1)Let \(x\) be the adult ticket price and \(y\) be the student ticket price.
Start with \(3x + 2y = 44\) and \(2x + 4y = 40\).
Eliminate \(y\): multiply the first equation by \(4\) and the second equation by \(2\), then subtract.
\((3)(4)x - (2)(2)x = (44)(4) - (40)(2)\), so \(8x = 96\).
Divide to get \(x = 12\).
Substitute into the first equation: \(3(12) + 2y = 44\), so \(36 + 2y = 44\) and \(y = 4\).
Answer: \((x, y) = (12, 4)\).
Adult tickets cost $12, and student tickets cost $4.
2)Let \(x\) be the pretzel price and \(y\) be the drink price.
Start with \(4x + 3y = 25\) and \(2x + 5y = 23\).
Eliminate \(y\): multiply the first equation by \(5\) and the second equation by \(3\), then subtract.
\((4)(5)x - (2)(3)x = (25)(5) - (23)(3)\), so \(14x = 56\).
Divide to get \(x = 4\).
Substitute into the first equation: \(4(4) + 3y = 25\), so \(16 + 3y = 25\) and \(y = 3\).
Answer: \((x, y) = (4, 3)\).
A pretzel costs $4, and a drink costs $3.
3)Let \(x\) be chickens and \(y\) be goats.
Start with \(x + y = 18\) and \(2x + 4y = 50\).
Eliminate \(y\): multiply the first equation by \(4\) and the second equation by \(1\), then subtract.
\((1)(4)x - (2)(1)x = (18)(4) - (50)(1)\), so \(2x = 22\).
Divide to get \(x = 11\).
Substitute into the first equation: \(1(11) + 1y = 18\), so \(11 + 1y = 18\) and \(y = 7\).
Answer: \((x, y) = (11, 7)\).
There are \(11\) chickens and \(7\) goats.
4)Let \(x\) be the small-shirt price and \(y\) be the large-shirt price.
Start with \(5x + 3y = 102\) and \(2x + 6y = 108\).
Eliminate \(y\): multiply the first equation by \(6\) and the second equation by \(3\), then subtract.
\((5)(6)x - (2)(3)x = (102)(6) - (108)(3)\), so \(24x = 288\).
Divide to get \(x = 12\).
Substitute into the first equation: \(5(12) + 3y = 102\), so \(60 + 3y = 102\) and \(y = 14\).
Answer: \((x, y) = (12, 14)\).
A small shirt costs $12, and a large shirt costs $14.
5)Let \(x\) be the first number and \(y\) be the second number.
Start with \(x + y = 31\) and \(3x + 2y = 82\).
Eliminate \(y\): multiply the first equation by \(2\) and the second equation by \(1\), then subtract.
\((1)(2)x - (3)(1)x = (31)(2) - (82)(1)\), so \(-1x = -20\).
Divide to get \(x = 20\).
Substitute into the first equation: \(1(20) + 1y = 31\), so \(20 + 1y = 31\) and \(y = 11\).
Answer: \((x, y) = (20, 11)\).
The numbers are \(20\) and \(11\).
6)Let \(x\) be cars and \(y\) be motorcycles.
Start with \(x + y = 42\) and \(4x + 2y = 128\).
Eliminate \(y\): multiply the first equation by \(2\) and the second equation by \(1\), then subtract.
\((1)(2)x - (4)(1)x = (42)(2) - (128)(1)\), so \(-2x = -44\).
Divide to get \(x = 22\).
Substitute into the first equation: \(1(22) + 1y = 42\), so \(22 + 1y = 42\) and \(y = 20\).
Answer: \((x, y) = (22, 20)\).
There are \(22\) cars and \(20\) motorcycles.
7)Let \(x\) be the muffin price and \(y\) be the bagel price.
Start with \(6x + 4y = 30\) and \(3x + 8y = 33\).
Eliminate \(y\): multiply the first equation by \(8\) and the second equation by \(4\), then subtract.
\((6)(8)x - (3)(4)x = (30)(8) - (33)(4)\), so \(36x = 108\).
Divide to get \(x = 3\).
Substitute into the first equation: \(6(3) + 4y = 30\), so \(18 + 4y = 30\) and \(y = 3\).
