0.2 Algebraic structure, sets, and inequalities - Courses of Computing4All
Precalculus for High School Students 0.2 Algebraic structure, sets, and inequalities
Precalculus for High School Students
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Learning goals: Use equivalent expressions carefully; solve equations and inequalities; describe solution sets with interval notation.

Structure before calculation

An expression names a quantity; an equation states that two expressions are equal. Simplifying \(3(x-2)+2x\) gives \(5x-6\), but it does not determine \(x\). Solving \(5x-6=14\) asks which values make the statement true. Here \(5x=20\), so \(x=4\). Substitute into the original expression to check: \(3(4-2)+2(4)=14\).

The distributive property connects expansion and factoring. For example, \(6x+9=3(2x+3)\), and \(x^2-9=(x-3)(x+3)\). Factoring is not decoration: a product equal to zero lets us identify zero factors. Conversely, a sum equal to zero does not require each summand to be zero.

Expansion and factoring describe the same rectangleExpansion and factoring describe the same rectangle2x333 · 2x = 6x3 · 3 = 9Total width: 2x + 3One whole area: 3(2x + 3). Two partial areas: 6x + 9.Expand to add the parts; factor to recover the common height.
The distributive property says \(3(2x+3)=6x+9\). This area model assumes \(x>0\) so its side lengths are positive; the algebraic identity itself is valid for every real x. The rectangle is schematic.

What makes a transformation legitimate?

Adding the same quantity to both sides of an equation preserves its solutions. Multiplying by a nonzero constant does too. Dividing by an expression requires it to be nonzero. In \(\frac{x^2-1}{x-1}=x+1\), cancellation is valid only for \(x\ne1\). The simplified formula does not restore an input excluded by the original expression. Similarly, \(\sqrt{x^2}=|x|\), because the principal square root is nonnegative.

Canceling a factor does not restore an excluded inputCanceling a factor does not restore an excluded input−1012301234xyOriginal: (x² − 1)/(x − 1)Factor: (x − 1)(x + 1)/(x − 1)Simplify: y = x + 1Restriction: x ≠ 1At x = 1 the denominator is zero.The open circle at (1, 2) is a hole in the original graph.Keep the restriction even after the factor has been canceled.
Every permitted input gives the same output as \(y=x+1\). The original quotient still has no value at \(x=1\), so its graph excludes the point \((1,2)\).

Ordered numbers and intervals

The number line turns inequalities into sets. The notation \([-2,5)\) includes \(-2\) but excludes \(5\); it means \(-2\le x\lt5\). Parentheses always accompany infinity because infinity is not a real-number endpoint. Intersection means “and,” so both conditions must hold. Union means “or,” so either condition suffices.

Solve \(-3x+7\ge16\): subtract seven to get \(-3x\ge9\), then divide by \(-3\) and reverse the inequality to obtain \(x\le-3\). Multiplication by a negative reverses order on the number line. Check a value such as \(-4\), which gives \(19\ge16\), and a rejected value such as zero, which gives the false statement \(7\ge16\).

Number lines show endpoints and reversed orderNumber lines show endpoints and reversed orderInterval [−2, 5) means −2 ≤ x < 5−205Filled: includedOpen: excludedMultiplying by −1 reflects order across zero−5−20252 < 5−2 > −5A negative multiplier reverses which value lies farther to the right.Thus −3x ≥ 9 becomes x ≤ −3 when both sides are divided by −3.
Use a filled endpoint when equality is allowed and an open endpoint when it is not. Reflection sends 2 to −2 and 5 to −5, reversing their order. This is why multiplication or division by a negative reverses an inequality.

Absolute value is distance

The expression \(|x-a|\) is the distance from \(x\) to \(a\). Thus \(|x-2|\le3\) describes all points at most three units from two: \[-3\le x-2\le3\quad\Longrightarrow\quad -1\le x\le5.\] In contrast, \(|x-2|\gt3\) describes the two exterior regions \(x\lt-1\) or \(x\gt5\). Do not join these exterior conditions with “and”: no number lies in both regions.

Absolute value describes distance on a number lineAbsolute value describes distance on a number line|x − 2| ≤ 3: at most 3 units from 2−1253 units3 units−1 ≤ x ≤ 5 → [−1, 5]|x − 2| > 3: farther than 3 units from 2−125x < −1 or x > 5 → (−∞, −1) ∪ (5, ∞)“And” selects the inside interval; “or” joins the two outside regions.
The center is 2 and the boundary points are three units away, at −1 and 5. “At most” includes the boundary points; “farther than” excludes them. A union joins the two separated exterior regions.

