throwback maths 4

A simultaneous equation!

throwback simultaneous
Solution

To solve, let $c$ denote the number of coffees and $t$ denote the number of teas.

Then from monday, we have

$$3c + 2t = 8.97$$

From Tuesday, we have

$$2c + 3t = 8.48$$

The most efficient method for solving the problem is to now use the elimination method to determine the values of $c$ and $t$.

From Mondays statement, we can get $9c + 6t = 26.91$ (by multiplying both sides of the equation by 3). From Tuesdays statement, we can get $4c + 6t = 16.98$ (by multiplying both sides of the equation by 2).

subtraction of one equation from the other gives $5c = 9.95$ and hence $c = 1.99$.

We now have $3(1.99) + 2t = 8.97$ which gives $2t = 3$, so $t = 1.50$

To answer the question, $5c + t = 5(1.99) + 1.50 = 11.45$. The cost of five coffees and a tea is $£11.45$