Methods of Solving Equations
An equation has a letter, an ‘equals' sign and has two sides, as in this example:
To solve an equation we find the value of the letter.
Method
Remove all the terms from one side of the equation, leaving the letter by itself.
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y + 5 = 7 |
(-5 from both sides) |
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To remove the + 5 we subtract 5. To keep the two sides equal, we must take 5 from both sides.
So
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y = 7 – 5
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y = 2 is the solution to the equation. |
Example 1: solve y + 8 = 11
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(– 8 from both sides)
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y = 11 – 8
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y = 3
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Example 2: solve y + 2 = 7
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(– 2 from both sides)
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y = 7 – 2
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y = 5
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Note: a quick way of solving this equation is to move the +2 to the other side and change its sign to –2.
Example 3:
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y + 10 = 14
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(-10 from both sides)
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y = 14 – 10
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y = 4 |
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Example 4:
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y - 6 = 2
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(+ 6 to both sides)
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y = 2 + 6
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Note: we add 6 in this case
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y = 8
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Example 5:
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3y = 15
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(divide both sides by 3)
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y = 15 ÷ 3
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We change the multiply by 3 to divide by 3.
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y = 5 |
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Example 6:
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y = 2
7
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(multiply both sides by 7)
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y = 2 x 7
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We change the divide by 7 to multiply by 7.
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y = 14 |
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General Rule
An important general rule is that if we move a term to the other side of the equation we change its sign.
If there is more than one term to move, always move the + or – term first.
| Example 1: solve 2x + 5 = 15 |
(–5 from both sides) |
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2x = 15 - 5
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2x = 10
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(then divide by 2)
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x = 10 ÷ 2
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x = 5
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| Example 2: solve |
x – |
4 = 2 |
(+ 4 to both sides) |
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3 |
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x = 2 + 4 |
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3 |
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x = 6 |
(multiply both sides by 3) |
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3
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x = 6 x 3
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x = 18 |
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Brackets
Brackets are used to group terms together.
If we want to remove brackets, then everything inside the bracket must be multiplied by the term on the outside.
3 (y + 2) = 3 x y + 3 x 2 = 3y + 6
Both y and the + 2 must be multiplied by 3.
5 (y – 3) = 5 x y – 5 x 3
= 5y – 15
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Equations with Brackets
Example 1: solve 5 (y – 3) = 20
Remove the brackets
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5y – 15 = 20 |
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5y = 20 + 15 |
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5y = 35 |
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y = 35 divided by 5 |
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y = 7 |
| Example 2: solve |
p + 4 = 5 |
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Note: this line brackets p + 4 together. So we cannot (- 4) first.
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p + 4 = 5 |
(multiply by 3)
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p + 4 = 5 x 3
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p + 4 = 15
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(– 4 from both sides) |
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p = 15 – 4
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p = 11 |
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Equations with Letters on Both Sides
Example 1: solve
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3 (y -1) = y + 7
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3y – 3 = y + 7
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(+ 3 to both sides)
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3y = y + 10
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(– y from both sides)
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The y terms must be brought to the same side and simplified. |
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3y – y = 10
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2y = 10
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(divide by 2)
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y = 5
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Example 2: solve
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4 (p + 2) = 18 – p
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(add p to both sides)
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5p + 8 = 18
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(– 8 from both sides)
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5p = 10
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(divide by 5)
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p = 2 |
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