Answer:
Biscuit: 25p
Cake: 40p
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.