7.03 Substitution

After building an algebraic expression we can solve it by substitution, where we replace variables with numeric values.

Worked examples

EXAMPLE 1

Consider the following equation.

$P=6Q+2$P=6Q+2

What is the value of $P$P if $Q=4$Q=4?

Think: We are trying to solve the equation for $P$P and we are given the value of $Q$Q so we can substitute in $4$4 for $Q$Q to find the value of $P$P.

Do: Wherever $Q$Q appears in the equation we replace it with it's value of $4$4:

$P$P	$=$=	$6Q+2$6Q+2
$P$P	$=$=	$6\times4+2$6×4+2
$P$P	$=$=	$26$26

Reflect: The process of substitution is putting a number where the variable is, we can do this because the variable is equivalent to the number.

We had two unknown variables and we were given the value of one, which we used to find the value of the other. Notice that this value of $P$P is only when $Q=4$Q=4. As $Q$Q changes so does $P$P, so we can also think of this problem in terms of a table of values.

EXAMPLE 2

Inside a room there are $4$4 dogs and $3$3 birds. The dogs have $4$4 legs each, and the birds have $2$2 legs each. How many legs in total are there in this room?

Think: We create an expression relating the number of dogs, the number of birds, and the number of legs. Then we can substitute values into our expression to find the total.

Do: Let the total number of legs be $T$T, the number of dogs be $D$D, and the number of birds be $B$B.

Every dog will contribute $4$4 legs to the total and every bird will add $2$2, so our equation will be:

$T=4D+2B$T=4D+2B

Let's substitute $D=4$D=4 and $B=3$B=3 into the the equation above. In other words, replace $D$D with $4$4 and $B$B with $3$3:

$T$T	$=$=	$4D+2B$4D+2B
$T$T	$=$=	$4\times4+2\times3$4×4+2×3
$T$T	$=$=	$22$22

Reflect: We are now working with two variables. An algebraic expression can contain any number of variables and we can substitute any number of these values in at once.

Practice questions

Question 1

Question 2

Question 3