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Equations and inequalities

An equation is a balance: do the same thing to both sides and it stays level. Solve 3x + 2 = 14 step by step, rearrange y = wx + b for x, and see why multiplying an inequality by a negative number flips it, using the decision rule 2x - 6 >= 0.
M01.1.3L0 Foundation25 minIn review

Before this lesson

M01.1.2 Expressions and the rules of algebra

After this lesson you can

  • Solve a linear equation in one unknown by doing the same operation to both sides, and check the answer by substitution.
  • Rearrange a formula such as y = wx + b to make a different letter its subject.
  • Solve a linear inequality, flipping the sign when multiplying or dividing by a negative number, and explain the flip with a reflected number line.
  • Draw the solutions of an inequality on the number line and write them as an interval such as [a, b) or (a, b).

In review. The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

The question

A spam filter counts the exclamation marks \(x\) in a message and predicts spam when \({2x - 6 \geq 0}\). Which messages does it flag? A second filter predicts spam when \({-2x + 6 \geq 0}\). Does it flag the same ones? Answering takes one move that works for every equation, and one twist that only inequalities have.

From the last lesson

1. In what order is \({3 + 4 \times 2^2}\) worked out, and what is its value?

Show answer

The power first, \({2^2 = 4}\); then the multiplication, \({4 \times 4 = 16}\); then the addition, \({3 + 16 = 19}\).

2. Expand \({(a + b)^2}\).

Show answer

\({a^2 + 2ab + b^2}\): two squares and two rectangles \(ab\).

Intuition

An equation says that two expressions are equal. Picture it as a balance. Each box on a pan weighs the unknown number \(x\), and each small cube weighs 1. The equation \({3x + 2 = 14}\) says that three boxes and two cubes weigh the same as fourteen cubes, so the beam hangs level.

A two-pan balance hanging level. The left pan holds three blue boxes, each marked x, and two small orange cubes; under it is written 3x plus 2, the left side. The right pan holds fourteen orange cubes; under it is written 14, the right side. Notes point to a box, each box weighs x, and to a cube, each cube weighs 1. A green line at the top says level: the two sides weigh the same.
M01.1.3-F1

The equation \({3x + 2 = 14}\) as a balance. The beam is level, so the two sides weigh the same.

To find \(x\), take things off the balance until one box sits alone. There is one rule: whatever you do to one pan, do the same to the other. Then the beam stays level, and every new line of algebra is still true. Take something off one pan only, and the beam tips.

Build it up

Step 1. What a solution is. An equation is true for some values of \(x\) and false for others. A value that makes it true is a solution. Try \({x = 3}\): \({3 \times 3 + 2 = 11}\), not 14. Try \({x = 4}\): \({3 \times 4 + 2 = 14}\). So 4 is a solution. Guessing works here, but it does not scale.

Step 2. Moves that keep the balance level. You may:

  • add the same number to both sides, or subtract it from both;
  • multiply both sides by the same number, or divide both by it, as long as that number is not 0.

Why not 0? Multiplying both sides of any equation by 0 gives \({0 = 0}\), which is true for every \(x\): the information about \(x\) is thrown away. And dividing by 0 has no meaning at all.

Step 3. Undo in reverse order. \({3x + 2}\) was built by “times 3, then add 2”. Undo it the other way round: subtract 2, then divide by 3. That is how to choose the moves.

Step 4. Inequalities. An inequality compares two sides with \({<}\), \({\leq}\) (“less than or equal to”), \({>}\) or \({\geq}\). The same moves work, with one exception. Adding the same number to both sides slides both points along the line together, so their order stays. Multiplying by a positive number stretches or squeezes the line, and the order stays. Multiplying by a negative number turns the line round:

Two number lines from minus 6 to 6, one above the other, with a dotted mirror line at 0 between them. On the top line, labelled before: 2 less than 5, blue dots sit at 2 and 5. Green arrows labelled times minus 1 cross over each other and land on the bottom line at minus 2 and minus 5, marked with vermillion dots, labelled after: minus 5 less than minus 2, so minus 2 greater than minus 5.
M01.1.3-F2

Multiplying by \(-1\) sends every point to its mirror image across 0. The arrows cross: left and right swap, so \({2 < 5}\) becomes \({-2 > -5}\).

