1. **Problem 1: Children and Adults in a Group** We have two variables: $x$ = number of children, $y$ = number of adults. Given equations: $$x + y = 2200$$ $$1.5x + 4y = 5050$$ **Step 1:** From the first equation, express $y$ in terms of $x$: $$y = 2200 - x$$ **Step 2:** Substitute $y$ into the second equation: $$1.5x + 4(2200 - x) = 5050$$ **Step 3:** Expand and simplify: $$1.5x + 8800 - 4x = 5050$$ **Step 4:** Combine like terms: $$\cancel{1.5x} - \cancel{4x} = -2.5x$$ $$-2.5x + 8800 = 5050$$ **Step 5:** Subtract 8800 from both sides: $$-2.5x = 5050 - 8800$$ $$-2.5x = -3750$$ **Step 6:** Divide both sides by $-2.5$: $$x = \frac{-3750}{-2.5} = 1500$$ **Step 7:** Find $y$: $$y = 2200 - 1500 = 700$$ **Answer:** There are 1500 children and 700 adults. 2. **Problem 2: Investment in Two Accounts** Variables: $x$ = amount invested at 6%, $y$ = amount invested at 10%. Given: - $x = 2y$ (twice as much in the lower-yielding account) - Total interest = 3520 **Step 1:** Write the interest equation: $$0.06x + 0.10y = 3520$$ **Step 2:** Substitute $x = 2y$: $$0.06(2y) + 0.10y = 3520$$ **Step 3:** Simplify: $$0.12y + 0.10y = 3520$$ $$0.22y = 3520$$ **Step 4:** Solve for $y$: $$y = \frac{3520}{0.22} = 16000$$ **Step 5:** Find $x$: $$x = 2 \times 16000 = 32000$$ **Answer:** Invested 32000 at 6% and 16000 at 10%. 3. **Problem 3: Angles of a Triangle** Variables: $x$ = largest angle, $y$ and $z$ = other two angles. Given: - $x = y + z$ - $x + y + z = 180$ - $2y = x - 10$ **Step 1:** Substitute $x = y + z$ into the sum: $$(y + z) + y + z = 180$$ $$2y + 2z = 180$$ **Step 2:** Divide both sides by 2: $$y + z = 90$$ **Step 3:** From $x = y + z$, we have $x = 90$. **Step 4:** Use $2y = x - 10$: $$2y = 90 - 10 = 80$$ $$y = 40$$ **Step 5:** Find $z$: $$y + z = 90$$ $$40 + z = 90$$ $$z = 50$$ **Answer:** Largest angle $x = 90^6$, other angles $y = 40^6$, $z = 50^6$.