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Week 2 · Chapter

Chapter 2 — Linear Equations & Inequalities

College Algebra Free sample edition

The module's primary reading · ~30–40 minutes · Objective 2 (solve linear equations and linear inequalities in one variable, including absolute-value equations and inequalities, and express solution sets in interval notation) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you learn to find the number that makes a statement true. Some statements have exactly one such number, and you find it by solving an equation (undoing what was done to the unknown, one step at a time, until it stands alone). Other statements are true for a whole range of numbers, like "spend at most 45 dollars," and you describe that range with an inequality (a statement that one side is smaller or larger than the other). You will also meet absolute value (how far a number sits from zero, whichever direction it lies), which turns "within 2 miles" into algebra. Every answer you find, you check by putting it back in.

Why you already care about this

Say your phone plan costs 20 dollars a month plus 5 dollars for every gigabyte you use, and your budget is 45 dollars. How many gigabytes can you use? You are not hunting for one exact number. You are hunting for every amount that keeps the bill at or under 45 dollars. That question is an inequality, and by the end of Section 3 you will answer it in three lines.

Now say you are looking for an apartment within 2 miles of campus, and campus sits at mile marker 10 on a long straight road. Which mile markers work? Or say a greenhouse alarm should sound when the temperature drifts more than 4 degrees Celsius away from 20. When does it ring? Both are questions about distance, and distance is what absolute value measures.

You already solve problems like these in your head. Algebra writes the steps down so you can trust them, repeat them, and check them.

By the end of this chapter you will be able to:

  • solve a linear equation by distributing, clearing fractions, collecting, and dividing, then check it by substitution;
  • tell an identity from a contradiction;
  • solve a linear inequality and flip the sign when you divide by a negative;
  • write a solution set in interval notation;
  • solve an absolute-value equation with the split rule;
  • solve an absolute-value inequality as a middle interval or as two rays.

Carry this sentence with you: less than is the middle — greater than is the two rays.

Where we are

  • Already covered: Week 1, Real Numbers, Exponents & Algebraic Expressions, gave you the order of operations, the properties of real numbers, and the habit of distributing and combining like terms.
  • This week: solving linear equations, identities and contradictions, linear inequalities and interval notation, and absolute-value equations and inequalities.
  • How they connect: every solve starts by simplifying each side, and simplifying is exactly what Week 1 taught.
  • Where we're going next: Week 3, Functions: Notation, Domain & Range, uses these solving moves to find which inputs a function allows.

Week 1 is the week this chapter leans on hardest. Recall the distributive property: a number in front of parentheses multiplies every term inside, so 3(x − 2) is 3x − 6. Recall too that Week 1 distributed a negative across a sum, as in 5x − 2(3x − 4) = −x + 8, where the −2 reaches both terms. You will use that same move in nearly every equation below.

1. Solve a linear equation: one principle, four tactics

Picture a two-pan kitchen scale with a bag of flour on each side, perfectly level. Add a cup to the left pan only, and the scale tips. Add a cup to both pans, and it stays level. Solving an equation works the same way. You may do almost anything to an equation, as long as you do the same thing to both sides.

A linear equation is an equation in one variable that can be rewritten in the form ax + b = 0 with a not equal to zero. In plain words, the variable appears only to the first power and is never inside a denominator, a root, or an absolute value. A solution is a value of the variable that makes a statement true; for an equation, it is the number that makes the two sides equal.

Here is the one principle and the four tactics, in the order you usually need them:

  1. Distribute to remove parentheses.
  2. Clear fractions by multiplying every term by the least common denominator.
  3. Collect the x terms on one side and the plain numbers on the other.
  4. Divide by the number in front of x.

Then you finish with a check by substitution: you replace the variable in the original equation with your answer and evaluate each side to confirm the statement is true. An answer without its check is unfinished.

Worked example — distribute, collect, divide. Solve 3(x − 2) = 2x + 5.

