Modelling with systems of equations Problem set A Worked solutions

Three customers shop at a farm shop. Kai buys 3 jars of honey, 2 loaves and 1 block of cheese and pays $23.00. Lena buys 1 jar of honey, 4 loaves and 2 blocks of cheese and pays $22.00. Mo buys 2 jars of honey, 1 loaf and 3 blocks of cheese and pays $23.65. Find the cost of one loaf.

• Let \(x\) = cost of one jar of honey ($), \(y\) = cost of one loaf ($), \(z\) = cost of one block of cheese ($).
• \(3x + 2y + z=23\)
• \(x + 4y + 2z=22\)
• \(2x + y + 3z=23.65\)
• Simultaneous equation solver: x = 4.80, y = 2.35, z = 3.90
• One loaf costs $2.35.

Solve the system and find all possible values of \(x\), \(y\) and \(z\): \(x^2 + y^2 + z^2=53\), \(2x^2 + y^2 - z^2=-3\), \(x^2 - 2y^2 + 3z^2=122\).

• Let \(p=x^2\), \(q=y^2\), \(r=z^2\). The system is linear in \(p, q, r\).
• \(p + q + r=53\)
• \(2p + q - r=-3\)
• \(p - 2q + 3r=122\)
• Simultaneous equation solver: p = 16, q = 1, r = 36
• Take square roots, positive and negative: x = ±4, y = ±1, z = ±6

Jamal invests $15 000 in three accounts for one year. Account X pays 2% per year, account Y pays 3% per year and account Z pays 5% per year. There is $2 000 more in account Z than in account X. The total interest earned over the year is $530.00. How much is in account Y?

• Let \(x\) = amount in account X ($), \(y\) = amount in account Y ($), \(z\) = amount in account Z ($).
• \(x + y + z=15000\)
• \(0.02x + 0.03y + 0.05z=530\)
• \(-x + z=2000\)
• Simultaneous equation solver: x = 4000, y = 5000, z = 6000
• The amount in account Y is $5 000.

A pottery makes bowls, mugs and vases. A bowl uses 600 g of clay and 40 g of glaze. A mug uses 350 g of clay and 25 g of glaze. A vase uses 900 g of clay and 70 g of glaze. The bowls sell for $18.00 each, the mugs for $9.00 each and the vases for $35.00 each. In one day all the ingredients used are 21.4 kg of clay and 1.54 kg of glaze, and the total sales are $676.00. How many mugs are made?

• Let \(x\) = number of bowls, \(y\) = number of mugs, \(z\) = number of vases.
• Work in grams: 21.4 kg = 21 400 g and 1.54 kg = 1 540 g.
• \(600x + 350y + 900z=21400\)
• \(40x + 25y + 70z=1540\)
• \(18x + 9y + 35z=676\)
• Simultaneous equation solver: x = 12, y = 20, z = 8
• 20 mugs are made.

At a shop one bike costs the same as 4 scooters, and the same as 10 skateboards. 2 bikes and 6 scooters cost $120 more than 25 skateboards. Find the total cost of one bike, one scooter and one skateboard.

• Let \(p\) = cost of one bike ($), \(q\) = cost of one scooter ($), \(r\) = cost of one skateboard ($).
• \(p-4q=0\) (one bike is 4 scooters)
• \(p-10r=0\) (one bike is 10 skateboards)
• \(2p + 6q - 25r=120\)
• Simultaneous equation solver: p = 120, q = 30, r = 12
• Total = 120 + 30 + 12 = $162

A three-digit number has digits \(a\), \(b\) and \(c\), in that order, so the number is \(100a+10b+c\). The digits add up to 10. The middle digit is half of the first digit. When the digits are reversed, the number decreases by 495. Find the number.

• Let \(a\) = the first digit, \(b\) = the middle digit, \(c\) = the last digit.
• \(a + b + c=10\)
• \(a - 2b=0\)
• \(99c-99a=-495\) (the reversed number minus the number is −495)
• Simultaneous equation solver: a = 6, b = 3, c = 1
• The digits are 6, 3, 1: the number is 631.

Three groups buy snacks at a stadium. Group 1 buys 2 hot dogs, 3 drinks and 1 bag of popcorn and pays $24.50. Group 2 buys 4 hot dogs, 2 drinks and 2 bags of popcorn and pays $36.00. Group 3 buys 1 hot dog, 5 drinks and 3 bags of popcorn and pays $38.00. Find the total cost of one hot dog, one drink and one bag of popcorn.

• Let \(x\) = cost of one hot dog ($), \(y\) = cost of one drink ($), \(z\) = cost of one bag of popcorn ($).
• \(2x + 3y + z=24.5\)
• \(4x + 2y + 2z=36\)
• \(x + 5y + 3z=38\)
• Simultaneous equation solver: x = 4.50, y = 3.25, z = 5.75
• x + y + z = 4.50 + 3.25 + 5.75 = $13.50

A school canteen records the lunches chosen by pupils on one day. The pie chart shows how pupils chose between hot meals, sandwiches and salads. The sector for hot meals has angle 200°. There are 6 more in sandwiches than in salads. The number in hot meals and salads together is 9 more than the number in sandwiches. One person is chosen at random. Find the probability that they chose salads.

• Let \(x\), \(y\), \(z\) be the numbers who chose hot meals, sandwiches and salads.
• \(y-z=6\)
• \(x-y+z=9\)
• The sector for hot meals is 200/360 of the whole: \(\frac{x}{x+y+z}=\frac{200}{360}\), so \(4x - 5y - 5z=0\)
• Simultaneous equation solver: x = 15, y = 9, z = 3
• Total = 27. Probability = \(\frac{3}{27}\) = 0.111

Three numbers \(a\), \(b\), \(c\) are in order, \(a\le b\le c\). The range is twice the median. The mean is 2 more than the median. The largest number is 8 more than the median. Find the smallest of the three numbers.

• The median is \(b\), the mean is \(\frac{a+b+c}{3}\) and the range is \(c-a\).
• \(-a - 2b + c=0\)
• \(a - 2b + c=6\)
• \(-b + c=8\)
• Simultaneous equation solver: a = 3, b = 5, c = 13
• The answer is 3.