The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 3 (Intermediate Algebra) Linear Equations, Absolute Value Equations, and Applications

Overview

This lecture develops two big ideas and then applies algebra to word problems. First, it nails down the precise definition of absolute value and uses it to solve absolute-value equations — including the harder case of an absolute value equal to another absolute value. Second, it introduces variation (direct, inverse, and joint), where an unknown constant of proportionality is recovered from a known data point. The lecture closes with a toolkit for translating English into equations and for the three classic application families: consecutive integers, age problems, and rate–time–distance problems. The recurring theme is to reason from definitions rather than memorize tricks.

Key Concepts

Definition of absolute value. The absolute value function is piecewise: $|x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$ It returns the input unchanged when the input is non-negative, and negates it when the input is negative. The common shortcut "just drop the minus sign" fails when the inside is a variable: if $a < 0$, then $-a > 0$, so $|-a| = -a$ (a positive quantity). Always test against the definition rather than reaching for a trick.

Absolute-value equations (one absolute value).

• $|X| = c$ with $c > 0$ has exactly two solutions, because only two numbers have absolute value $c$: $X = c$ or $X = -c$.
• This works for any expression inside the bars: $|f(x)| = c \Rightarrow f(x) = c \text{ or } f(x) = -c$.
• $|X| = 0$ has the single solution $X = 0$.
• $|X| = c$ with $c < 0$ has no solution — an absolute value can never be negative.
• Isolate first: if the equation has extra terms (e.g. $5 + |x| = 8$ or $-4|x| = -20$), get the absolute value alone on one side before splitting into the two cases.

Absolute value equal to absolute value. For $|A| = |B|$ there appear to be four sign combinations, but they collapse to two, because $-A = -B$ is the same equation as $A = B$, and $-A = B$ is the same as $A = -B$. So solve only: $A = B \quad \text{or} \quad A = -B$ Drop the bars on one side (conventionally the right), distribute the negative carefully in the second case, and solve both equations.

Direct variation. $y$ varies directly as $x$ means $y = kx$ for a constant $k$. As $x$ grows, $y$ grows. Recover $k$ by substituting a known $(x, y)$ pair.

Inverse variation. $y$ varies inversely as $x$ means $y = \dfrac{k}{x}$. As $x$ grows, $y$ shrinks. To solve for $k$, multiply both sides by $x$.

Joint variation. $x$ jointly proportional to $y$ and $z$ means $x = kyz$ — direct proportion to two variables at once. Powers are allowed: "area varies directly as the square of the radius" gives $A = kr^2$.

Translating words into algebra.

• "a number" $\to$ a variable $x$
• "is" $\to$ $=$
• "more than" $\to$ addition, written back-to-front: "three more than a number" $\to$ $x + 3$
• "less than" $\to$ subtraction, back-to-front: "four less than a number" $\to$ $x - 4$
• "times" $\to$ multiplication: "five times a number" $\to$ $5x$

Consecutive integers. If the first is $x$, consecutive integers are $x, x+1, x+2, \dots$. Consecutive odd (or even) integers step by two: $x, x+2, x+4, \dots$.

Age problems. Introduce one variable per person, write one equation for the present-day relationship and one for the past/future relationship (add or subtract the same number of years from each age). Solve one equation for a variable and substitute into the other.

Rate–time–distance. The single governing formula is $d = r \cdot t.$ For two movers, set up $d = rt$ for each, then combine using the situation: travelers moving toward each other on a course of length $L$ satisfy $d_1 + d_2 = L$; one traveler catching another covers an equal distance $d_1 = d_2$; a head start is handled by relating the two times (e.g. $t_1 = t_2 + 6$).

Worked Example

Solve $|2x + 3|$ within $\;2 - 4\,|2x+3| = -18$. Isolate the absolute value, then split into two cases. The rule for an isolated absolute value is $|f(x)| = c \Rightarrow f(x) = c \text{ or } f(x) = -c$.

Check $x = 1$: $2 - 4|2(1)+3| = 2 - 4(5) = 2 - 20 = -18.$ Check $x = -4$: $2 - 4|2(-4)+3| = 2 - 4|-5| = 2 - 20 = -18.$ Both solutions work.

Variation example — recover the constant, then predict. The speed $r$ Josiah travels varies inversely as the time $t$, so $r = \dfrac{k}{t}$. Given $r = 35$ mph when $t = \tfrac{5}{2}$ hr, find how long the trip takes at $r = 55$ mph.

Summary

• $|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x < 0$; reason from this definition instead of blindly dropping minus signs.
• An isolated $|f(x)| = c$ splits into $f(x) = c$ or $f(x) = -c$ when $c > 0$; it has one solution when $c = 0$ and no solution when $c < 0$. Always isolate the absolute value first.
• $|A| = |B|$ reduces to just two cases, $A = B$ or $A = -B$, because the other two sign choices are duplicates.
• Variation models: direct $y = kx$, inverse $y = \dfrac{k}{x}$, joint $x = kyz$. Find $k$ from a known data point, then use the completed formula to answer the question.
• Translate carefully: "a number" $\to x$, "is" $\to =$, "more/less than" reverse the order of the operands.
• Every motion problem rests on $d = rt$; combine two movers' distances or times according to the scenario (meeting, catching up, opposite directions, head starts).