Substitution
Substitution is the name given to the process of swapping an algebraic letter for its value. Consider the expression 8\({z}\) + 4. This can take on a range of values depending on what number \({z}\) actually is.
If we are told \({z}\) = 5, we can work out the value of the expression by swapping the \({z}\) for the number 5. (Remember that 8\({z}\) means 8 multiply \({z}\)).
8\({z}\) + 4 = 8 × 5 + 4 = 44
When \({z}\) = 5, 8\({z}\) + 4 = 44
Example
Calculate the value of 3\({a}\) + 4\({b}\) when \({a}\) = 2 and \({b}\) = 6.
Solution
Firstly, we substitute the values we already know into the expression:
3\({a}\) + 4\({b}\) = 3 × 2 + 4 × 6
Remembering BODMASThis is the order in which arithmetic operations must be carried out. Brackets - Of - Divide - Multiply - Add - Subtract.:
3 × 2 + 4 × 6 = 6 + 24 = 30
Substituting negative numbers
We are also expected to be able to substitute negative numbers into expressions. The method is the same but once again, we must be careful with our signs.
Example
Find the value of 4\({i}\) – 2\({p}\) when \({i}\) = 3 and \({p}\) = -5.
Substituting the values into the formula gives:
(4 × 3) – (2 × -5) = 12 – -10 = 12 + 10 = 22
Question
Find the value of 8\({y}\) – 4\({z}\) when \({y}\) = 3 and \({z}\) = 7.
Substituting the values into the equation gives:
8\({y}\) – 4\({z}\) = (8 × 3) – (4 × 7)
Evaluating the terms leaves: 24 – 28 = -4
Question
Find the value of -2\({z}\) -4\({p}\) when \({z}\) = 3 and \({p}\) = -8.
Substituting the values into the expression gives:
-2\({z}\) – 4\({p}\) = (-2 × 3) + (-4 × -8)
Evaluating the terms leaves: -6 + 32 = 26
Equations written in words
Equations can also be written in words.
Example
Mr Loynd is having a party. He buys \({m}\) boxes of soft drink and \({n}\) packs of soft drink. Each box contains 18 cans and each pack contains 6 cans. Write an expression for the total number of cans of soft drink Mr Loynd has bought.
Solution
To solve this, we must break down the question into parts. We will consider the boxes first.
We know that he buys \({m}\) boxes, and that each box contains 18 cans. So the total number of cans he has from the boxes will be 18 × \({m}\) = 18\({m}\).
The packs contain 6 cans and there are \({n}\) of them: 6 × \({n}\) = 6\({n}\).
Adding these together, we obtain the expression 18\({m}\) + 6\({n}\) for the total number of cans of soft drink.
If we are later told that \({m}\) = 3 and \({n}\) = 2, we can evaluate the expression by using substitution:
18\({m}\) + 6\({n}\) when \({m}\) = 3 and \({n}\) = 2 gives (18 × 3) + (6 × 2) = 54 + 12 = 66.
Question
Mrs Scott is planting bulbs in her garden. Each row contains 6 bulbs and she plants \({p}\) rows. She also plants some bulbs in pots. She has \({q}\) pots which can each hold 2 bulbs. Write an expression for the number of bulbs Mrs Scott plants.
Each row contains 6 bulbs and there are \({p}\) rows so 6 × \({p}\) = 6\({p}\)
Each pot contains 2 bulbs and there are \({q}\) pots so 2 × \({q}\) = 2\({q}\)
Adding these together gives:
6\({p}\) + 2\({q}\)
Question
Mrs Scott decides to plant 4 rows of bulbs and also uses 5 pots. How many bulbs does she plant?
Using 6\({p}\) + 2\({q}\) and substituting \({p}\) = 4 and \({q}\) = 5 gives 6 × 4 + 2 × 5
Evaluating leaves:
24 + 10 = 34
More guides on this topic
- Equations of lines – WJEC
- Equations of curves - Intermediate & Higher tier – WJEC
- Equations and formulae – WJEC
- Factorising - Intermediate & Higher tier – WJEC
- Quadratic expressions - Intermediate & Higher tier – WJEC
- Sequences – WJEC
- Functions - Higher only – WJEC
- Inequalities - Intermediate & Higher tier – WJEC
- Simultaneous equations - Intermediate & Higher tier – WJEC