MTH1W Solving Linear Equations: Notes and Practice

Your phone plan costs 25 dollars a month plus 10 dollars for each extra gigabyte of data. This month's bill is 65 dollars. How many extra gigabytes did you use? You could guess: 3 GB gives 55 dollars, 4 GB gives 65 dollars. Guessing works when the answer is a small whole number, but if the bill were 61.40 dollars you would be guessing for a long time. Writing the situation as an equation, $25 + 10 g = 65$ , and solving it gives the answer directly, whether it is a whole number, a decimal or a negative. This lesson builds that method one step at a time.

What solving means

A linear equation is an equation in which the variable appears only to the first power, like $3 x + 2 = 11$ . A solution is a value of the variable that makes the left side equal the right side. Think of the equation as a balance scale that is level: the left pan holds $3 x + 2$ and the right pan holds $11$ . You can do anything you like to the scale as long as you do the same thing to both sides, and it stays level (the one exception: never multiply or divide both sides by zero). Your goal is to end up with $x$ alone on one side; whatever is on the other side is then the solution. Finally, you verify (check) the solution by substituting it into the original equation.

Fig 1Three balance scales: 3x + 2 = 11, then 3x = 9 after removing 2 from each side, then x = 3 after splitting each side into 3 equal groups.

Quick checkTry it before you read on

A level balance shows

3 x + 2 = 11

. Which move keeps it level?

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Which value of

x

is the solution of

2 x + 7 = 15

Inverse operations

Operation done to x	Inverse that undoes it	Example
add a number	subtract the same number	$x + 7 = 12 \Rightarrow x = 12 - 7 = 5$
subtract a number	add the same number	$x - 4 = 9 \Rightarrow x = 9 + 4 = 13$
multiply by a non-zero number	divide by the same number	$6 x = 42 \Rightarrow x = 42 \div 6 = 7$
divide by a number	multiply by the same number	$\frac{x}{5} = 3 \Rightarrow x = 3 \times 5 = 15$

When more than one operation has been done to $x$ , undo them in reverse order, like taking off your shoes before your socks. In $3 x + 2 = 11$ , the variable was first multiplied by 3 and then 2 was added. So first subtract 2 from both sides, then divide both sides by 3. Undoing the addition first keeps the numbers whole; dividing first would also work, but only if you divide every term, which creates fractions.

3 x + 2 3 x + 2 - 2 3 x \frac{3 x}{3} x = 11 = 11 - 2 = 9 = \frac{9}{3} = 3

Subtract 2 from both sides, then divide both sides by 3.

Quick checkTry it before you read on

To solve

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

in reverse order, what should you do first?

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Solve

\frac{x}{3} + 4 = 10

Variables on both sides and brackets

When $x$ appears on both sides, first gather all the $x$ terms on one side and all the constants on the other, using the same inverse operations. A helpful habit is to move the $x$ term with the smaller coefficient, so the $x$ term that remains is positive. In $5 x - 4 = 2 x + 11$ , subtract $2 x$ from both sides to get $3 x - 4 = 11$ , then add 4 to get $3 x = 15$ , so $x = 5$ . If an equation has brackets, expand them first with the distributive property, then collect like terms on each side before moving anything across. In $3 (x - 2) = 2 (x + 1) + 5$ , expanding gives $3 x - 6 = 2 x + 2 + 5$ , which simplifies to $3 x - 6 = 2 x + 7$ .

Fig 25x − 4 = 2x + 11 as two bars of equal length, drawn to scale with x = 5: crossing out 2x from both bars leaves 3x − 4 = 11, and adding 4 to both gives 3x = 15, so x = 5.

5 x - 4 3 x - 4 3 x x = 2 x + 11 = 11 = 15 = 5 subtract 2 x add 4 divide by 3

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7 x + 3 = 4 x + 18

, which first step gathers the

x

terms and leaves a positive

x

term?

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Solve

2 (x + 3) = x + 10

Equations with fractions

Fractions are easier to remove than to work with. Multiply every term on both sides by the lowest common denominator (LCD), the smallest number that every denominator divides into. For $\frac{x + 2}{3} = \frac{x - 1}{2}$ the LCD is 6. Multiplying both sides by 6 gives $2 (x + 2) = 3 (x - 1)$ , because $6 \div 3 = 2$ and $6 \div 2 = 3$ . Notice the brackets: the whole numerator gets multiplied. Now it is an ordinary equation: $2 x + 4 = 3 x - 3$ , so $x = 7$ . For $\frac{x}{4} + 3 = 7$ , multiply every term by 4: $x + 12 = 28$ , so $x = 16$ . The constant 3 must be multiplied too.

Fig 3Top: every term of x/4 + 3 = 7 is multiplied by 4, giving x + 12 = 28. Bottom: the multiples of 3 and of 2 first meet at 6, and multiplying both sides by 6 turns (x + 2)/3 = (x − 1)/2 into 2(x + 2) = 3(x − 1).

Always verify in the original equation

Substitute your answer into the equation you were given, not into a line you wrote later (a later line may already contain your mistake). Work out each side separately. For

x = 7

\frac{x + 2}{3} = \frac{x - 1}{2}

: left side

\frac{9}{3} = 3

, right side

\frac{6}{2} = 3

. Both sides equal 3, so

x = 7

is correct.

When there is no single answer

Most linear equations have exactly one solution, but two special cases happen when the

x

terms cancel. In

2 x + 3 = 2 x + 5

, subtracting

2 x

leaves

3 = 5

, which is false, so no value of

x

works: there is no solution. In

2 (x + 1) = 2 x + 2

, expanding gives

2 x + 2 = 2 x + 2

, and subtracting

2 x

leaves

2 = 2

, which is true for every

x

: there are infinitely many solutions, because the two sides are equivalent expressions.

If there are fractions, multiply every term on both sides by the LCD.
Expand any brackets.
Collect like terms on each side separately.
Move the variable terms to one side and the constants to the other with inverse operations.
Divide both sides by the coefficient of the variable (including its sign).
Verify by substituting into the original equation.

Quick checkTry it before you read on

What is the lowest common denominator for

\frac{x}{4} = \frac{x + 1}{6}

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Multiply every term of

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

by 3. Which equation do you get?

Solving Linear Equations

Key things to remember

What solving means

Inverse operations

Variables on both sides and brackets

Equations with fractions

Worked examples

Where marks get lost

Flashcards 8

Practice