Simultaneous equations and inequalities

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Formula

Explanation

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Worked solution

1. The system of equations is given: 120x + 200y = 3200 (budget constraint) and 800x + 1500y = 18000 (cargo constraint). We are to solve for x, the number of small vans, exactly as an integer.
2. Simplify both equations by dividing by the greatest common divisor of the coefficients. For the budget equation, divide all terms by 40: 120/40 = 3, 200/40 = 5, 3200/40 = 80. New equation: 3x + 5y = 80.
3. Simplify the cargo equation by dividing all terms by 100: 800/100 = 8, 1500/100 = 15, 18000/100 = 180. New equation: 8x + 15y = 180.
4. Use the elimination method. Multiply the simplified budget equation (3x + 5y = 80) by 3 to align the y-coefficient with the cargo equation: 3*(3x + 5y) = 3*80 → 9x + 15y = 240.
5. Subtract the cargo equation (8x + 15y = 180) from the modified budget equation (9x + 15y = 240): (9x + 15y) - (8x + 15y) = 240 - 180 → x = 60.
6. Substitute x = 60 into the simplified budget equation: 3(60) + 5y = 80 → 180 + 5y = 80.
7. Solve for y: 5y = 80 - 180 = -100 → y = -100 / 5 = -20.
8. The value y = -20 is not valid because the number of large trucks cannot be negative. This indicates that the solution must be re-examined for integer feasibility under constraints.
9. Re-evaluate the system: since y must be non-negative, test integer values of y starting from 0 to find a valid x that satisfies both equations and gives integer x.
10. Use the simplified budget equation: 3x + 5y = 80. Rearranged: 3x = 80 - 5y → x = (80 - 5y)/3. For x to be integer, (80 - 5y) must be divisible by 3.
11. Test values of y from 0 to 16 (since 5y ≤ 80). Try y = 4: 80 - 5*4 = 60; 60/3 = 20 → x = 20. Check in cargo equation: 8(20) + 15(4) = 160 + 60 = 220 ≠ 180. Too high. Try y = 8: 80 - 40 = 40; 40/3 ≈ 13.33 → not integer. Try y = 2: 80 - 10 = 70; 70/3 ≈ 23.33. Try y = 1: 80 - 5 = 75; 75/3 = 25 → x = 25. Check cargo: 8(25) + 15(1) = 200 + 15 = 215 ≠ 180. Try y = 5: 80 - 25 = 55; 55/3 ≈ 18.33. Try y = 7: 80 - 35 = 45; 45/3 = 15 → x = 15. Check cargo: 8(15) + 15(7) = 120 + 105 = 225 ≠ 180. Try y = 10: 80 - 50 = 30; 30/3 = 10 → x = 10. Check cargo: 8(10) + 15(10) = 80 + 150 = 230 ≠ 180. Try y = 14: 80 - 70 = 10; 10/3 ≈ 3.33. Try y = 16: 80 - 80 = 0; 0/3 = 0 → x = 0. Check cargo: 8(0) + 15(16) = 240 ≠ 180. No valid solution found? Recheck original equations.
12. Recheck original equations: 120x + 200y = 3200 and 800x + 1500y = 18000. Use elimination again with original coefficients. Multiply first equation by 15 and second by 2: 15*(120x + 200y) = 15*3200 → 1800x + 3000y = 48000; 2*(800x + 1500y) = 2*18000 → 1600x + 3000y = 36000. Subtract: (1800x + 3000y) - (1600x + 3000y) = 48000 - 36000 → 200x = 12000 → x = 60. Then substitute x = 60 into 120(60) + 200y = 3200 → 7200 + 200y = 3200 → 200y = 3200 - 7200 = -4000 → y = -20. Same result. But y = -20 is invalid. Therefore, no non-negative integer solution exists that satisfies both equations exactly. But the question asks for the exact integer value of x that meets the cargo requirement while staying within budget. Re-express: the company must transport at least 18,000 kg and stay under or at $3200. The equations are for exact match. But no valid (x,y) with y ≥ 0 satisfies both. Therefore, the only way to meet the cargo requirement exactly and stay within budget is to find integer x such that for some y ≥ 0, both constraints hold and cargo = 18000. Try x = 15: 800*15 = 12000; 18000 - 12000 = 6000; 6000 / 1500 = 4 → y = 4. Check cost: 120*15 = 1800; 200*4 = 800; total = 2600 ≤ 3200. Valid. But not exact match. Try x = 0: 18000 / 1500 = 12 → y = 12; cost = 200*12 = 2400 ≤ 3200. Valid. But not exact. Try x = 10: 8000 kg; 10000 left; 10000 / 1500 ≈ 6.66 → y = 6 or 7. y = 6: 9000; total 17000 < 18000. y = 7: 10500; total 18500 > 18000. Cost: 120*10 = 1200; 200*7 = 1400; total 2600 ≤ 3200. But cargo exceeds. Try x = 12: 9600 kg; 8400 left; 8400 / 1500 = 5.6 → y = 5: 7500; total 17100 < 18000. y = 6: 9000; total 18600 > 18000. Cost: 120*12 = 1440; 200*6 = 1200; total 2640 ≤ 3200. But not exact. Try x = 15, y = 4: cargo 12000 + 6000 = 18000; cost 1800 + 800 = 2600 ≤ 3200. This satisfies both constraints exactly for cargo and within budget. But the system says 800x + 1500y = 18000 and 120x + 200y = 3200. Plug x = 15, y = 4: 120*15 = 1800; 200*4 = 800; sum 2600 ≠ 3200. So not on budget line. But the question says: 'the company wants to determine the exact number of small vans (x) that must be used to exactly meet the cargo requirement while staying within the daily budget'. So it does not require the budget to be exactly $3200, only to stay within it. Therefore, x = 15, y = 4 is valid. But is there a solution where cargo = 18000 and cost ≤ 3200, and x is integer? Yes. But the system of equations given is for exact budget and exact cargo. But no solution with y ≥ 0. Therefore, the only way to satisfy the question is to find x such that for some y ≥ 0, 800x + 1500y = 18000 and 120x + 200y ≤ 3200. But the question says 'solve exactly for x and provide the exact integer value' using the given system. But the system has no non-negative solution. Therefore, the only way to resolve this is to realise that the system is inconsistent under non-negative constraints. But the question says 'solve exactly for x' using the system. The algebraic solution gives x = 60, y = -20. But y = -20 is invalid. Therefore, no valid solution exists. But the question asks for the exact integer value. So either the question is flawed, or we must accept x = 60 as the algebraic solution even if y is negative. But the context says 'must transport at least 18,000 kg' and 'staying within budget'. So the system is for exact match, but it's not feasible. However, the question says: 'solve exactly for x and provide the exact integer value'. So it wants the algebraic solution, even if not feasible. Therefore, x = 60 is the solution to the system, even if not realistic.

Reference answer: 60

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