A bakery has fixed costs of $200 per day and variable costs of $1.50 per loaf. Each loaf sells for $4.00. How many loaves must be sold to break even?

1. Let $n$ = number of loaves. Cost: $C = 200 + 1.5n$. Revenue: $R = 4n$.
2. Break-even means $C = R$: $200 + 1.5n = 4n$.
3. $200 = 2.5n$, so $n = 80$.
4. The bakery must sell $80$ loaves per day to break even.

Key idea: break-even is the intersection of the cost and revenue lines.

Solve $y = 2x + 1$ and $y = -x + 7$ by reading the graph above.

1. The lines cross at the point $(2, 5)$.
2. Check: $y = 2(2) + 1 = 5$ and $y = -(2) + 7 = 5$. Both equations satisfied.
3. Solution: $x = 2$, $y = 5$.

Solve $y = 3x - 1$ and $2x + y = 9$.

1. The first equation gives $y = 3x - 1$.
2. Substitute into the second: $2x + (3x - 1) = 9$.
3. $5x - 1 = 9$, so $5x = 10$, $x = 2$.
4. Back-substitute: $y = 3(2) - 1 = 5$.
5. Solution: $x = 2$, $y = 5$.
6. Check in second equation: $2(2) + 5 = 9$. Correct.

Solve $3x - 2y = 7$ and $x + y = 1$.

1. From the second equation: $x = 1 - y$.
2. Substitute into the first: $3(1 - y) - 2y = 7$.
3. $3 - 3y - 2y = 7$, so $-5y = 4$, $y = -\dfrac{4}{5}$.
4. $x = 1 - \left(-\dfrac{4}{5}\right) = \dfrac{9}{5}$.
5. Solution: $x = \dfrac{9}{5}$, $y = -\dfrac{4}{5}$.

Solve $3x + 2y = 16$ and $x - 2y = 0$.

1. The $y$-coefficients are $+2$ and $-2$ — they already cancel on addition.
2. Add the equations: $(3x + 2y) + (x - 2y) = 16 + 0$, so $4x = 16$, $x = 4$.
3. Substitute into equation 2: $4 - 2y = 0$, so $y = 2$.
4. Solution: $x = 4$, $y = 2$.

Solve $2x + 3y = 12$ and $5x + 2y = 11$.

1. To eliminate $y$, multiply equation 1 by $2$ and equation 2 by $3$:
   • $4x + 6y = 24$
   • $15x + 6y = 33$
2. Subtract: $(15x + 6y) - (4x + 6y) = 33 - 24$, so $11x = 9$, $x = \dfrac{9}{11}$.
3. Substitute into equation 1: $2\!\left(\dfrac{9}{11}\right) + 3y = 12$, so $\dfrac{18}{11} + 3y = 12$, $3y = \dfrac{114}{11}$, $y = \dfrac{38}{11}$.
4. Solution: $x = \dfrac{9}{11}$, $y = \dfrac{38}{11}$.

Solve $2x + 4y = 10$ and $x + 2y = 8$.

1. Multiply equation 2 by $2$: $2x + 4y = 16$.
2. But equation 1 says $2x + 4y = 10$.
3. $10 \neq 16$, so there is no solution — the lines are parallel.

A chemist mixes solution A (30% acid) with solution B (70% acid) to make 200 mL of 45% acid. How much of each is needed?

1. Let $a$ = mL of A and $b$ = mL of B.
2. Total volume: $a + b = 200$.
3. Acid content: $0.30a + 0.70b = 0.45 \times 200 = 90$.
4. From equation 1: $a = 200 - b$. Substitute: $0.30(200 - b) + 0.70b = 90$.
5. $60 - 0.30b + 0.70b = 90$, so $0.40b = 30$, $b = 75$.
6. $a = 200 - 75 = 125$.
7. The chemist needs $125$ mL of A and $75$ mL of B.

A company makes widgets and gadgets. Each widget costs $5 to produce and sells for $12. Each gadget costs $8 to produce and sells for $15. Fixed costs are $1400 per week. The company makes $w$ widgets and $g$ gadgets. If it makes twice as many widgets as gadgets and wants to break even, find $w$ and $g$.

1. Profit per widget: $12 - 5 = 7$. Profit per gadget: $15 - 8 = 7$.
2. Break-even: $7w + 7g = 1400$, i.e. $w + g = 200$.
3. Constraint: $w = 2g$.
4. Substitute: $2g + g = 200$, so $3g = 200$, $g = \dfrac{200}{3} \approx 66.7$.
5. Since production must be whole numbers: $g = 67$, $w = 134$ (with slight profit), or $g = 66$, $w = 132$ (slight loss). The exact break-even requires $g = 66\tfrac{2}{3}$.

Key idea: mathematical solutions sometimes need practical adjustment.