Equations & Inequalities

Study Sheet

Equations & Inequalities

Everything you need to solve, rearrange, and graph.

What Is an Equation?

Concept
The Big Idea: Balance

An equation says that two things are equal, joined by an equals sign, like x+3=7x + 3 = 7. Think of the == sign as the middle of a balance scale. Whatever you do to one side you must do to the other, or the scale tips and the equation is no longer true.

To solve an equation means to find the value of the variable that makes it true. We do this by using inverse operations (opposite operations) to undo what has been done to the variable, one step at a time, until the variable is alone.

Tip

Inverse pairs: addition \leftrightarrow subtraction, and multiplication \leftrightarrow division. To undo +5+5, subtract 55. To undo ×4\times 4, divide by 44.

Example
Check by substituting

Is x=4x = 4 the solution of x+3=7x + 3 = 7? Replace xx with 44:   4+3=7  \;4 + 3 = 7 \;\checkmark. Yes! Always check your answer by plugging it back in.

One-Step Equations

Concept
One inverse operation

The variable has one operation attached to it. Undo that operation on both sides.

Example
Addition and subtraction

Solve x8=5x - 8 = 5.

x8=5x8+8=5+8(add 8 to both sides)x=13\begin{aligned} x - 8 &= 5\\ x - 8 + 8 &= 5 + 8 &&\text{(add 8 to both sides)}\\ x &= 13 \end{aligned}

Solve x+6=10x + 6 = 10: subtract 6 from both sides to get x=4x = 4.

Example
Multiplication and division

Solve x3=6\dfrac{x}{3} = 6.

x33=63x=18.\frac{x}{3}\cdot 3 = 6 \cdot 3 \quad\Rightarrow\quad x = 18.

Solve 5x=355x = 35: divide both sides by 5 to get x=7x = 7.

Two-Step Equations

Concept
Undo in reverse order

When two operations act on the variable, undo them in the reverse of the order of operations: first undo addition/subtraction, then undo multiplication/division.

Example
A two-step solve

Solve 2x+7=192x + 7 = 19.

2x+7=192x=12(subtract 7)x=6(divide by 2)\begin{aligned} 2x + 7 &= 19\\ 2x &= 12 &&\text{(subtract 7)}\\ x &= 6 &&\text{(divide by 2)} \end{aligned}

Check: 2(6)+7=12+7=19  2(6) + 7 = 12 + 7 = 19\;\checkmark

Multi-Step Equations

Concept
Simplify first

Before you isolate the variable, clean up each side: use the distributive property to remove parentheses, then combine like terms. Then solve as usual.

Tip

Distributing a negative: 3(x4)=3x+12-3(x - 4) = -3x + 12. The minus sign changes the sign of every term inside the parentheses. This is a very common place to make a mistake, so slow down here.

Example
Distribute, then combine

Solve 3(x+2)+4x=273(x + 2) + 4x = 27.

3(x+2)+4x=273x+6+4x=27(distribute)7x+6=27(combine like terms)7x=21(subtract 6)x=3(divide by 7)\begin{aligned} 3(x+2) + 4x &= 27\\ 3x + 6 + 4x &= 27 &&\text{(distribute)}\\ 7x + 6 &= 27 &&\text{(combine like terms)}\\ 7x &= 21 &&\text{(subtract 6)}\\ x &= 3 &&\text{(divide by 7)} \end{aligned}

Variables on Both Sides

Concept
Gather the variables

Move all variable terms to one side and all number terms to the other by adding or subtracting. A good habit: move the smaller variable term so the coefficient stays positive.

Example
Both sides

Solve 5x3=2x+95x - 3 = 2x + 9.

5x3=2x+93x3=9(subtract 2x)3x=12(add 3)x=4(divide by 3)\begin{aligned} 5x - 3 &= 2x + 9\\ 3x - 3 &= 9 &&\text{(subtract $2x$)}\\ 3x &= 12 &&\text{(add 3)}\\ x &= 4 &&\text{(divide by 3)} \end{aligned}

Equations with Fractions & Decimals

Concept
Clear the denominators

Multiply every term on both sides by the least common denominator (LCD). This turns a fraction equation into a whole-number equation. For decimals, multiply every term by 10, 100, \dots to clear the decimal points.

Example
Clearing fractions

Solve x2+13=56\dfrac{x}{2} + \dfrac{1}{3} = \dfrac{5}{6}. The LCD of 2,3,62, 3, 6 is 66.