Answer: \((x, y) = (3, 3)\).
A muffin costs $3, and a bagel costs $3.
8)Let \(x\) be the base fee and \(y\) be the cost per mile.
Start with \(x + 5y = 18\) and \(x + 9y = 30\).
Eliminate \(y\): multiply the first equation by \(9\) and the second equation by \(5\), then subtract.
\((1)(9)x - (1)(5)x = (18)(9) - (30)(5)\), so \(4x = 12\).
Divide to get \(x = 3\).
Substitute into the first equation: \(1(3) + 5y = 18\), so \(3 + 5y = 18\) and \(y = 3\).
Answer: \((x, y) = (3, 3)\).
The base fee is $3, and the rate is $3 per mile.
9)Let \(x\) be the calculator price and \(y\) be the notebook price.
Start with \(7x + 10y = 155\) and \(4x + 15y = 135\).
Eliminate \(y\): multiply the first equation by \(15\) and the second equation by \(10\), then subtract.
\((7)(15)x - (4)(10)x = (155)(15) - (135)(10)\), so \(65x = 975\).
Divide to get \(x = 15\).
Substitute into the first equation: \(7(15) + 10y = 155\), so \(105 + 10y = 155\) and \(y = 5\).
Answer: \((x, y) = (15, 5)\).
A calculator costs $15, and a notebook costs $5.
10)Let \(x\) be nickels and \(y\) be dimes.
Start with \(x + y = 46\) and \(5x + 10y = 340\).
Eliminate \(y\): multiply the first equation by \(10\) and the second equation by \(1\), then subtract.
\((1)(10)x - (5)(1)x = (46)(10) - (340)(1)\), so \(5x = 120\).
Divide to get \(x = 24\).
Substitute into the first equation: \(1(24) + 1y = 46\), so \(24 + 1y = 46\) and \(y = 22\).
Answer: \((x, y) = (24, 22)\).
There are \(24\) nickels and \(22\) dimes.
11)Let \(x\) be length and \(y\) be width.
Start with \(2x + 2y = 58\) and \(x - y = 5\).
Eliminate \(y\): multiply the first equation by \(-1\) and the second equation by \(2\), then subtract.
\((2)(-1)x - (1)(2)x = (58)(-1) - (5)(2)\), so \(-4x = -68\).
Divide to get \(x = 17\).
Substitute into the first equation: \(2(17) + 2y = 58\), so \(34 + 2y = 58\) and \(y = 12\).
Answer: \((x, y) = (17, 12)\).
The length is \(17\) inches, and the width is \(12\) inches.
12)Let \(x\) be the balcony price and \(y\) be the floor price.
Start with \(12x + 8y = 680\) and \(5x + 10y = 550\).
Eliminate \(y\): multiply the first equation by \(10\) and the second equation by \(8\), then subtract.
\((12)(10)x - (5)(8)x = (680)(10) - (550)(8)\), so \(80x = 2400\).
Divide to get \(x = 30\).
Substitute into the first equation: \(12(30) + 8y = 680\), so \(360 + 8y = 680\) and \(y = 40\).
Answer: \((x, y) = (30, 40)\).
A balcony ticket costs $30, and a floor ticket costs $40.
13)Let \(x\) be the amount at \(4\%\) and \(y\) be the amount at \(7\%\).
Start with \(x + y = 900\) and \(4x + 7y = 5100\).
Eliminate \(y\): multiply the first equation by \(7\) and the second equation by \(1\), then subtract.
\((1)(7)x - (4)(1)x = (900)(7) - (5100)(1)\), so \(3x = 1200\).
Divide to get \(x = 400\).
Substitute into the first equation: \(1(400) + 1y = 900\), so \(400 + 1y = 900\) and \(y = 500\).
Answer: \((x, y) = (400, 500)\).
$400 is invested at \(4\%\), and $500 is invested at \(7\%\).
14)Let \(x\) be boat speed and \(y\) be current speed.
Start with \(x + y = 18\) and \(x - y = 8\).
Eliminate \(y\): multiply the first equation by \(-1\) and the second equation by \(1\), then subtract.
\((1)(-1)x - (1)(1)x = (18)(-1) - (8)(1)\), so \(-2x = -26\).
