Algebra 1 · Unit 4
Create two linear equations with two variables that represent a real-world situation.
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Name both variables first
Say what each letter stands for, with units. Two unknowns need two letters.
Two facts, two equations
Each sentence that relates the two quantities becomes one equation. A system is the pair of equations together.
Count and value
A common pair is (number of items) and (total cost or total amount). Keep the units the same in each equation.
Check that you used both facts
If one equation uses only one letter, you may have combined the facts too soon. Write both relationships first.
Read the story. Name the letters, then write two equations. You do not have to solve yet.
Let p = the number of popcorns.
Let d = the number of drinks.
Count: p + d = 7
Cost: 4p + 2d = 20
Answer: p + d = 74p + 2d = 20
Let a = the number of adult tickets.
Let c = the number of child tickets.
a + c = 9
12a + 7c = 88
Answer: a + c = 912a + 7c = 88
Let a = the number of apples.
Let o = the number of oranges.
a + o = 10
2a + 3o = 24
Answer: a + o = 102a + 3o = 24
Let ℓ = the larger number.
Let s = the smaller number.
ℓ + s = 18
ℓ − s = 4
Answer: ℓ + s = 18ℓ − s = 4
Let s = the number of shirts.
Let h = the number of hats.
s + h = 6
15s + 8h = 76
Answer: s + h = 615s + 8h = 76
Let w = hours walking.
Let r = hours running.
w + r = 5
3w + 6r = 21
Answer: w + r = 53w + 6r = 21
The two facts may be two different combinations, or a total plus a comparison.
Let c = the price of one bag of chips in dollars.
Let s = the price of one soda in dollars.
3c + 2s = 13
c + 4s = 11
Answer: 3c + 2s = 13c + 4s = 11
Let c = the number of coffees.
Let t = the number of teas.
c = 2t
4c + 3t = 33
Answer: c = 2t4c + 3t = 33
Let ℓ = the length.
Let w = the width.
ℓ = w + 5
2ℓ + 2w = 38
Answer: ℓ = w + 52ℓ + 2w = 38
Let n = the number of nickels.
Let d = the number of dimes.
n + d = 18
5n + 10d = 150
Answer: n + d = 185n + 10d = 150
Let w = the weekday price in dollars.
Let e = the weekend price in dollars.
4w + 2e = 52
3w + 5e = 61
Answer: 4w + 2e = 523w + 5e = 61
Downstream speed is b + c. Upstream speed is b − c.
Distance = (rate)(time)
2(b + c) = 24
3(b − c) = 18
Or, after dividing: b + c = 12 and b − c = 6
Answer: b + c = 12b − c = 6
Percents, two investments, and mixed units still become two linear equations.
Let t = liters of 15% solution.
Let h = liters of 40% solution.
t + h = 20
0.15t + 0.40h = 0.25(20)
0.15t + 0.40h = 5
Answer: t + h = 200.15t + 0.40h = 5
Let x = dollars in the 4% account.
Let y = dollars in the 7% account.
x + y = 5000
0.04x + 0.07y = 275
Answer: x + y = 50000.04x + 0.07y = 275
Let a = pounds of \$6 nuts.
Let b = pounds of \$9 nuts.
a + b = 12
6a + 9b = 7.50(12)
6a + 9b = 90
Answer: a + b = 126a + 9b = 90
Let L = the price of a large pizza.
Let S = the price of a small pizza.
2L + 3S = 31
4L + S = 33
Answer: 2L + 3S = 314L + S = 33
With the wind, speed is p + 30.
Against the wind, speed is p − 30.
(p + 30)t = 900
(p − 30)t = 750
Answer: (p + 30)t = 900(p − 30)t = 750
Let x = the first number.
Let y = the second number.
3x + y = 28
x + 4y = 31
Answer: 3x + y = 28x + 4y = 31
Let a = adult tickets and s = student tickets.
a + s = 40
11a + 6s = 355
A negative s would mean the ticket counts cannot both be true with these prices. In the story, ticket counts cannot be negative, so that pair would not be a real sale.
Answer: a + s = 4011a + 6s = 355
Let t = liters of 10% and h = liters of 50%.
t + h = 8
0.10t + 0.50h = k100(8)
You still have three letters. You need a value for k (or one of the amounts) before the system has a unique pair (t, h).
Answer: t + h = 80.10t + 0.50h = 0.08k
Let g = gigabytes and C = the common cost in dollars.
C = 20 + 5g
C = 8g
The solution is the break-even pair (g, C).
Answer: C = 20 + 5gC = 8g
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