  1. Distribute. The 3 multiplies both terms inside. 3 times x is 3x, and 3 times −2 is −6. So 3(x − 2) = 2x + 5 becomes 3x − 6 = 2x + 5.
  2. Collect the x terms. Subtract 2x from both sides. The left side becomes 3x − 2x − 6, which is x − 6. The right side becomes 5. So x − 6 = 5.
  3. Isolate x. Add 6 to both sides. Then x = 5 + 6, so x = 11.
  4. Check in the original. The left side is 3(11 − 2) = 3 · 9 = 27. The right side is 2 · 11 + 5 = 22 + 5 = 27. Both sides are 27.

In words: the only number that balances this equation is 11, and the check proves it.

Fractions slow people down, so the second tactic removes them in one step. The least common denominator (the LCD) is the smallest number that every denominator in an equation divides into evenly; multiplying every term by it clears the fractions in one step.

Worked example — clear the fractions first. Solve x ⁄ 2 + 1 ⁄ 3 = 5 ⁄ 6.

  1. Find the LCD. The denominators are 2, 3, and 6. The smallest number all three divide into is 6, so the LCD is 6.
  2. Multiply every term by 6. 6 · (x ⁄ 2) = 3x. 6 · (1 ⁄ 3) = 2. 6 · (5 ⁄ 6) = 5. So the equation becomes 3x + 2 = 5, with no fractions left.
  3. Collect. Subtract 2 from both sides to get 3x = 3.
  4. Divide. Divide both sides by 3 to get x = 1.
  5. Check in the original. 1 ⁄ 2 + 1 ⁄ 3 = 3 ⁄ 6 + 2 ⁄ 6 = 5 ⁄ 6. That matches the right side.

So x = 1. Notice that the check uses the original fractions, not the cleaned-up equation. A check on a later line can hide an earlier slip.

Misconception check — the 3 that only reaches one term. Students often think 3(x − 2) is 3x − 2. It is tempting because the 3 sits right next to the x. But the distributive property says the 3 multiplies every term inside the parentheses, so 3(x − 2) is 3x − 6. The same slip shows up with fractions: multiplying only the left side by the LCD and leaving 5 ⁄ 6 alone. The cure is one habit. Draw an arrow from the outside number to every term it must reach, and count the arrows.

In practice (your tools). In Desmos, type y = 3(x - 2) on one line and y = 2x + 5 on the next. Click the point where the two lines cross. Desmos labels it (11, 27): the x-value 11 is your solution, and 27 is the value both sides share, the same 27 your check produced.

2. Spot an identity or a contradiction

Most equations you meet have exactly one solution. Two kinds do not, and both announce themselves the same way: the x terms vanish. Think of two friends describing the same lunch order in different words. Whatever the price turns out to be, the two descriptions always agree. Now think of two receipts for the same order that list different totals. No price can make them agree.

An identity is an equation that is true for every real number; solving it collapses to a true statement such as 0 = 0, so its solution set is all real numbers. A contradiction is an equation that is true for no value of the variable; solving it collapses to a false statement such as 3 = 7, so it has no solution.

Worked example — an identity. Solve 2(x + 3) = 2x + 6.

  1. Distribute. 2 times x is 2x, and 2 times 3 is 6. So the left side is 2x + 6, and the equation reads 2x + 6 = 2x + 6.
  2. Collect. Subtract 2x + 6 from both sides. You get 0 = 0.
  3. Read the result. The variable is gone, and 0 = 0 is always true. Every real number works.

Try any value to see it. With x = 5, the left side is 2(5 + 3) = 2 · 8 = 16, and the right side is 2 · 5 + 6 = 10 + 6 = 16. The solution set is all real numbers.

Worked example — a contradiction. Solve 2(x + 3) − 3 = 2x + 7.

  1. Distribute. The left side becomes 2x + 6 − 3, which is 2x + 3. The equation reads 2x + 3 = 2x + 7.
  2. Collect. Subtract 2x from both sides. You get 3 = 7.
  3. Read the result. The variable is gone, and 3 = 7 is never true. No number works, so there is no solution.

