To substitute a number into an algebraic expression, all you need to do is re-write the expression in exactly the same way, except replacing the variable (letter) with the number. It always makes it clearer to put the number in brackets too. Then you can simplify your new expression and you have your answer!

Let's have a look at an example.

3x+7 where x = 5

So here, x is the variable and you are substituting in the number 5. All you do is write back the expression but with 5 instead of x. And don’t forget those brackets!

3(5) + 7

Now to simplify, just multiply out the brackets and add the 7:

3 x 5 + 7

= 15 + 7

= 22

And that’s your answer!

How about a harder example:

2y² – 3y + 4 where y = 2

You do this one in exactly the same way, but this time the variable is y and the number is 2. So, write back the expression with 2 (in brackets!) instead of y:

2(2)²– 3(2) + 4

And simplify… remember, always do the brackets first:

2 x 2²– 3 x 2 + 4

= 2 x 4 – 3 x 2 + 4

Multiplication always comes before addition and subtraction:

= 8 – 6 + 4

= 6

One more example, this time with two variables.

4x²– y²+ 2xy where x = - 3 and y = 4

You have to be a bit more careful with this one – make sure you include the negative sign in the brackets when you replace x with (-3) and instead of y, write 4.

4(-3)²– (4)²+ 2(-3)(4)

Start simplifying…

= 4 x (-3)²– (4)²+ 2 x -3 x 4

Remember when you square a negative number you get a positive solution:

= 4 x 9 – 16 + 2 x -3 x 4

= 36 – 16 + (-24)

= - 4

What is the Pythagoras Equation?

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