Kai buys 5 drinks and 3 cakes for £16.35 - OCR - GCSE Maths - Question 22 - 2021 - Paper 1

Let's manipulate the second equation to isolate $d$ in terms of $c$:

From $2d + 6c = 14.70$, we can rewrite it as:

$d = \frac{14.70 - 6c}{2}$

Now substitute this expression for $d$ back into the first equation:

$5\left(\frac{14.70 - 6c}{2}\right) + 3c = 16.35$

Now simplifying this:

$\frac{5(14.70) - 30c + 6c}{2} = 16.35$

Multiply through by 2 to eliminate the fraction:

$5(14.70) - 30c + 6c = 32.70$

Calculating $5(14.70)$ gives:

$73.5 - 24c = 32.70$

Now rearranging gives:

$73.5 - 32.70 = 24c$

$40.80 = 24c$

Solving for $c$, we find:

$c = \frac{40.80}{24} = 1.70$

Now that we have the cost of one cake ($c = 1.70$), we can substitute this back into the expression for $d$:

$d = \frac{14.70 - 6(1.70)}{2}$

This simplifies to:

$d = \frac{14.70 - 10.20}{2}$

$d = \frac{4.50}{2} = 2.25$