and 16 < 10 is false.Check in the original, bars and all, not in the two problems you made.