Misconception check — writing x = 0. Students often write "x = 0" when the x terms disappear. It is tempting because a zero appeared on the page. But x = 0 claims that exactly one number, zero, works. For the identity, every number works, so you write "all real numbers." For the contradiction, nothing works, so you write "no solution." The cure: when x vanishes, stop solving and read the leftover statement. True means all real numbers; false means no solution.

Honesty note: some courses write the empty answer with a special symbol for the empty set. This course writes the words "no solution," and either is correct as long as you never write x = 0 for it.

3. Solve a linear inequality and write it in interval notation

Back to the phone plan. It costs 20 dollars plus 5 dollars per gigabyte, and you can spend at most 45 dollars. "At most" does not mean "exactly." Spending 30 dollars is fine. Spending 45 is fine. Spending 46 is not. So you need a tool that captures a whole range of answers at once.

A linear inequality is a statement that compares two linear expressions with <, >, ≤, or ≥; its solution is usually a whole range of numbers rather than a single value. The symbol < means "is less than," and ≤ means "is less than or equal to." Likewise > means "is greater than," and ≥ means "is greater than or equal to."

Worked example — the budget ceiling. Let g be the number of gigabytes you use in a month. The bill is 20 + 5g dollars, and it must be at most 45.

  1. Write the inequality. 20 + 5g ≤ 45.
  2. Subtract 20 from both sides. 5g ≤ 25.
  3. Divide both sides by 5. g ≤ 5.

You can use up to 5 gigabytes. Check the edge: with g = 5, the bill is 20 + 5 · 5 = 20 + 25 = 45 dollars, which is allowed. With g = 6, the bill is 20 + 30 = 50 dollars, which breaks the budget.

So far an inequality behaves exactly like an equation. There is one new rule. Start with a true statement: 2 < 5. Multiply both sides by −1, and you get −2 and −5. On a number line, −2 sits to the right of −5, so −2 > −5. The order reversed. That is the flip rule: when both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign reverses, so < becomes > and ≤ becomes ≥.

Worked example — dividing by a negative. Solve −2x + 1 < 9.

  1. Subtract 1 from both sides. −2x < 8.
  2. Divide both sides by −2, and flip the sign. 8 divided by −2 is −4, and < becomes >. So x > −4.
  3. Check a number inside. Try x = 0. Then −2 · 0 + 1 = 1, and 1 < 9 is true.
  4. Check the endpoint. Try x = −4. Then −2 · (−4) + 1 = 8 + 1 = 9, and 9 < 9 is false. So −4 itself is not included.

Every number greater than −4 makes the inequality true.

Misconception check — forgetting to flip. Students often solve −3x > 9 by dividing by −3 and writing x > −3. It is tempting because the steps look identical to an equation. But you divided by a negative, so the sign must flip: x < −3. Test it. With x = −4, you get −3 · (−4) = 12, and 12 > 9 is true, so −4 should be in the answer. It is less than −3, not greater. The cure: circle the number you divide by. If it is negative, flip before you write anything else.

Now you need a compact way to write "every number greater than −4." Interval notation (a shorthand for a range of numbers) writes a set of real numbers by its two endpoints. It uses a square bracket for an endpoint that is included and a parenthesis for an endpoint that is excluded or infinite, as in [2, 5) for 2 ≤ x < 5. Infinity is the symbol ∞ (or −∞), which marks an interval that keeps going without end to the right (or the left); it is not a number, so it is always written with a parenthesis.

The table below translates this chapter's answers from inequality form into interval form.

This chapter's solution sets in inequality form and in interval notation, with whether each endpoint is included
Inequality Interval notation Endpoints
x > −4 (−4, ∞) −4 excluded
g ≤ 5 (−∞, 5] 5 included
−5 < x < 5 (−5, 5) both excluded
8 ≤ m ≤ 12 [8, 12] both included

So the answer to −2x + 1 < 9 is written (−4, ∞). The parenthesis beside −4 says −4 is left out, which matches the endpoint check. The parenthesis beside ∞ is required every time.

Misconception check — a bracket on infinity. Students sometimes write (−4, ∞]. It is tempting to "close" the interval on both ends. But a bracket means "this endpoint is included," and infinity is not a number you can reach or include. The cure: infinity always gets a parenthesis, with no exceptions.

