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Chapter Review · a rapid pre-test review (speedrun)

Equations and Inequalities: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Solution set
The set of reals making a statement true; R\mathbb{R} is all reals, \varnothing none. Solving describes that set, not just one value.
Conditional, identity, contradiction
True for some values but not all; true for every real; true for none. Every equation is exactly one of the three.
Reversible step
A step you can undo. It guarantees an EQUIVALENT equation, one with the same solution set, but guarantees nothing for an inequality: x<3x < 3 times 1-1 is reversible, yet x<3-x < -3 names a different set.
Extraneous and lost solutions
A candidate solving a derived statement but not the original is extraneous; a solution the original had and the derived one dropped is lost.
Excluded value
A value making a denominator of the ORIGINAL zero, so never a solution whatever the algebra says later.
Unknown and parameter
In a literal equation (more than one letter), the letter solved for is the unknown, each letter held fixed a parameter; the question assigns the roles, not the alphabet.
Compound inequality
Two conditions joined by "and", the intersection of their solution sets, or by "or", the union; a chain a<x<ba < x < b is an "and".
Breakpoint
A value where an expression inside bars changes sign; breakpoints cut the line into intervals on which every bar resolves to one formula.

Formulas and theorems

  • The trichotomy of ax=bax = b

    a0: x=baa=0, b0: a=0, b=0: R\begin{gathered} a \neq 0:\ x = \frac{b}{a} \\ a = 0,\ b \neq 0:\ \varnothing \\ a = 0,\ b = 0:\ \mathbb{R} \end{gathered}

    Use when Any reals aa, bb, numbers or parameters. Exhaustive, so a linear equation never has exactly two solutions; a parameter aa hides which case holds, so carry all three.

  • Safe moves on an equation

    Add any expression defined for every xx to both sides; multiply or divide both sides by a nonzero constant.

    Use when An expression that can vanish ADDS roots at its zeros when you multiply and LOSES them when you divide; multiplying by 00 destroys the equation.

  • Solving for a letter in several terms

    ax+b=cx+d (ac)x=db x=dbac\begin{gathered} ax + b = cx + d \\ \Longrightarrow\ (a - c)x = d - b \\ \Longrightarrow\ x = \frac{d - b}{a - c} \end{gathered}

    Use when aca \neq c, a condition on the whole factor divided by, not its letters: a=c=5a = c = 5 fails though neither is zero. At a=ca = c the trichotomy takes over.

  • Where a rearranged formula gains a condition

    A rearrangement gains a NEW restriction exactly where it divides by an expression that can be zero, and none where it divides only by constants.

    Use when Scope: rearrangements built from adding constants and multiplying or dividing; the original's own restrictions still apply.

    e.g. 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} gives v=fuufv = \frac{fu}{u - f}, adding ufu \neq f.

  • Order rules for inequalities

    c>0: a<b    ca<cbc<0: a<b    ca>cb\begin{gathered} c > 0:\ a < b \iff ca < cb \\ c < 0:\ a < b \iff ca > cb \end{gathered}

    Use when Adding any expression defined for every xx keeps set and direction. Only the sign of what multiplies BOTH sides flips the symbol; multiplying by 00 destroys the inequality; an unknown-sign expression is a fork, not a step.

    e.g. 3x>12-3x > 12 gives x<4x < -4; at x=5x = -5, 15>1215 > 12.

  • The four cases of ax>bax > b

    a>0: x>baa<0: x<baa=0, b<0: Ra=0, b0: \begin{gathered} a > 0:\ x > \tfrac{b}{a} \\ a < 0:\ x < \tfrac{b}{a} \\ a = 0,\ b < 0:\ \mathbb{R} \\ a = 0,\ b \ge 0:\ \varnothing \end{gathered}

    Use when Any reals aa, bb. The strict boundary falls opposite to the equation's: at a=b=0a = b = 0 this is \varnothing, since 0>00 > 0 is false, while ax=bax = b is R\mathbb{R}. For axbax \ge b the R\mathbb{R} branch becomes b0b \le 0.

  • Absolute value, defined and squared

    x={xif x0,xif x<0,x=x2\begin{gathered} \lvert x \rvert = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0, \end{cases} \\ \lvert x \rvert = \sqrt{x^2} \end{gathered}

    Use when Both hold for every real xx. The bars are a hidden if, so a bar holding an unknown carries an unresolved case.

  • X=c\lvert X \rvert = c for a constant cc

    c>0: X=c  or  X=cc=0: X=0c<0: \begin{gathered} c > 0:\ X = c \ \text{ or } \ X = -c \\ c = 0:\ X = 0 \\ c < 0:\ \varnothing \end{gathered}

    Use when Check the sign of cc before splitting: only c>0c > 0 gives two cases.

  • X=Y\lvert X \rvert = Y with an unknown on the right

    X=Y    Y0  and (X=Y  or  X=Y)\begin{gathered} \lvert X \rvert = Y \iff \\ Y \ge 0 \ \text{ and } \\ \bigl(X = Y \ \text{ or } \ X = -Y\bigr) \end{gathered}

    Use when Any expressions XX, YY. The bare split drops Y0Y \ge 0 and runs one way only, so test each candidate against Y0Y \ge 0; both can fail, leaving \varnothing.