6x2+613=6563x+2=53x=3x=1\begin{aligned} 6\cdot\frac{x}{2} + 6\cdot\frac{1}{3} &= 6\cdot\frac{5}{6}\\ 3x + 2 &= 5\\ 3x &= 3 \quad\Rightarrow\quad x = 1 \end{aligned}
Example
Clearing decimals

Solve 0.2x+0.5=1.10.2x + 0.5 = 1.1. Multiply every term by 10:

2x+5=112x=6x=3.2x + 5 = 11 \quad\Rightarrow\quad 2x = 6 \quad\Rightarrow\quad x = 3.

Literal Equations (Solving for a Variable)

Concept
Rearranging formulas

A literal equation has several letters. To “solve for” one letter means to get that letter alone, treating all the other letters as if they were numbers. Use the same inverse operations as always.

Example
Solve a formula for one letter

Solve A=wA = \ell w for ww.

A=wA=w(divide both sides by ).A = \ell w \quad\Rightarrow\quad \frac{A}{\ell} = w \quad\text{(divide both sides by $\ell$).}

Solve P=2+2wP = 2\ell + 2w for ww:

P2=2ww=P22.P - 2\ell = 2w \quad\Rightarrow\quad w = \frac{P - 2\ell}{2}.

Special Cases: No Solution or Infinitely Many

Concept
When the variable disappears

Sometimes, while solving, all the variable terms cancel out. Look at what is left:

  • [leftmargin=*]
  • A false statement (like 3=73 = 7) means no solution. No value of xx works.
  • A true statement (like 5=55 = 5) means infinitely many solutions. Every value of xx works; the two sides are really the same expression.
Example
Both special cases

Solve 2x+4=2x12x + 4 = 2x - 1: subtract 2x2x to get 4=14 = -1, which is false. No solution.

Solve 3(x+2)=3x+63(x + 2) = 3x + 6: distribute to get 3x+6=3x+63x + 6 = 3x + 6; subtract 3x3x to get 6=66 = 6, always true. Infinitely many solutions.

Solving Inequalities

Concept
Almost like equations

Inequalities use <,>,,<, >, \le, \ge instead of ==. Solve them the same way you solve equations, with one special rule.

Tip

The Flip Rule: When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. For example, 2x<6-2x < 6 becomes x>3x > -3 (divide by 2-2 and flip << to >>).

Example
An inequality with a flip

Solve 3x+110-3x + 1 \ge 10.

3x+1103x9(subtract 1)x3(divide by 3, flip  to )\begin{aligned} -3x + 1 &\ge 10\\ -3x &\ge 9 &&\text{(subtract 1)}\\ x &\le -3 &&\text{(divide by $-3$, flip $\ge$ to $\le$)} \end{aligned}

Graphing on a Number Line

Concept
Circles and arrows
-5-4-3-2-1012345

Graph the solution of a one-variable inequality on a number line:

  • [leftmargin=*]
  • Use an open circle \circ for << or >> (the endpoint is not included).
  • Use a closed (filled) circle \bullet for \le or \ge (the endpoint is included).
  • Shade an arrow to the right for “greater than” and to the left for “less than.”

An open circle means the endpoint is not included.

Example
Describing a graph

The graph of x3x \ge -3 has a closed circle at 3-3 and an arrow shaded to the right. The graph of x<2x < 2 has an open circle at 22 and an arrow shaded to the left.

Compound Inequalities (And / Or)

Concept
Two conditions at once

A compound inequality joins two inequalities.

  • [leftmargin=*]
  • And (written a<x<ba < x < b): xx must satisfy both. The graph is the overlap between the two endpoints.
  • Or: xx satisfies either one. The graph shades outward in two directions.

For an “and” inequality, do the same operation to all three parts.

Example
An “and” inequality

Solve 4<2x6-4 < 2x \le 6.

42<2x2622<x3.\frac{-4}{2} < \frac{2x}{2} \le \frac{6}{2} \quad\Rightarrow\quad -2 < x \le 3.

This means xx is between 2-2 (open) and 33 (closed).

Absolute Value Equations & Inequalities

Concept
Distance from zero

x|x| means the distance of xx from 00, which is never negative. So x=5|x| = 5 has two answers: x=5x = 5 or x=5x = -5. Get the absolute value alone first, then split into two cases.

Tip

Two patterns for inequalities:

  • [leftmargin=*]
  • Less than” (x<a|x| < a) becomes an and: a<x<a-a < x < a.
  • Greater than” (x>a|x| > a) becomes an or: x<ax < -a or x>ax > a.

If the absolute value equals a negative number, there is no solution (distance can't be negative).

Example
Absolute value equation

Solve x2=5|x - 2| = 5. Split into two cases:

x2=5    x=7orx2=5    x=3.x - 2 = 5 \;\Rightarrow\; x = 7 \qquad\text{or}\qquad x - 2 = -5 \;\Rightarrow\; x = -3.

