1. **Problem Statement:** The company produces two gadgets, x and y. We have two equations: Revenue: $$4x + 5y = 200$$ Cost: $$3x + 2y = 100$$ We need to find the values of x and y that satisfy both. 2. **Formula and Rules:** We solve simultaneous linear equations using substitution or elimination. 3. **Step A: Solve the system** Multiply the cost equation by 5: $$5(3x + 2y) = 5(100) \Rightarrow 15x + 10y = 500$$ Multiply the revenue equation by 2: $$2(4x + 5y) = 2(200) \Rightarrow 8x + 10y = 400$$ Subtract the second from the first: $$ (15x + 10y) - (8x + 10y) = 500 - 400 \Rightarrow 7x = 100 \Rightarrow x = \frac{100}{7} \approx 14.29$$ Substitute $x$ into revenue equation: $$4(\frac{100}{7}) + 5y = 200 \Rightarrow \frac{400}{7} + 5y = 200$$ $$5y = 200 - \frac{400}{7} = \frac{1400 - 400}{7} = \frac{1000}{7}$$ $$y = \frac{1000}{35} = \frac{200}{7} \approx 28.57$$ 4. **Step B: Verify solution** Revenue: $$4(14.29) + 5(28.57) = 57.16 + 142.85 = 200.01 \approx 200$$ Cost: $$3(14.29) + 2(28.57) = 42.87 + 57.14 = 100.01 \approx 100$$ Values satisfy both equations. 5. **Step C: Business meaning** The company should produce approximately 14 units of Gadget x and 29 units of Gadget y weekly to meet revenue and cost targets. 6. **Step D: New revenue target 250** New system: $$4x + 5y = 250$$ $$3x + 2y = 100$$ Multiply cost by 5: $$15x + 10y = 500$$ Multiply revenue by 2: $$8x + 10y = 500$$ Subtract: $$7x = 0 \Rightarrow x = 0$$ Substitute into revenue: $$4(0) + 5y = 250 \Rightarrow y = 50$$ **Final answer:** - Original solution: $$x = \frac{100}{7} \approx 14.29$$, $$y = \frac{200}{7} \approx 28.57$$ - New solution with revenue 250: $$x=0$$, $$y=50$$