Setting up and solving systems from context

Step 1 — Define variables

Let $c$ = price of one coffee ($) and $m$ = price of one muffin ($).

Step 2 — Write one equation per fact

$3c + 2m = 23 \quad (1)$ $2c + 4m = 26 \quad (2)$

Step 3 — Solve

Halve (2): $c + 2m = 13$, so $c = 13 - 2m$. Substitute into (1):

$3(13 - 2m) + 2m = 23 \;\Rightarrow\; 39 - 6m + 2m = 23 \;\Rightarrow\; -4m = -16 \;\Rightarrow\; m = 4$

Then $c = 13 - 2(4) = 5$.

Step 4 — Answer in context

Step 1 — Define variables

Let $a$, $s$, $c$ be the numbers of adult, student and child tickets.

Step 2 — Write the three equations

$a + s + c = 100 \quad (1)$ $15a + 10s + 5c = 1000 \quad (2)$ $s = c \quad (3)$

Step 3 — Substitute (3) and simplify

Put $s = c$ into (1) and (2):

$a + 2c = 100 \quad (1')$ $15a + 15c = 1000 \;\Rightarrow\; 3a + 3c = 200 \ldots$

Divide (2) by 5: $3a + 2s + c = 200$, and with $s = c$: $3a + 3c = 200 \quad (2')$.

Step 4 — Solve (1′) and (2′)

From (1′): $a = 100 - 2c$. Substitute into (2′):

$3(100 - 2c) + 3c = 200 \;\Rightarrow\; 300 - 6c + 3c = 200 \;\Rightarrow\; -3c = -100 \;\Rightarrow\; c = \tfrac{100}{3}$

$c \approx 33.3$, which is not a whole number of tickets.

Step 5 — Interpret