UK Secondary (7-11)
Substitute into algebraic expressions I
Lesson

In algebra, we use letters to represent unknown values, and we call these unknowns variables or variables. For example if we know that $x+2=5$x+2=5, then we can work out that $x=3$x=3 since we also know that $3+2=5$3+2=5.

Sometimes we want to do this process in reverse, however, and we substitute numbers into equations in place of variables to determine a final value. We can substitute in any kinds of numbers, including whole numbers, decimals and fractions.

#### Worked Example

##### Example 1

Evaluate: If $x=3$x=3, evaluate $6x-4$6x4.

Think: This means that everywhere the letter $x$x has been written, we will replace it with the number $3$3.

Do:

 $6x-4$6x−4 $=$= $6\times3-4$6×3−4 $=$= $18-4$18−4 $=$= $14$14

The same process applies even if there is more than one unknown value.

Evaluate: If $x=6$x=6 and $y=0.5$y=0.5, evaluate $6x-2y-12$6x2y12.

Think: Just like before, we will replace the letter $x$x with the number $6$6, and the letter $y$y with the number $0.5$0.5. We also need to keep the order of operations in mind when we do these kinds of calculations!

Do:

 $6x-2y-12$6x−2y−12 $=$= $6\times6-2\times0.5-12$6×6−2×0.5−12 replacing $x$x with $6$6, and $y$y with $0.5$0.5. $=$= $36-1-12$36−1−12 evaluating multiplication before subtraction. $=$= $23$23

Now let's watch some worked solutions to the following questions.

#### Practice Questions

##### Question 1

Evaluate $8x+4$8x+4 when $x=2$x=2.

##### Question 2

For  $x=10$x=10 and  $y=6$y=6, evaluate $8x+6y+4$8x+6y+4