  • X=Y\lvert X \rvert = \lvert Y \rvert

    X=Y     X=Y  or  X=Y     X2=Y2\begin{gathered} \lvert X \rvert = \lvert Y \rvert \\ \iff\ X = Y \ \text{ or } \ X = -Y \\ \iff\ X^2 = Y^2 \end{gathered}

    Use when An equivalence for any XX, YY: no side condition, since both sides are nonnegative, so nothing is discarded and squaring is exact.

    e.g. x6=x+2\lvert x - 6 \rvert = \lvert x + 2 \rvert: one case gives 6=2-6 = 2, so only x=2x = 2.

  • Absolute-value inequalities

    X<c    c<X<cX>c    X<c  or  X>c\begin{gathered} \lvert X \rvert < c \iff -c < X < c \\ \lvert X \rvert > c \iff X < -c \ \text{ or } \ X > c \end{gathered}
    Less than gives one band between the points; greater than gives two rays outward from themTwo horizontal number lines, each marked with minus c, zero, and c. On the upper one a thick segment runs from minus c across zero to c, ending in an open circle at each end, and is captioned one band. On the lower one two thick arrows start at those same open circles and run outward off each side, leaving the stretch between them bare, and it is captioned two rays.|X| < cone band−c0c|X| > ctwo rays−c0c
    Text description

    A number line on which the absolute value of X is less than c, giving one band between minus c and c, above a second number line on which the absolute value of X is greater than c, giving two rays running outward from those same two points.

    Use when Both hold for EVERY real cc: the band is nonempty only when c>0c > 0, since at c0c \le 0 its ends cross, and the rays cover R\mathbb{R} when c<0c < 0 and say X0X \neq 0 at c=0c = 0. Also X0\lvert X \rvert \le 0 forces X=0X = 0, and xac\lvert x - a \rvert \le c is [ac,a+c][a - c, a + c].

    e.g. 3x5<2\lvert 3x - 5 \rvert < -2 is \varnothing; 47x>1\lvert 4 - 7x \rvert > -1 is R\mathbb{R}.

  • Sum of two distances

    xp+xq=dg=pq\begin{gathered} \lvert x - p \rvert + \lvert x - q \rvert = d \\ g = \lvert p - q \rvert \end{gathered}
    Beyond the gap g, the two solutions step out equally from the anchorsA horizontal number line with anchors p and q marked as plain dots and a measuring bar above them spanning the two, labelled g. Two highlighted dots, each labelled x, sit outside the anchors, one to the left of p and one to the right of q. Under each, a highlighted measuring bar runs from the anchor to that solution, and a single caption below states both steps equal d minus g over two.gxpqxeach step (d − g)/2
    Text description

    The case where d is greater than g, on a number line: one solution sits outside each anchor, both the same step of d minus g over 2 beyond it, which is why no solution exists once d is less than g.

    Use when Anchors pp, qq, any real dd: d<gd < g gives \varnothing, d=gd = g the closed interval between the anchors, and d>gd > g two solutions, one step of dg2\frac{d - g}{2} outside each anchor.

    e.g. x6+x+2=12\lvert x - 6 \rvert + \lvert x + 2 \rvert = 12: g=8g = 8, so x=4x = -4 or x=8x = 8.

Problem types, step by step

Solve a linear equation and classify its solution set

  1. Distribute, combine like terms, and clear numeric fractions with the LCD.
  2. Collect xx terms one side, constants the other, reaching ax=bax = b.
  3. If a0a \neq 0, divide and report {ba}\left\{\frac{b}{a}\right\}.
  4. If the variable cancels, the bare statement decides: true gives R\mathbb{R}, false \varnothing.

e.g. 5x3(x2)=2x+15x - 3(x - 2) = 2x + 1 collapses to 6=16 = 1, so \varnothing.

Solve an equation with the variable in a denominator

  1. List the excluded values, the zeros of every denominator, BEFORE any algebra.
  2. Multiply by the LCD and solve the cleared equation.
  3. Discard candidates on the excluded list; confirm survivors in the ORIGINAL.
  4. A cleared identity means every real except the excluded values.

e.g. 6x=2x4\frac{6}{x} = \frac{2}{x - 4}: exclude 00, 44; 6x24=2x6x - 24 = 2x gives x=6x = 6.

Rearrange a formula for a named variable

  1. Name the unknown; treat every other letter as a fixed number.
  2. Clear fractions, then gather all terms holding the target on one side.
  3. Factor the target out and divide by the whole remaining factor.
  4. State the nonzero condition on that factor; none if you divided only by constants.

e.g. A=h(b1+b2)2A = \frac{h(b_1 + b_2)}{2} gives b1=2Ahb2b_1 = \frac{2A}{h} - b_2, needing h0h \neq 0.

