1. **Problem statement:** We have two equations from the costs of sodas and burgers: $$3s + 4b = 11$$ $$s + 7b = 15$$ where $s$ is the cost of one soda and $b$ is the cost of one burger. We want to find the cost of: a) 1 soda ($s$) b) 1 burger ($b$). 2. **Formula and rules:** We will use the method of substitution or elimination to solve this system of linear equations. 3. **Step 1: Express $s$ from the second equation:** $$s = 15 - 7b$$ 4. **Step 2: Substitute $s$ into the first equation:** $$3(15 - 7b) + 4b = 11$$ 5. **Step 3: Simplify:** $$45 - 21b + 4b = 11$$ $$45 - 17b = 11$$ 6. **Step 4: Isolate $b$:** $$-17b = 11 - 45$$ $$-17b = -34$$ 7. **Step 5: Divide both sides by $-17$:** $$b = \frac{-34}{-17}$$ $$b = 2$$ 8. **Step 6: Substitute $b=2$ back into $s = 15 - 7b$:** $$s = 15 - 7(2)$$ $$s = 15 - 14$$ $$s = 1$$ **Answer for problem 1:** a) Cost of 1 soda $s = 1$ b) Cost of 1 burger $b = 2$ --- 1. **Problem statement:** We have two equations from the costs of pens and erasers: $$2p + 3e = 9.50$$ $$4p + 4e = 16$$ where $p$ is the cost of one pen and $e$ is the cost of one eraser. We want to find the cost of a pen and a ruler. 2. **Note:** The problem asks for the cost of a pen and a ruler, but we only have pens and erasers here. The ruler cost will be found in the next problem. 3. **Step 1: Simplify the second equation by dividing both sides by 4:** $$\frac{4p + 4e}{\cancel{4}} = \frac{16}{\cancel{4}}$$ $$p + e = 4$$ 4. **Step 2: Express $e$ from this equation:** $$e = 4 - p$$ 5. **Step 3: Substitute $e$ into the first equation:** $$2p + 3(4 - p) = 9.50$$ 6. **Step 4: Simplify:** $$2p + 12 - 3p = 9.50$$ $$-p + 12 = 9.50$$ 7. **Step 5: Isolate $p$:** $$-p = 9.50 - 12$$ $$-p = -2.50$$ 8. **Step 6: Multiply both sides by $-1$:** $$p = 2.50$$ 9. **Step 7: Substitute $p=2.50$ back into $e = 4 - p$:** $$e = 4 - 2.50 = 1.50$$ **Answer for problem 2:** Cost of a pen $p = 2.50$ --- 1. **Problem statement:** We have two equations from the costs of rulers and pens: $$r + 3p = 10$$ $$3r + 3p = 12$$ where $r$ is the cost of one ruler and $p$ is the cost of one pen. We want to find: a) Cost of a ruler ($r$) b) Cost of 5 pens ($5p$) 2. **Step 1: Subtract the first equation from the second:** $$(3r + 3p) - (r + 3p) = 12 - 10$$ $$3r + 3p - r - 3p = 2$$ $$2r = 2$$ 3. **Step 2: Divide both sides by 2:** $$r = \frac{2}{2} = 1$$ 4. **Step 3: Substitute $r=1$ into the first equation:** $$1 + 3p = 10$$ 5. **Step 4: Isolate $p$:** $$3p = 10 - 1 = 9$$ 6. **Step 5: Divide both sides by 3:** $$p = \frac{9}{3} = 3$$ 7. **Step 6: Calculate cost of 5 pens:** $$5p = 5 \times 3 = 15$$ **Answer for problem 3:** a) Cost of a ruler $r = 1$ b) Cost of 5 pens $= 15$