So multiplying or dividing both sides of an inequality by a negative number flips its sign: \({<}\) becomes \({>}\), and \({\leq}\) becomes \({\geq}\).

Step 5. Solutions fill a stretch of the line. An inequality usually has infinitely many solutions. Together they form an interval:

Four rows, each with an inequality, its interval and a number line from 0 to 6. Row 1: 1 less than or equal to x less than or equal to 4, the interval closed bracket 1, 4 closed bracket, a thick vermillion stretch from 1 to 4 with solid dots at both ends. Row 2: 1 less than or equal to x less than 4, written square bracket 1, 4 round bracket, solid dot at 1 and hollow dot at 4. Row 3: 1 less than x less than 4, written round bracket 1, 4 round bracket, hollow dots at both ends. Row 4: x greater than or equal to 3, written square bracket 3, infinity round bracket, a solid dot at 3 and a thick ray with an arrowhead going right. A note reads solid dot: included, hollow dot: left out.
M01.1.3-F3

Four intervals. A square bracket and a solid dot mean the end is included; a round bracket and a hollow dot mean it is left out.

The formal statement

NoteLinear equations

A linear equation in one unknown \(x\) is one that can be written \({ax + b = c}\), where \(a\), \(b\) and \(c\) are numbers and \(a\) is not 0. It has exactly one solution:

\[ \begin{aligned} ax + b &= c \\ ax &= c - b \\ x &= \frac{c - b}{a} . \end{aligned} \]

The second line subtracts \(b\) from both sides. The third divides both sides by \(a\), which is allowed because \(a\) is not 0. Putting the answer back in checks it:

\[ a \times \frac{c - b}{a} + b = (c - b) + b = c . \]

Rearranging a formula uses the same moves on letters. The model \({y = wx + b}\) gives the output \(y\) for an input \(x\). To ask which input gives a chosen output, make \(x\) the subject: subtract \(b\), then divide by \(w\) (which must not be 0):

\[ x = \frac{y - b}{w} . \]

With \({w = 1.4}\) and \({b = 0.5}\), the output 4.7 comes from the input

\[ x = \frac{4.7 - 0.5}{1.4} = \frac{4.2}{1.4} = 3 . \]

NoteRules for inequalities

For all numbers \(a\), \(b\) and \(c\), if \({a < b}\), then:

  • \({a + c < b + c}\) and \({a - c < b - c}\);
  • \({ca < cb}\) when \({c > 0}\);
  • \({ca > cb}\) when \({c < 0}\): the sign flips.

The same holds with \({\leq}\), which flips to \({\geq}\).

Why the sign flips. That \({a < b}\) means \({b - a}\) is positive: \(b\) lies to the right of \(a\). Now compare \(-a\) and \(-b\). Their difference is

\[ (-a) - (-b) = b - a , \]

which is positive, so \(-a\) lies to the right of \(-b\): \({-a > -b}\). A negative \(c\) is \(-d\) for a positive \(d\). Multiplying by \(d\) keeps the order, \({da < db}\), and multiplying by \(-1\) then flips it: \({-da > -db}\), which is \({ca > cb}\).

NoteIntervals
Inequality Interval Ends
\({a \leq x \leq b}\) \([a, b]\) both included
\({a \leq x < b}\) \([a, b)\) \(a\) in, \(b\) out
\({a < x < b}\) \((a, b)\) both left out
\({x \geq a}\) \([a, \infty)\) \(a\) in, no end
\({x \leq b}\) \({(-\infty, b]}\) no end, \(b\) in

The sign \(\infty\), “infinity”, is not a number: it says the interval goes on for ever, so it always takes a round bracket.

Worked example

Solve \({3x + 2 = 14}\).