Find the parameter values giving one, no, or infinitely many solutions

  1. Reduce to (coefficient) x=x = (constant), both depending on the parameter.
  2. Solve coefficient =0= 0 for the boundaries; elsewhere there is exactly one solution.
  3. At each boundary evaluate the constant: nonzero gives \varnothing, zero R\mathbb{R}.
  4. An outcome no boundary produces never occurs for that family.

e.g. kx+6=2x+3kkx + 6 = 2x + 3k gives (k2)x=3(k2)(k - 2)x = 3(k - 2): x=3x = 3 unless k=2k = 2, where R\mathbb{R}.

Solve a linear inequality, including one with a parameter coefficient

  1. Collect the variable with additions, which never touch the direction.
  2. Divide by the coefficient: keep the symbol if positive, reverse it if negative.
  3. A parameter coefficient forces three cases, a>0a > 0, a<0a < 0, and a=0a = 0.
  4. With a=0a = 0 read the bare statement: true gives R\mathbb{R}, false \varnothing.

e.g. ax>5ax > -5: x>5ax > -\frac{5}{a} if a>0a > 0, x<5ax < -\frac{5}{a} if a<0a < 0, R\mathbb{R} at a=0a = 0.

Solve an inequality whose multiplier has unknown sign

  1. Exclude any value making a denominator zero.
  2. Split the line by the sign of the expression you must multiply by.
  3. Multiply in each case, keeping the direction where positive and reversing it where negative.
  4. Intersect each result with its own case assumption, then union the pieces.

e.g. 3x<1\frac{3}{x} < 1: x>0x > 0 gives x>3x > 3, x<0x < 0 gives every negative, so (,0)(3,)(-\infty, 0) \cup (3, \infty).

Solve a compound inequality

  1. Solve each piece to its own solution set.
  2. Intersect for "and", union for "or"; a chain a<x<ba < x < b is an "and".
  3. A piece losing its variable is R\mathbb{R} if the leftover statement is true, else \varnothing.

e.g. 2x1>52x - 1 > 5 and 3x>03 - x > 0 give x>3x > 3 and x<3x < 3, so \varnothing.

Solve an absolute-value equation

  1. Isolate the bars so one side reads X\lvert X \rvert alone.
  2. Right side a number: split only when it is positive; 00 gives X=0X = 0, a negative gives \varnothing.
  3. Right side an expression YY: solve X=±YX = \pm Y, then keep only candidates with Y0Y \ge 0.
  4. Right side another absolute value: solve both cases, or square, and keep every root.

e.g. x7=2x5\lvert x - 7 \rvert = 2x - 5: candidates 2-2, 44; only 44 meets 2x502x - 5 \ge 0.

Solve an absolute-value inequality

  1. Isolate the bars, reversing the symbol if you divide by a negative.
  2. Check the right side: below a nonpositive number is \varnothing, above a negative number is R\mathbb{R}.
  3. Otherwise convert to the band or the two rays, working a band on all three parts at once.

e.g. 2x+15\lvert 2x + 1 \rvert \ge 5 gives x3x \le -3 or x2x \ge 2.

Solve an equation with two absolute values added

  1. Find the breakpoints, where each inside is zero, cutting the line into three intervals.
  2. On each interval resolve every bar by its inside's sign there, then solve the linear equation.
  3. Keep a root only if it lies in its own interval; an interval reducing to a true statement contributes all its points.

e.g. x1+x5=4\lvert x - 1 \rvert + \lvert x - 5 \rvert = 4: the target equals the gap 44, so every point of [1,5][1, 5] works.

Exam traps

  • Trap Dividing both sides by a variable: from x2=xx^2 = x, dividing by xx reports {1}\{1\}.

    Fix That destroys x=0x = 0. Factor instead: x(x1)=0x(x - 1) = 0 gives {0,1}\{0, 1\}.

  • Trap Confirming a candidate in the CLEARED equation, so x+3x4=7x4\frac{x + 3}{x - 4} = \frac{7}{x - 4} is answered x=4x = 4.

    Fix x=4x = 4 is the excluded value, so it is extraneous and the solution set is \varnothing.

  • Trap Reversing the inequality symbol because a negative number appears somewhere in the problem.

    Fix Only the sign of what multiplies or divides BOTH sides flips it; a negative constant elsewhere, or a subtraction, never does.

  • Trap Multiplying an inequality by the variable to clear a fraction, so 1x<1\frac{1}{x} < 1 is answered x>1x > 1.

    Fix That treats xx as positive and loses the ray x<0x < 0. Splitting on the sign gives x<0x < 0 or x>1x > 1.

  • Trap Reporting both cases of X=Y\lvert X \rvert = Y, so 2x+1=x3\lvert 2x + 1 \rvert = x - 3 is answered 4-4 and 23\frac{2}{3}.

    Fix Both fail x30x - 3 \ge 0, so both are extraneous and the solution set is \varnothing.

  • Trap Simplifying x2\sqrt{x^2} to xx.

    Fix It equals x\lvert x \rvert: at x=5x = -5 the value is 55, since a square root is never negative.

  • Trap Treating X>0\lvert X \rvert > 0 as true for every real number.

    Fix It holds everywhere except where X=0X = 0; the one true for every real number is X0\lvert X \rvert \ge 0.

Chapter test Questions from across the chapter