Efficient checking

Use exact fractions until the last step, especially when rounding could move an endpoint. Label exclusions and check boundary points separately. A decimal approximation can support a calculation, but it does not replace an exact solution when the question asks for one. Explain both the algebraic transformation and its effect on the solution set.

Practice: attempt before reading the solutions

Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.

  1. Simplify \(4(2x-3)-3(x+2)\).
  2. Solve \(\frac{x}{3}+2=7\).
  3. Solve \(5-2x\lt11\) and give interval notation.
  4. Solve \(-1\le2x+3\lt9\).
  5. Solve \(|x+4|\le2\).
  6. Solve \(|2x-1|\gt5\).
  7. Simplify \(\frac{x^2-4}{x-2}\) and state the restriction.
  8. Evaluate \(\sqrt{x^2}\) when \(x=-7\).
Worked solutions and reasoning

  1. Apply the distributive property to each bracket. The second multiplier is −3, so both terms in that bracket change accordingly: \[4(2x-3)-3(x+2)=8x-12-3x-6.\]

    Combine like terms: \(8x-3x=5x\) and \(-12-6=-18\). The simplified expression is \(5x-18\). This is an equivalent expression, not an equation that determines x. A quick check at x = 2 gives −8 in both forms: \(4(1)-3(4)=-8\) and \(10-18=-8\).

  2. Undo the operations surrounding x in reverse order. First subtract two from both sides: \(x/3+2-2=7-2\), so \(x/3=5\). Then multiply both sides by the nonzero constant three: \(x=15\).

    Both operations preserve the solution set because they can be reversed. Check in the original equation: \(15/3+2=5+2=7\). Therefore the solution is \(x=15\). Multiplying the entire original equation by three would also work, provided every term is multiplied: \(x+6=21\).

  3. Subtract five from both sides to isolate the variable term: \(5-2x\lt11\) becomes \(-2x\lt6\). Dividing by −2 reverses the inequality, giving \(x\gt-3\).

    The solution set is \((-3,\infty)\). The endpoint −3 is excluded because the original inequality is strict: substituting it gives \(11\lt11\), which is false. A test inside the interval, x = 0, gives \(5\lt11\); a test outside, x = −4, gives \(13\lt11\), which is false.

  4. A compound inequality imposes both bounds at once. Perform the same operation on all three parts: subtract three to obtain \(-4\le2x\lt6\), then divide by the positive number two to obtain \(-2\le x\lt3\).

    Thus the solution interval is \([-2,3)\). The left bracket includes −2, because it makes \(2x+3=-1\), satisfying the non-strict lower bound. The right parenthesis excludes 3, because it makes \(2x+3=9\), violating the strict upper bound. Dividing by a positive number does not reverse either inequality.

  5. Interpret \(|x+4|\) as \(|x-(-4)|\): the distance from x to −4. A distance at most two means x lies between two units left and two units right of −4.

    Algebraically, \(-2\le x+4\le2\). Subtract four from all three parts to get \(-6\le x\le-2\), or \([-6,-2]\). Both endpoints are included because their distance from −4 is exactly two. The center x = −4 has distance zero and is included as well.

  6. An absolute value greater than five means the inside expression is either greater than five or less than negative five. Therefore solve the two branches \(2x-1\gt5\) or \(2x-1\lt-5\).

    Adding one and dividing by two gives \(x\gt3\) or \(x\lt-2\). The set is \((-\infty,-2)\cup(3,\infty)\). At x = −2 and x = 3 the absolute value is exactly five, so both endpoints are excluded. Use “or,” because a solution belongs to either exterior region; no x can satisfy both inequalities simultaneously.

  7. Record the original restriction first: the denominator x − 2 cannot be zero, so \(x\ne2\). Next use the difference-of-squares factorization \(x^2-4=(x-2)(x+2)\).

    For the allowed inputs, \[\frac{(x-2)(x+2)}{x-2}=x+2.\] The simplified expression is therefore \(x+2\), with \(x\ne2\). At x = 2 the original expression would be 0/0, which is undefined; cancellation does not create a value there. Its graph follows the line y = x + 2 with a hole at (2, 4).

  8. Substitute before simplifying: with x = −7, \(x^2=(-7)^2=49\). The square-root symbol denotes the nonnegative square root, so \(\sqrt{x^2}=\sqrt{49}=7\).

    This agrees with the identity \(\sqrt{x^2}=|x|\), since \(|-7|=7\). Although both 7 and −7 square to 49, the expression \(\sqrt{49}\) has the single value 7. The two values arise when solving an equation such as \(y^2=49\), which is a different task.

Reflection

Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.

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