A trace in three numbered rows, each a balance beside a line of algebra. Row 1: three boxes and two cubes balance fourteen cubes, beside 3x plus 2 equals 14, the equation as given: the pans balance. Row 2: the two cubes on the left and two of the cubes on the right are drawn as dashed green outlines, taken away; three boxes balance twelve cubes, beside 3x equals 12, take 2 cubes off each pan: subtract 2 from both sides. Row 3: two of the boxes and eight of the twelve cubes are dashed outlines; one box balances four cubes, beside x equals 4 in vermillion, keep a third of each pan: divide both sides by 3. A line at the bottom says check by putting 4 back in: 3 times 4 plus 2 equals 12 plus 2 equals 14, the right side.
M01.1.3-F4

Each move is made on both pans, so the beam stays level and each line is still true. The dashed outlines show what was just taken away.

\[ \begin{aligned} 3x + 2 &= 14 \\ 3x &= 12 \\ x &= 4 . \end{aligned} \]

Line 2 subtracts 2 from both sides. Line 3 divides both sides by 3. Check by putting 4 back into the original equation: \({3 \times 4 + 2}\) is \({12 + 2}\), which is 14, the right side. So \({x = 4}\).

The two spam filters. Solve each rule for \(x\), writing every move:

\[ \begin{aligned} 2x - 6 &\geq 0 \\ 2x &\geq 6 \\ x &\geq 3 . \end{aligned} \]

Add 6 to both sides, then divide both sides by 2. Since 2 is positive, the sign stays. The first filter flags every message with 3 or more exclamation marks: the interval \([3, \infty)\).

\[ \begin{aligned} -2x + 6 &\geq 0 \\ -2x &\geq -6 \\ x &\leq 3 . \end{aligned} \]

Subtract 6 from both sides, then divide both sides by \(-2\). Since \(-2\) is negative, the sign flips. The second filter flags messages with 3 or fewer marks: \({(-\infty, 3]}\), the mirror image.

Two number lines from 0 to 6 with the score written above each whole number. Top, predict yes when 2x minus 6 is at least 0: the scores minus 6, minus 4, minus 2, 0, 2, 4, 6, the last four bold and vermillion; a vermillion ray starts with a solid dot at 3 and runs right, labelled yes, with no on the left; below, x at least 3, the interval 3 to infinity. Bottom, predict yes when minus 2x plus 6 is at least 0: the scores 6, 4, 2, 0, minus 2, minus 4, minus 6; the ray starts at 3 and runs left, labelled yes, with no on the right; below, x at most 3, the interval minus infinity to 3.
M01.1.3-F5

The score of each rule at \({x = 0, 1, \dots, 6}\). A rule says yes where its score is at least 0. The sign in front of \(x\) decides which side of 3 says yes.

Check with points. At \({x = 4}\) the first score is \({2 \times 4 - 6 = 2}\), at least 0, so yes; the second is \({-8 + 6 = -2}\), so no. At \({x = 3}\) both scores are 0, so both say yes: the end 3 belongs to both intervals. A machine-learning classifier works just like this. It computes a score \({wx + b}\) and says yes when the score is at least 0. Its decision boundary is where the score is 0, here \({x = 3}\), and the sign of the weight \(w\) decides which side says yes.

Code

A function that solves \({ax + b = c}\) and prints each step, the way you would write it by hand:

def solve(a, b, c):
    """Solve a*x + b = c step by step. a must not be 0."""
    print(a, "* x +", b, "=", c)
    # subtract b from both sides
    c = c - b
    print(a, "* x =", c)
    # divide both sides by a
    x = c / a
    print("x =", x)
    return x

x = solve(3, 2, 14)
# put the answer back into the left side
print("check:", 3 * x + 2)
3 * x + 2 = 14
3 * x = 12
x = 4.0
check: 14.0

The same function on \({3x + 2 = 13}\) gives a decimal that cannot be stored exactly. Passing fractions instead keeps every step exact:

from fractions import Fraction

# decimals: the answer is rounded
x = solve(3, 2, 13)
# fractions: the answer is exact
x = solve(Fraction(3), Fraction(2), Fraction(13))
print("check:", 3 * x + 2)
3 * x + 2 = 13
3 * x = 11
x = 3.6666666666666665
3 * x + 2 = 13
3 * x = 11
x = 11/3
check: 13

The rearranged formula \({x = (y - b)/w}\) is a function too. With decimals the answer misses 3 in the last place; with fractions made from the written decimals, it is exactly 3:

def input_for(y, w, b):
    """The input x for which w*x + b equals y."""
    return (y - b) / w

print(input_for(4.7, 1.4, 0.5))
print(input_for(Fraction("4.7"), Fraction("1.4"), Fraction("0.5")))
3.0000000000000004
3

Last, the two filters of the worked example, tested on 0 to 6 exclamation marks:

for marks in range(0, 7):
    print(marks, 2 * marks - 6 >= 0, -2 * marks + 6 >= 0)
0 False True
1 False True
2 False True
3 True True
4 True False
5 True False
6 True False

The first filter says True from 3 upwards, the second up to 3, and both say True at 3.