Solutions: x=7x = 7 or x=3x = -3.

Example
Absolute value inequality

Solve x+14|x + 1| \le 4. This is a “less than,” so use and:

4x+145x3.-4 \le x + 1 \le 4 \quad\Rightarrow\quad -5 \le x \le 3.

Going Deeper: Advanced Equations & Inequalities

Concept
Parameter equations: when does a solution exist?

A parameter is a letter that stands for a fixed but unknown number (often aa, bb, kk, mm). When a linear equation contains a parameter, the number of solutions can change depending on the parameter's value. Every linear equation can be rearranged into the form

(coefficient)x=(constant),that isAx=B.(\,\text{coefficient}\,)\cdot x = (\,\text{constant}\,), \qquad\text{that is}\qquad Ax = B .

Everything then depends on whether the coefficient AA is zero:

  • [leftmargin=*]
  • If A0A \ne 0: exactly one solution, x=BAx = \dfrac{B}{A}.
  • If A=0A = 0 and B=0B = 0 (you get 0=00 = 0): infinitely many solutions.
  • If A=0A = 0 and B0B \ne 0 (you get 0=B0 = B, a contradiction): no solution.

The whole strategy is: collect all xx-terms on one side, factor xx out, then study its coefficient.

Tip

The key move: write the equation as Ax=BAx = B where AA and BB are expressions in the parameter. The “interesting” values of the parameter are exactly those that make the coefficient A=0A = 0. Test each of them separately.

Example
All three cases at once (worked example)

For each value of kk, describe the number of solutions of

k2x+5=9x+k+2.k^2 x + 5 = 9x + k + 2 .

Gather the xx-terms on the left and the constants on the right:

k2x+5=9x+k+2k2x9x=k+25(subtract 9x and 5)(k29)x=k3(factor x out)(k3)(k+3)x=k3(difference of squares)\begin{aligned} k^2 x + 5 &= 9x + k + 2\\ k^2 x - 9x &= k + 2 - 5 &&\text{(subtract $9x$ and $5$)}\\ (k^2 - 9)\,x &= k - 3 &&\text{(factor $x$ out)}\\ (k-3)(k+3)\,x &= k - 3 &&\text{(difference of squares)} \end{aligned}

Now the coefficient is A=(k3)(k+3)A = (k-3)(k+3), zero only when k=3k = 3 or k=3k = -3.

  • [leftmargin=*]
  • k=3k = 3: the equation becomes 0x=00\cdot x = 0, true for every xx     \;\Rightarrow\; infinitely many solutions.
  • k=3k = -3: the equation becomes 0x=60\cdot x = -6, impossible     \;\Rightarrow\; no solution.
  • any other kk: one solution, x=k3(k3)(k+3)=1k+3x = \dfrac{k-3}{(k-3)(k+3)} = \dfrac{1}{k+3}.
Concept
Nested absolute values: peel from the outside in

An expression like x32\bigl||x-3| - 2\bigr| has one absolute value inside another. Treat the whole inner absolute value as a single block and undo the outer bars first, exactly as you would solve block=c|\,\text{block}\,| = c. Each branch then leaves you with an ordinary absolute-value equation to finish. Always discard any branch that asks an absolute value to equal a negative number.

Example
A nested absolute-value equation (worked example)

Solve x32=4\bigl||x-3| - 2\bigr| = 4. Let the block be B=x3B = |x-3|. The outer bars give two branches:

x32=4orx32=4.|x-3| - 2 = 4 \qquad\text{or}\qquad |x-3| - 2 = -4 .

That is x3=6|x-3| = 6 or x3=2|x-3| = -2. The second is impossible (an absolute value is never negative), so discard it. From x3=6|x-3| = 6:

x3=6    x=9orx3=6    x=3.x - 3 = 6 \;\Rightarrow\; x = 9 \qquad\text{or}\qquad x - 3 = -6 \;\Rightarrow\; x = -3 .

Solutions: x=9x = 9 or x=3x = -3. (Check: 932=62=4  \bigl||9-3|-2\bigr| = |6-2| = 4\;\checkmark.)

Concept
Sums of absolute values as distances

Read xa|x - a| as “the distance between xx and aa on the number line.” Then xa+xb|x-a| + |x-b| is the total distance from xx to the two fixed points aa and bb. If aa and bb are dd units apart, that total is smallest exactly when xx sits between them, where it equals dd and can go no lower. So:

  • [leftmargin=*]
  • xa+xb=c|x-a| + |x-b| = c has no solution if c<dc < d, and infinitely many (a whole segment) if c=dc = d.
  • If c>dc > d, there are exactly two solutions, one on each side of the pair.