Caveat: in a scenario, the algebra can hand you numbers that make no sense. The inequality g ≤ 5 includes −3, but nobody uses −3 gigabytes. In context, the useful answers run from 0 up to 5. Always read your interval back into the story.

In practice (your tools). In Desmos, type y = -2x + 1 and y = 9. The slanted line sits below the flat line y = 9 for every x to the right of −4. Hover over the crossing point to see (−4, 9): that crossing is the excluded endpoint.

4. Solve an absolute-value equation with the split rule

Stand at a bus stop and walk 7 steps. You might end up 7 steps east or 7 steps west. Either way, you are 7 steps away. The direction is lost; only the distance is kept. That is exactly what absolute value does.

Absolute value is the distance of a number from 0 on the number line, written |x|; because a distance is never negative, |7| = 7 and |−7| = 7. So if |X| = 7, the expression X could be 7 or −7. That gives the split rule: for a positive number a, the equation |X| = a is solved by splitting it into the two equations X = a or X = −a; if a is negative there is no solution.

Worked example — the split rule. Solve |2x − 1| = 7.

  1. Split. Either 2x − 1 = 7 or 2x − 1 = −7.
  2. Solve the first. Add 1 to both sides: 2x = 8. Divide by 2: x = 4.
  3. Solve the second. Add 1 to both sides: 2x = −6. Divide by 2: x = −3.
  4. Check both. For x = 4: |2 · 4 − 1| = |7| = 7. For x = −3: |2 · (−3) − 1| = |−7| = 7. Both check.

The equation has two solutions, x = 4 or x = −3. Each one puts 2x − 1 exactly 7 units from zero, one on each side.

Worked example — the no-solution case. Solve |x| = −5. The left side is a distance, and a distance is never negative. No number sits −5 units from zero. So |x| = −5 has no solution, and there is nothing to split.

Misconception check — splitting a negative. Students often "solve" |x| = −5 as x = 5 or x = −5. It is tempting because the split rule has become a reflex. But check it: |5| = 5, not −5, and |−5| = 5, not −5. Neither candidate works. The cure: before you split, look at the right side. If it is negative, write "no solution" and stop.

In practice (your tools). In Desmos, type y = abs(2x - 1) and y = 7. The V-shaped graph crosses the flat line twice, at (−3, 7) and at (4, 7). Those two crossings are your two solutions. Now change the second line to y = -5. The V never dips below zero, so the graphs never meet, which is the picture of "no solution."

5. Solve an absolute-value inequality: the middle or the two rays

Return to the apartment search. Campus sits at mile marker 10, and you want a place within 2 miles. Walk 2 miles one way and you reach mile 8. Walk 2 miles the other way and you reach mile 12. Every marker between those two works. "Within" a distance always describes one stretch in the middle.

Now flip the question. The greenhouse alarm sounds when the temperature is more than 4 degrees away from 20. That happens when the room is too cold or too hot. "More than" a distance always describes two pieces, one off each end.

An absolute-value inequality is an inequality with an absolute value on one side; for a positive a, |X| < a is the bounded middle interval −a < X < a, and |X| > a is the two outward rays X < −a or X > a. The middle needs both conditions at once, so it is an AND. The two rays need only one condition or the other, so they form an OR. The symbol ∪ is a union: it joins two sets into one that contains every number in either, read as or.

Worked example — less than, the middle. Solve |x| < 5.

  1. Read it as distance. The numbers within 5 units of zero.
  2. Write the middle. −5 < x < 5.
  3. Write the interval. (−5, 5). Both endpoints are excluded, because the sign is strict.
  4. Check. Try x = 0: |0| = 0, and 0 < 5 is true.

Worked example — greater than or equal, the two rays. Solve |x| ≥ 3.

  1. Read it as distance. The numbers at least 3 units from zero.
  2. Write the two rays. x ≤ −3 or x ≥ 3.
  3. Write the interval. (−∞, −3] ∪ [3, ∞). The brackets include −3 and 3, because the sign is ≥.
  4. Check one inside. Try x = 4: |4| = 4, and 4 ≥ 3 is true.
  5. Check one outside. Try x = 0: |0| = 0, and 0 ≥ 3 is false, so 0 is not in the set.