Explore

M01.1-X1

Equation balance

Each button does one operation to both pans and writes the new line of algebra underneath. Turn on Change the left pan only to break the rule and watch the beam tip.

A static picture in place of the interactive: three balances, one above the other. Three boxes and two cubes balance fourteen cubes, written 3x plus 2 equals 14; after taking 2 cubes off each pan, three boxes balance twelve cubes, 3x equals 12; after keeping a third of each pan, one box balances four cubes, x equals 4.

Things to try:

  1. Solve \({3x + 2 = 14}\) with the buttons in two moves. Then press Next step on a fresh start and compare.
  2. Choose \({5x + 3 = 2x + 12}\), which has \(x\) on both sides. Which move gathers the boxes on one pan?
  3. Turn on Change the left pan only, subtract 2, and read what the read-out says about the two sides. Then press Undo.
  4. Try to multiply both sides by 0. Why does the balance refuse?

Pitfalls

Two panels, each solving minus 2x plus 6 at least 0, which becomes minus 2x at least minus 6. The left panel, marked with a cross and titled sign kept, divides by minus 2 to get x at least 3, drawn as a purple ray to the right from 3; a blue test point at 5 lies on it, with the note test x equals 5: minus 2 times 5 plus 6 equals minus 4, and minus 4 is not at least 0. The right panel, marked with a tick and titled sign flipped, gets x at most 3, drawn as a vermillion ray to the left from 3; a blue test point at 1 lies on it, with the note test x equals 1: minus 2 times 1 plus 6 equals 4, and 4 is at least 0, as it should be.
M01.1.3-F6

Wrong beside right. Testing one point from the answer catches a missing flip at once.

“Dividing \({-2x \geq -6}\) by \(-2\) gives \({x \geq 3}\).”

Dividing by a negative number flips the sign, so the answer is \({x \leq 3}\). A quick test catches the slip: \({x = 5}\) lies in the wrong answer, but \({-2 \times 5 + 6 = -4}\) is not at least 0.

“I subtracted 2 from the left side; the right side can stay as it is.”

Then the two sides are no longer equal, and the new line has a different solution: \({3x = 14}\) is solved by \(14/3\), not by 4. Every move goes on both sides, every time.

“Multiplying both sides by the same number is always allowed, so 0 is fine.”

Multiplying by 0 turns any equation into \({0 = 0}\), which every \(x\) satisfies: the answer is lost. Dividing by 0 is not defined. Both moves need a number that is not 0.

“\({x \geq 3}\) is the interval \((3, \infty)\).”

The round bracket leaves 3 out, so \((3, \infty)\) is \({x > 3}\). For \({x \geq 3}\) write \([3, \infty)\). In a classifier the end matters: it decides whether a message with exactly 3 marks is flagged.

Check yourself

Recall

1. What one rule keeps an equation true while you solve it?

Show answer

Do the same thing to both sides: add or subtract the same number, or multiply or divide both sides by the same number that is not 0.

2. Solve \({5x - 4 = 11}\), and check your answer.

Show answer

Add 4 to both sides: \({5x = 15}\). Divide both sides by 5: \({x = 3}\). Check: \({5 \times 3 - 4 = 11}\).

3. When does the sign of an inequality flip?

Show answer

When both sides are multiplied or divided by a negative number, because that turns the number line round.

4. Write \({2 < x \leq 7}\) as an interval. Which end is included?

Show answer

\((2, 7]\). The end 7 is included and the end 2 is left out.

5. Rearrange \({y = wx + b}\) to make \(x\) the subject.

Show answer

\({x = (y - b)/w}\), which needs \(w\) to be not 0.

