A cake costs 15p more than a biscuit.
Megan bought a cake and two biscuits for 90p.
How much do a cake and a biscuit each cost?

Biscuit: 25p

Cake: 40p

Step-by-step explanation:

You can represent the problem using a system of equations.

State your variables.

let c be the cost of a cake

let b be the cost of a biscuit

c = (15p) + b         A cake is 15p more than a biscuit

c + 2b = (90p)         Megan bought a cake and two biscuits for 90p

I put brackets around 15p and 9p so the units won't be confused for variables.

Solve using the method substitution, where you replace a variable with an equivalent equation. Since "c" equals "(15p) + b" and there is "c" in the second equation, you can replace it.

c + 2b = (90p)                        Take the second equation

(15p) + b + 2b = (90p)                 Replace "c" with the first expression

(15p) + 3b = (90p)                 Combine like terms (b + 2b = 3b)

Start isolating "b"

(15p) - (15p) + 3b = (90p) - (15p)                  Subtract 15p from both sides

3b = (75p)                      

3b/3 = (75p)/3                       Divide both sides by 3

b = (25p)            Value of "b", Cost of one biscuit

Now we can substitute "b" for 25p in any of the equations to find "c".

c = (15p) + b              Take the first equation

c = (15p) + (25p)                Replace "b" with its value

c = (40p)                 Value of "c", Cost of one cake

Therefore a biscuit costs 25p and a cake costs 40p.

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.

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