Worked example — within 2 miles of campus. Let m be a mile marker. Being within 2 miles of marker 10 means |m − 10| ≤ 2.

  1. Write the middle. −2 ≤ m − 10 ≤ 2.
  2. Add 10 to all three parts. 8 ≤ m ≤ 12.
  3. Write the interval. [8, 12]. Markers 8 and 12 are exactly 2 miles away, so they count.

Worked example — the temperature alarm. Let T be the temperature in degrees Celsius. The alarm rule is |T − 20| > 4.

  1. Write the two rays. T − 20 < −4 or T − 20 > 4.
  2. Add 20 to each piece. T < 16 or T > 24.
  3. Write the interval. (−∞, 16) ∪ (24, ∞). At exactly 16 or 24 the alarm stays quiet, because the rule says "more than."

Misconception check — squeezing the two rays into a middle. Students often write |x| > 3 as −3 < x < 3. It is tempting because the middle form is the one they practiced first. But test x = 0: |0| = 0, and 0 > 3 is false, yet 0 sits inside −3 < x < 3. The middle form describes the numbers close to zero, and "greater than" asks for the numbers far from zero. The cure: say the inequality as a distance out loud. "Less than" is close, so it is the middle. "Greater than" is far, so it is the two rays.

In practice (your tools). In Desmos, type y = abs(x) and y = 5. The V sits below the flat line only between x = −5 and x = 5, which pictures (−5, 5). Change the flat line to y = 3. The V sits on or above it to the left of −3 and to the right of 3, which pictures the two rays.

Pause and think, choosing one of these two options:

  • (a) Describe a situation from your own life that is a "within" range, and say why it is an AND.
  • (b) Describe a situation from your own life that sets off an alarm when you are "too far" either way, and say why it is an OR.

Key terms

  • linear equation — An equation in one variable that can be rewritten in the form ax + b = 0 with a not equal to zero, so the variable appears only to the first power and is never inside a denominator, a root, or an absolute value. (first used: 1. Solve a linear equation: one principle, four tactics)
  • solution — A value of the variable that makes a statement true; for an equation, the number that makes the two sides equal, confirmed by substituting it into the original equation. (first used: 1. Solve a linear equation: one principle, four tactics)
  • check by substitution — Replacing the variable in the original equation or inequality with a proposed answer and evaluating each side to confirm the statement is true. (first used: 1. Solve a linear equation: one principle, four tactics)
  • least common denominator — The LCD: the smallest number that every denominator in an equation divides into evenly; multiplying every term by it clears the fractions in one step. (first used: 1. Solve a linear equation: one principle, four tactics)
  • identity — An equation that is true for every real number; solving it collapses to a true statement such as 0 = 0, so its solution set is all real numbers. (first used: 2. Spot an identity or a contradiction)
  • contradiction — An equation that is true for no value of the variable; solving it collapses to a false statement such as 3 = 7, so it has no solution. (first used: 2. Spot an identity or a contradiction)
  • linear inequality — A statement that compares two linear expressions with <, >, ≤, or ≥; its solution is usually a whole range of numbers rather than a single value. (first used: 3. Solve a linear inequality and write it in interval notation)
  • flip rule — When both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign reverses, so < becomes > and ≤ becomes ≥. (first used: 3. Solve a linear inequality and write it in interval notation)
  • interval notation — A way to write a set of real numbers by its two endpoints, using a square bracket for an endpoint that is included and a parenthesis for an endpoint that is excluded or infinite, as in [2, 5) for 2 ≤ x < 5. (first used: 3. Solve a linear inequality and write it in interval notation)
  • infinity — The symbol ∞ (or −∞) marks an interval that keeps going without end to the right (or the left); it is not a number, so it is always written with a parenthesis. (first used: 3. Solve a linear inequality and write it in interval notation)
  • absolute value — The distance of a number from 0 on the number line, written |x|; because a distance is never negative, |7| = 7 and |−7| = 7. (first used: 4. Solve an absolute-value equation with the split rule)
  • split rule — For a positive number a, the equation |X| = a is solved by splitting it into the two equations X = a or X = −a; if a is negative there is no solution. (first used: 4. Solve an absolute-value equation with the split rule)
  • absolute-value inequality — An inequality with an absolute value on one side; for a positive a, |X| < a is the bounded middle interval −a < X < a, and |X| > a is the two outward rays X < −a or X > a. (first used: 5. Solve an absolute-value inequality: the middle or the two rays)
  • union — The symbol ∪ joins two sets into one that contains every number in either, read as or; it writes two separate rays as a single solution set. (first used: 5. Solve an absolute-value inequality: the middle or the two rays)

