SPI Mathematics · Linear Equations

Solving Linear Equations

A linear equation is solved by doing the same thing to both sides until the unknown stands alone; a pair in x and y is solved by removing one unknown first.

All Linear Equations notes

Why this matters

Eight questions in the bank, 6 EASY and 2 MODERATE. Each is solved by balancing both sides: one equation in one unknown, a linear inequality, or a pair of equations in x and y. The marks are lost to sign slips and to fractions cleared carelessly.

Concept 1 of 3: Solving an equation in one variable

An equation is a balance. Whatever you do to one side you must do to the other: add, subtract, multiply or divide by the same non-zero number. The aim is to gather the unknown on one side and the numbers on the other.

Definition

Moving a term across the equals sign changes its sign: $x + 4 = 10$ becomes $x = 10 - 4$ .
Multiplying or dividing both sides by the same non-zero number does not change the solution.
Fractions: multiply every term by the LCM of the denominators first. Put brackets round each numerator, because a minus sign in front of a fraction applies to the whole numerator.
Check the answer by putting it back into both sides.

One unknown on each side

ax + b = c \;\Rightarrow\; x = \dfrac{c - b}{a} \qquad ax + b = cx + d \;\Rightarrow\; x = \dfrac{d - b}{a - c} \quad (a \ne c)

Worked example

Solve

7x - 4 = 3x + 12

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Linear Equations · Solving Linear Equations

Solve:

a - \frac{a-1}{2} = 1 - \frac{a-2}{3}

Not changing the sign when a term moves

5x - 2 = 3x + 6

gives

2x = 8

, so

x = 4

. Moving

-2

across without changing its sign gives

2x = 4

and a wrong

x = 2

The minus sign in front of a fraction

-\frac{x - 1}{2}

multiplied by 2 is

-(x - 1) = -x + 1

, not

-x - 1

. Bracket the numerator before you multiply.

Multiplying only some of the terms

When clearing fractions, every term is multiplied, including a number standing alone.

\frac{x}{3} + 2 = 5

times 3 is

x + 6 = 15

, so

x = 9

Drill 3 more on solving an equation in one variable

Concept 2 of 3: Solving a linear inequality

An inequality is solved like an equation, with one extra rule: multiplying or dividing both sides by a negative number turns the sign round. The answer is a range of values, not a single number.

Definition

Adding or subtracting the same thing on both sides keeps the sign.
Multiplying or dividing by a positive number keeps the sign.
Multiplying or dividing by a negative number flips the sign: $<$ becomes $>$ , and $\le$ becomes $\ge$ .
Test the answer with one value inside the range.

Flipping rule

a < b \text{ and } k < 0 \;\Rightarrow\; ka > kb

Worked example

Solve

5 - 3x \le 14

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Linear Equations · Solving Linear Equations

Which one of the following is a solution of

\frac{3x - 7}{3} > \frac{4x + 2}{5}

Forgetting to flip

-2x > 6

gives

x < -3

. Dividing by

-2

without turning the sign gives

x > -3

, which fails: at

x = 0

-2x = 0

is not more than 6.

Flipping when the number is positive

Multiplying by 15 to clear the denominators 3 and 5 keeps the sign as it is, because 15 is positive. Only a negative multiplier flips it.

Concept 3 of 3: Solving a pair of equations in two variables

One equation in x and y has endless solutions. Two equations usually pin down a single pair. Make the number in front of one unknown the same in both equations, then add or subtract to remove it.

Definition

A pair $(x, y)$ is a solution of $ax + by = c$ when it makes the equation true. To find a missing constant, put the given $x$ and $y$ in.
Elimination: multiply the equations so that one unknown has equal or opposite coefficients, then subtract or add.
Substitution: write one unknown in terms of the other from one equation and put it into the other.
With fractions, clear the denominators in each equation first.
Check the answer in both equations. In a multiple-choice question you can also test each option.
Shortcut formula: for $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ with $a_1b_2 - a_2b_1 \ne 0$ , $x = \dfrac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1}$ and $y = \dfrac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1}$ .

Solution of a pair, and the solution check

x = \dfrac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1}, \quad y = \dfrac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1} \qquad (x_0, y_0) \text{ solves } ax + by = c \text{ when } ax_0 + by_0 = c

Worked example

Solve

3x + 2y = 16

and

x - y = 2

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Linear Equations · Solving Linear Equations

\frac{x}{3} + \frac{y}{4} = 11

and

\frac{5x}{6} - \frac{y}{3} = -7

then

Checking in only one equation

A pair that fits one equation can miss the other.

(5, 2)

fits

x - y = 3

but not

x + 2y = 11

, since

5 + 4 = 9

Subtracting when the signs are opposite

If one equation has

+6y

and the other

-6y

, add the equations to remove

y

. Subtracting them doubles the y-term instead.

Drill 2 more on solving a pair of equations in two variables

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Solving an equation in one variable
One unknown on each side
$ax + b = c \;\Rightarrow\; x = \dfrac{c - b}{a} \qquad ax + b = cx + d \;\Rightarrow\; x = \dfrac{d - b}{a - c} \quad (a \ne c)$
Solving a linear inequality
Flipping rule
$a < b \text{ and } k < 0 \;\Rightarrow\; ka > kb$
Solving a pair of equations in two variables
Solution of a pair, and the solution check
$x = \dfrac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1}, \quad y = \dfrac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1} \qquad (x_0, y_0) \text{ solves } ax + by = c \text{ when } ax_0 + by_0 = c$

Watch out for (7)

Not changing the sign when a term moves
→ Solving an equation in one variable
The minus sign in front of a fraction
→ Solving an equation in one variable
Multiplying only some of the terms
→ Solving an equation in one variable
Forgetting to flip
→ Solving a linear inequality
Flipping when the number is positive
→ Solving a linear inequality
Checking in only one equation
→ Solving a pair of equations in two variables
Subtracting when the signs are opposite
→ Solving a pair of equations in two variables

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