Divide to get \(x = 13\).
Substitute into the first equation: \(1(13) + 1y = 18\), so \(13 + 1y = 18\) and \(y = 5\).
Answer: \((x, y) = (13, 5)\).
The boat speed is \(13\) mph, and the current speed is \(5\) mph.
15)Let \(x\) be liters of \(20\%\) solution and \(y\) be liters of \(50\%\) solution.
Start with \(x + y = 30\) and \(20x + 50y = 1140\).
Eliminate \(y\): multiply the first equation by \(50\) and the second equation by \(1\), then subtract.
\((1)(50)x - (20)(1)x = (30)(50) - (1140)(1)\), so \(30x = 360\).
Divide to get \(x = 12\).
Substitute into the first equation: \(1(12) + 1y = 30\), so \(12 + 1y = 30\) and \(y = 18\).
Answer: \((x, y) = (12, 18)\).
Use \(12\) liters of \(20\%\) solution and \(18\) liters of \(50\%\) solution.
16)Let \(x\) be regular tickets and \(y\) be premium tickets.
Start with \(x + y = 85\) and \(18x + 30y = 1830\).
Eliminate \(y\): multiply the first equation by \(30\) and the second equation by \(1\), then subtract.
\((1)(30)x - (18)(1)x = (85)(30) - (1830)(1)\), so \(12x = 720\).
Divide to get \(x = 60\).
Substitute into the first equation: \(1(60) + 1y = 85\), so \(60 + 1y = 85\) and \(y = 25\).
Answer: \((x, y) = (60, 25)\).
They sold \(60\) regular tickets and \(25\) premium tickets.
17)Let \(x\) be \(2\)-point questions and \(y\) be \(5\)-point questions.
Start with \(x + y = 25\) and \(2x + 5y = 83\).
Eliminate \(y\): multiply the first equation by \(5\) and the second equation by \(1\), then subtract.
\((1)(5)x - (2)(1)x = (25)(5) - (83)(1)\), so \(3x = 42\).
Divide to get \(x = 14\).
Substitute into the first equation: \(1(14) + 1y = 25\), so \(14 + 1y = 25\) and \(y = 11\).
Answer: \((x, y) = (14, 11)\).
There are \(14\) two-point questions and \(11\) five-point questions.
18)Let \(x\) be Printer A rate and \(y\) be Printer B rate.
Start with \(6x + 6y = 570\) and \(4x + 3y = 320\).
Eliminate \(y\): multiply the first equation by \(3\) and the second equation by \(6\), then subtract.
\((6)(3)x - (4)(6)x = (570)(3) - (320)(6)\), so \(-6x = -210\).
Divide to get \(x = 35\).
Substitute into the first equation: \(6(35) + 6y = 570\), so \(210 + 6y = 570\) and \(y = 60\).
Answer: \((x, y) = (35, 60)\).
Printer A prints \(35\) pages per minute, and Printer B prints \(60\) pages per minute.
19)Let \(x\) be the basic laptop price and \(y\) be the advanced laptop price.
Start with \(4x + 3y = 4700\) and \(2x + 5y = 5500\).
Eliminate \(y\): multiply the first equation by \(5\) and the second equation by \(3\), then subtract.
\((4)(5)x - (2)(3)x = (4700)(5) - (5500)(3)\), so \(14x = 7000\).
Divide to get \(x = 500\).
Substitute into the first equation: \(4(500) + 3y = 4700\), so \(2000 + 3y = 4700\) and \(y = 900\).
Answer: \((x, y) = (500, 900)\).
A basic laptop costs $500, and an advanced laptop costs $900.
20)Let \(x\) be still-water speed and \(y\) be current speed.
Start with \(x + y = 21\) and \(x - y = 13\).
Eliminate \(y\): multiply the first equation by \(-1\) and the second equation by \(1\), then subtract.
\((1)(-1)x - (1)(1)x = (21)(-1) - (13)(1)\), so \(-2x = -34\).
Divide to get \(x = 17\).
Substitute into the first equation: \(1(17) + 1y = 21\), so \(17 + 1y = 21\) and \(y = 4\).
Answer: \((x, y) = (17, 4)\).
The still-water speed is \(17\) mph, and the current speed is \(4\) mph.