To solve algebraically, split the line at the critical points x=ax = a and x=bx = b and handle each region with casework.

Example
Summed absolute values with casework (worked example)

Solve x1+x+2=5|x - 1| + |x + 2| = 5. The critical points are x=1x = 1 and x=2x = -2, which are d=3d = 3 apart. Since 5>35 > 3, expect two solutions. Break the line into three regions.

x<2:(x1)(x+2)=5    2x1=5    x=3  2x1:(x1)+(x+2)=5    3=5  (false, no solution here)x>1:(x1)+(x+2)=5    2x+1=5    x=2  \begin{aligned} x < -2:\quad &-(x-1) - (x+2) = 5 \;\Rightarrow\; -2x - 1 = 5 \;\Rightarrow\; x = -3 \;\checkmark\\ -2 \le x \le 1:\quad &-(x-1) + (x+2) = 5 \;\Rightarrow\; 3 = 5 \;\text{(false, no solution here)}\\ x > 1:\quad &(x-1) + (x+2) = 5 \;\Rightarrow\; 2x + 1 = 5 \;\Rightarrow\; x = 2 \;\checkmark \end{aligned}

The middle region gives the constant 33 (the minimum total distance), confirming the sum can never dip below 33. Solutions: x=3x = -3 or x=2x = 2.

Concept
Literal equations that need factoring

When you solve a formula for a letter that appears in more than one term, you cannot simply divide --- you must first collect every term containing that letter on one side, then factor the letter out, and only then divide by the leftover expression. This is the same “Ax=BAx = B” idea used for parameters, applied to rearranging formulas.

Example
Factoring to isolate a variable

Solve xy+3x=5xy + 3x = 5 for xx. Both terms on the left contain xx, so factor it out:

x(y+3)=5x=5y+3(y3).x(y + 3) = 5 \quad\Rightarrow\quad x = \frac{5}{y + 3} \quad (y \ne -3).

Similarly, solve ax+b=cx+dax + b = cx + d for xx:

axcx=db    x(ac)=db    x=dbac(ac).ax - cx = d - b \;\Rightarrow\; x(a - c) = d - b \;\Rightarrow\; x = \frac{d - b}{a - c}\quad (a \ne c).
Concept
Counting integer solutions of an inequality

Some problems ask not for the solution set but for how many integers lie in it. First solve the inequality to get a range, then count the whole numbers inside, watching the endpoints:

  • [leftmargin=*]
  • A strict endpoint (<< or >>) does not include that boundary value.
  • A non-strict endpoint (\le or \ge) does include it.

For a run of consecutive integers from mm to nn inclusive, the count is nm+1n - m + 1.

Example
How many integer solutions?

How many integers satisfy 2x1<7|2x - 1| < 7? Rewrite the “less than” absolute value as an “and”:

7<2x1<7    6<2x<8    3<x<4.-7 < 2x - 1 < 7 \;\Rightarrow\; -6 < 2x < 8 \;\Rightarrow\; -3 < x < 4 .

Both ends are strict, so 3-3 and 44 are excluded. The integers are 2,1,0,1,2,3-2, -1, 0, 1, 2, 3 --- that is 6\mathbf{6} integers.

Example
Modeling a harder word problem (worked example)

Plan A costs $30 per month plus $0.10 for each text message. Plan B costs a flat $50 per month. For how many text messages is Plan A the cheaper choice?

Let mm be the number of texts. “Plan A cheaper” means its cost is less than Plan B's:

30+0.10m<500.10m<20(subtract 30)m<200(divide by 0.10)\begin{aligned} 30 + 0.10m &< 50\\ 0.10m &< 20 &&\text{(subtract 30)}\\ m &< 200 &&\text{(divide by $0.10$)} \end{aligned}

Since mm counts texts, it must be a whole number, so Plan A is cheaper for 0m1990 \le m \le 199 texts (fewer than 200200). At exactly 200200 texts the two plans tie at $50, and beyond that Plan B wins.

Formulas, Proofs & Tips

Tip
Solving equations and inequalities
Do the same thing to both sides;multiplying an inequality by a negative flips it\text{Do the same thing to both sides};\qquad \text{multiplying an inequality by a negative flips it}

What it means. Equations stay balanced under identical operations; inequalities have one extra rule.

Example. 2x+3=112x=8x=42x+3=11\Rightarrow 2x=8\Rightarrow x=4.

Why it works. If a=ba=b then a+c=b+ca+c=b+c and ac=bcac=bc — equality is preserved. But if a<ba<b, multiplying by 1-1 reverses their order on the number line, since a-a now lies to the right of b-b.

Tip. Undo operations in reverse PEMDAS order: addition first, multiplication last.