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An equation is a statement with an \(=\) sign. It tells us that the expression on the left hand side of the \(=\) sign is equal to the expression on the right hand side.

An equation contains an unknown variable (often represented by \(x\))

An equation is solved by finding the value of \(x\).

Knowledge of Module 1 - solving equations is useful

Example

Solve the equation \(3x + 2 = x + 5\)

Solution

Rearrange to have all \(x\) terms on the left hand side and all numbers on the right.

Subtract \(2\) from both sides

\(3x + 2 - 2 = x + 4 - 2\)

\(3x = x + 3\)

Subtract \(x\) from both sides

\(3x - x = x + 3 -x\)

Simplify \(2x = 3\)

Solve by dividing

\(x = 3 \div 2\)\(= 1.5\)

Answer

\(x = 1.5\)

When rearranging, always carry out just one step at a time.

Do not try to subtract \(2\) from both sides as well as subtracting \(x\) in the same step.

This can lead to errors, especially with \(+\) and \(-\) numbers.

Example

Solve the equation \(x + 2 = 3x - 10\)

Solution

Rearrange one step at a time

Subtract \(2\) from each side

\(x + 2 - 2\)

\(= 3x - 10 - 2\)

Simplify \(x = 3x -12\)

Simplify, subtract \(3x\) from both sides

\(x - 3x = 3x - 12 - 3x\)

\(x - 3x = -12\)

Simplify \(-2x = -12\)

Solve by dividing

\(-12 \div -2 = +6\)

Answer

\(x = +6\)

or just \(6\)

Remember

Dividing two negative numbers gives a positive result.

\(-12 \div -2 = +6\)

Example

Solve the equation \(\dfrac{t}{4} = 7\)

Solution

In this example, there is no need to rearrange. The unknown is on the left hand side and the numer is on the right hand side.

Solve by multiplying both sides by \(4\)

\(\dfrac{t}{4} \times 4 = 7 \times 4\)

\(t = 28\)

Answer

\(t = 28\)

Remember

\( \dfrac{t}{4}\) is the same as \( \frac{1}{4}t\)

\( \frac{1}{4}t \times 4 = t\)

Example

Paper cups are sold in boxes of different sizes.

A medium box holds \(p\) paper cups, a small box holds \(p - 20\) paper cups and a large box holds twice as many as a small one.

Karen buys one large, one medium and one small box. Altogether she buys 340 paper cups.

a. Form an equation and solve it to find the value of \(p\).

b. Jaxon buys 3 small boxes. How many paper cups has he bought?

Solution

a. A small box has \(p - 20\) paper cups and a large box has twice as many, \(2 (p - 20)\)

To form an equation, add together the number of paper cups in one large, one medium and one small box and put this equal to \(340\)

\(p - 20 + p + 2 (p - 20) = 340\)

\(small + medium + large = total\)

To solve, first remove the bracket

\(p - 20 + p + 2 (p - 20) = 340\)

\(p - 20 + p + 2p - 40 = 340\)

Simplify \(4p - 60 = 340\)

\(4p - 60 + 60 = 340 + 60\)

\(4p = 400\)

\(p = 100\)

b. When \(p = 100\)

\(p - 20 = 80\)

A small box holds 8 paper cups.

Jaxon buys 3 of them \(3 \times 80 = 240\)

He buys 240 paper cups.

In this question, you must form an equation.

No marks would be given for just finding the value of \(p\) even if it is correct.