Summary

  • Solving an equation means doing the same thing to both sides until the variable stands alone.
  • The four tactics are distribute, clear fractions with the LCD, collect, and divide.
  • Every answer gets a check by substitution in the original equation.
  • If x vanishes and the leftover is true, write all real numbers; if it is false, write no solution.
  • Solve an inequality like an equation, but flip the sign when you multiply or divide by a negative.
  • In interval notation, a bracket includes an endpoint, a parenthesis excludes it, and infinity always gets a parenthesis.
  • Absolute value is distance from zero, so an absolute-value equation splits into two cases, unless the right side is negative.
  • Less than a distance is the middle, an AND; greater than a distance is the two rays, an OR.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. Solve 4(x + 1) = x + 13, and check your answer.
  2. Solve x ⁄ 3 + 1 ⁄ 2 = 7 ⁄ 6 by clearing the fractions first, and check your answer.
  3. Solve 5 − 3x ≥ 20 and write the answer in interval notation.
  4. A basketball coach wants practice to last within 15 minutes of 90 minutes. Write an absolute-value inequality for the practice length t in minutes, solve it, and write the interval.
  5. A classmate writes the solution of |x| > 6 as −6 < x < 6. Explain what went wrong and write the right solution set.

Answers (worked reasoning — read after attempting)

  1. Distribute the 4: 4x + 4 = x + 13. Subtract x from both sides: 3x + 4 = 13. Subtract 4: 3x = 9. Divide by 3: x = 3. Check in the original: the left side is 4(3 + 1) = 4 · 4 = 16, and the right side is 3 + 13 = 16. Both sides are 16, so x = 3.
  2. The denominators are 3, 2, and 6, so the LCD is 6. Multiply every term by 6: 6 · (x ⁄ 3) = 2x, 6 · (1 ⁄ 2) = 3, and 6 · (7 ⁄ 6) = 7. So 2x + 3 = 7. Subtract 3: 2x = 4. Divide by 2: x = 2. Check: 2 ⁄ 3 + 1 ⁄ 2 = 4 ⁄ 6 + 3 ⁄ 6 = 7 ⁄ 6, which matches.
  3. Subtract 5 from both sides: −3x ≥ 15. Divide both sides by −3 and flip the sign: x ≤ −5. In interval notation that is (−∞, −5], with a bracket because −5 is included. Check x = −5: 5 − 3 · (−5) = 5 + 15 = 20, and 20 ≥ 20 is true. Check x = 0: 5 − 0 = 5, and 5 ≥ 20 is false, so 0 is correctly left out. Forgetting the flip would give x ≥ −5, which wrongly includes 0.
  4. "Within 15 minutes of 90" is |t − 90| ≤ 15. Write the middle: −15 ≤ t − 90 ≤ 15. Add 90 to all three parts: 75 ≤ t ≤ 105. The interval is [75, 105], so practice runs from 75 to 105 minutes, ends included.
  5. The classmate used the middle form, which describes numbers close to zero, but "greater than 6" asks for numbers far from zero. Test x = 0: |0| = 0, and 0 > 6 is false, yet 0 is inside −6 < x < 6. The right solution set is x < −6 or x > 6, written (−∞, −6) ∪ (6, ∞). Check x = 10: |10| = 10, and 10 > 6 is true.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, and the quiz — and next week these solving moves meet function notation, the rule that names one output for each input.

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