Exercises

1. ★ Solve \({7 - 2x = 1}\) and \({4(x - 1) = 2x + 6}\). Check both answers.

Show solution

\({7 - 2x = 1}\). Subtract 7 from both sides: \({-2x = -6}\). Divide both sides by \(-2\): \({x = 3}\). Check: \({7 - 2 \times 3 = 1}\).

\({4(x - 1) = 2x + 6}\). Expand the bracket: \({4x - 4 = 2x + 6}\). Subtract \(2x\) from both sides: \({2x - 4 = 6}\). Add 4: \({2x = 10}\). Divide by 2: \({x = 5}\). Check: the left side is \({4 \times 4 = 16}\) and the right side is \({10 + 6 = 16}\).

2. ★★ Solve \({-3 < 2x + 1 \leq 7}\). Write the answer as an interval and test both ends.

Show solution

Do each move to all three parts. Subtract 1: \({-4 < 2x \leq 6}\). Divide by 2, which is positive, so no sign flips: \({-2 < x \leq 3}\). The interval is \({(-2, 3]}\).

Test the ends. At \({x = 3}\): \({2 \times 3 + 1 = 7}\), and \({7 \leq 7}\) is true, so 3 is included. At \({x = -2}\): \({2 \times (-2) + 1 = -3}\), and \({-3 < -3}\) is false, so \(-2\) is left out.

3. ★★★ Model A predicts \({1.4x + 0.5}\) and model B predicts \({-0.6x + 4.5}\). For which inputs \(x\) does A predict more than B? Write the answer as an interval, find the input where the two agree, and check one input on each side.

Show solution

We want the inputs for which

\[ 1.4x + 0.5 > -0.6x + 4.5 . \]

Add \(0.6x\) to both sides: \({2x + 0.5 > 4.5}\). Subtract 0.5: \({2x > 4}\). Divide by 2, which is positive: \({x > 2}\). The answer is the interval \((2, \infty)\).

They agree where the two sides are equal, and the same moves with \({=}\) in place of \({>}\) give \({x = 2}\). There A predicts \({2.8 + 0.5 = 3.3}\) and B predicts \({-1.2 + 4.5 = 3.3}\).

Checks. At \({x = 3}\): A gives 4.7 and B gives 2.7, so A is larger. At \({x = 0}\): A gives 0.5 and B gives 4.5, so B is larger.

Summary card

An equation is a balance: do the same thing to both sides and it stays level.

The anchor figure again: a level balance with three boxes and two cubes on the left pan, 3x plus 2, and fourteen cubes on the right pan, 14.
M01.1.3-F1

Same move, both pans.

  • \({ax + b = c}\) with \(a\) not 0 gives \({x = (c - b)/a}\). Check by putting it back.
  • \({y = wx + b}\) rearranged for \(x\) is \({x = (y - b)/w}\).
  • Multiplying or dividing an inequality by a negative number flips its sign.
  • \([a, b)\) includes \(a\) and leaves out \(b\); \((a, b)\) leaves out both.

Sources and next steps

Sources. Each was opened and checked against this lesson on 12 November 2026.

  • Jay Abramson and others, Algebra and Trigonometry 2e, OpenStax. Section 2.2, Linear Equations in One Variable, writes a linear equation as \({ax + b = 0}\) with \(a\) not 0, solves it by doing the same thing to both sides of the equals sign, and notes that we cannot divide by zero. Section 2.7 gives the properties of inequalities, with the exception that multiplying or dividing by a negative number reverses the inequality symbol, and interval notation with brackets and parentheses.
  • Richard Hammack, Book of Proof, third edition, Section 1.1. It lists the closed, open, half-open and infinite intervals in set notation and draws each as a darkened segment, with a solid circle for an end that is included and a hollow circle for one that is not.
  • Python Software Foundation, fractions: Rational numbers (Python 3.12). Fraction(numerator, denominator), the string form Fraction("4.7"), and exact arithmetic on the results.

Next. M01.1.4 Reading notation reads formulas with subscripts, hats, bars and Greek letters aloud, using the four data points of this course.

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M01.1.4 Reading notation: subscripts, Greek letters and formulas

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