Equations and Inequalities

An equation says that two expressions have the same value. Solving it means finding every value of the letter that makes that true. Every step has one justification: do the same thing to both sides, and the two sides stay equal. The one exception is an inequality, where dividing by a negative number reverses the sign.

How do you solve a linear equation?

Undo the operations around the letter, starting with the one applied last. In 3x + 7 = 22, subtract 7 from both sides: 3x = 15. Divide both sides by 3: x = 5.

A bracket is expanded first, and the number in front multiplies every term inside it. See Equations with Brackets.

When the number in front divides the other side, divide both sides by it instead: 3(x + 1) = 12 becomes x + 1 = 4. See Treating a Bracket as a Single Quantity.

When the letter appears on both sides, subtract the smaller x term from both sides so that it appears only once. Removing the smaller term keeps the coefficient positive. See Equations with the Unknown on Both Sides.

When the equation has fractions, multiply every term by the common denominator, including any term that is not a fraction. In $\frac{x}{3}+\frac{x}{4}=7$x/3 + x/4 = 7, multiply by 12: 4x + 3x = 84, so x = 12. See Equations with Fractions.

How many solutions does it have?

Usually the x terms survive and there is exactly one solution. Sometimes they cancel, and what is left decides. If it is false, such as 1 = 5, no value of x works. If it is true, such as 4 = 4, every value works, and the equation is an identity. See How Many Solutions an Equation Has.

In ax + 3 = 5x + 7, the x terms collect to (a − 5)x = 4. Choose a = 5 and the x term is zero, and zero lots of x can never equal 4. See Choosing a Coefficient to Fix the Solution Count.

How do you change the subject of a formula?

The subject of a formula is the letter that stands alone on one side. To change it, undo the operations around the letter you want, doing the same to both sides. The answer is an expression in the other letters. To make r the subject of $A=\pi r^2$A = πr², divide both sides by $\pi$π, then take the square root: $r=\sqrt{\frac{A}{\pi }}$r = √(A/π). See Changing the Subject of a Formula.

When the letter appears twice, bring every term containing it to one side, take it out as a common factor, then divide both sides by the bracket. See Making a Twice-Appearing Letter the Subject.

Turning words into equations

A word problem becomes an equation once the unknown has a name. Write the definition down: let n be the smaller number. See Forming Equations.

A formula answers every version of a question at once, and a case you can count checks it. See Writing Formulas from Words.

How do you solve simultaneous equations?

Remove one unknown, solve for the other, then substitute back. On a graph each equation is a line, and the solution is the point where they cross, because both equations hold there.

Elimination adds or subtracts the equations so that one letter cancels. See Simultaneous by Elimination.

When no letter has the same coefficient in both equations, multiply one equation first so that a pair of coefficients match. See Solving Simultaneous Equations by Scaling.

Substitution rearranges one equation for one letter and puts that expression into the other. If y = x + 1 and 2x + y = 10, then 2x + x + 1 = 10, so x = 3 and y = 4. Use it when one equation already gives a letter on its own. See Simultaneous by Substitution.

A word problem with two unknowns needs two facts. Name both unknowns, write one equation from each fact, and read the answer back in the words of the question. See Writing a Pair of Equations from a Word Problem.

Two lines can cross once, run parallel, or be the same line: one solution, none, or infinitely many. Scaling one equation to match the other shows which. See How Many Solutions a Pair of Equations Has and Choosing a Coefficient in a Pair of Equations.

Sometimes the question asks only for x + y. If 3x + 2y = 11 and 2x + 3y = 9, adding gives 5x + 5y = 20, so x + y = 4. See Solving for x + y Without Finding x and y.

What does an inequality mean?

An inequality describes a range of values rather than one value. x > 4 means every number greater than 4. On a number line it is a ray, with a hollow circle at 4 because 4 is not greater than 4. $x\ge 4$x ≥ 4 includes 4, and the circle is filled. As a set it is written {x : x > 4}. See Inequalities and Solution Sets of Inequalities.

Solve an inequality the way you solve an equation. From 2x + 3 < 11, subtract 3, then divide by 2: x < 4. See Two-Step Inequalities.

There is one exception. Multiplying or dividing both sides by a negative number reverses the sign. From −2x > 6, divide by −2: x < −3. A negative sign reflects the number line: 2 < 3, but −2 > −3. Adding or subtracting moves both sides the same way and leaves their order alone.

A compound inequality has two ends. $-3<x\le 5$−3 < x ≤ 5 has infinitely many solutions but only eight integer solutions, because −3 is out and 5 is in. See Integer Solutions of Inequalities.

Inequalities in two variables

A limit on a total of two things is written with an inequality sign: pens at $2 and pads at $5 with $30 to spend is $2p+5d\le 30$2p + 5d ≤ 30. "At most" gives $\le$≤ and "at least" gives $\ge$≥. See Forming an Inequality in Two Variables.

To graph it, draw the boundary line, dashed when the sign is strict and solid when it is $\le$≤ or $\ge$≥. Test a point off the line, usually the origin, and shade the side where the inequality holds. See Graphing an Inequality in Two Variables.

Two inequalities together are solved by the region where both shadings overlap. See The Overlap of Two Inequality Regions.

Equations with a modulus

The modulus |x| is the distance of x from zero, so it is never negative, and |x| = 5 has two solutions: 5 and −5. An equation with a modulus splits into two cases, because the expression inside the bars can be positive or negative. From |x − 3| = 5, either x − 3 = 5, giving x = 8, or x − 3 = −5, giving x = −2. Substitute both back to check. |x + 1| = −4 has no solution. See Solving Absolute Value Equations.

An inequality with a modulus is a statement about distance. |x − 3| < 5 means x is within 5 of 3, one stretch of the line: −2 < x < 8. |x − 3| > 5 means more than 5 away from 3, which is two separate rays: x < −2 or x > 8. See Solving Absolute Value Inequalities.

How do you know a solution is right?

Substitute it into the original equation, not into a line you wrote along the way. Solving 5x + 4 = 2x + 19 gave x = 5. Check: 5 × 5 + 4 = 29 and 2 × 5 + 19 = 29. For simultaneous equations, check both equations.

For an inequality, test one value from inside the answer and one from outside. If x < −3 solves −2x > 6, then x = −4 should work and x = 0 should fail. This catches a sign that was not reversed.

The four mistakes worth naming

• Multiplying only some terms. Clearing fractions or expanding a bracket multiplies every term.
• Not reversing the sign. Dividing an inequality by a negative number reverses it every time.
• Leaving the letter undefined. "Let x be the number of adult tickets" is part of the answer.
• Not checking a modulus case. Substitute both cases back into the original equation.

Where this leads next

The algebra behind these methods is in algebraic expressions, indices and surds, and the arithmetic in arithmetic methods. Quadratic equations need factoring or the quadratic formula. Proportional relationships written as equations start in ratio and proportion.

Practice it in the game

Math Challenge teaches each method above as an illustrated lesson with a try-it question, and the Linear Equations topic drills the solving itself.