Math problem set Grade 8 Math
Junegrass Valley Water Supply: Systems of Equations Break-Even
A grade 8 problem set where students write and solve a system of linear equations to find where a new well and imported water cost a fictional community the same amount.
Objectives
- Students will write a system of two linear equations from a real supply-cost decision.
- Students will solve a system algebraically and interpret the intersection point in context.
Vocabulary
Key words: system of equations, break-even point, intersection
Teacher overview
Communities often compare a source with a high fixed cost but a low running cost against one with no fixed cost but a higher running cost, and the two lines cross at a break-even point. This set has students write both cost equations, solve the system, and read what the intersection means for a real decision. A common mistake to watch for is stopping at the x-value; the break-even is a point, so students should also find the cost at that usage and check it in both equations.
Problems
Junegrass Valley is a fictional community deciding how to supply water. Option A is a new well with a fixed monthly cost plus an energy cost for each 1,000 gallons pumped. Option B buys water from a neighbor at a flat price per 1,000 gallons. Let x be the number of thousands of gallons used in one month.
- Option A has a fixed cost of $1,200 per month plus $2 for each 1,000 gallons pumped. Write Option A's cost equation.
- Option B charges $5 for each 1,000 gallons and has no fixed cost. Write Option B's cost equation.
- Solve the system to find the value of x where the two options cost the same.
- Find the monthly cost at that break-even point, and check it in both equations.
- At 250,000 gallons per month, where x = 250, which option is cheaper, and by how much?
- At 600,000 gallons per month, where x = 600, which option is cheaper, and by how much?
- Explain what the break-even point tells Junegrass Valley about when to choose each option.
- A vendor offers Option C with the equation C = 4x + 300. Solve the system to find where Option A and Option C cost the same.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: A = 2x + 1,200.
The rate is $2 per thousand gallons and the fixed cost is $1,200, so A = 2x + 1,200. - Answer: B = 5x.
The rate is $5 per thousand gallons with no fixed cost, so B = 5x. - Answer: x = 400.
Set 2x + 1,200 = 5x. Subtract 2x: 1,200 = 3x, so x = 400 thousand gallons. - Answer: $2,000.
Using B = 5x: 5(400) = $2,000. Check with A: 2(400) + 1,200 = 800 + 1,200 = $2,000. Both agree. - Answer: Option B is cheaper by $450.
A = 2(250) + 1,200 = 1,700. B = 5(250) = 1,250. B costs $1,700 - $1,250 = $450 less. - Answer: Option A is cheaper by $600.
A = 2(600) + 1,200 = 2,400. B = 5(600) = 3,000. A costs $3,000 - $2,400 = $600 less. - Answer: Below 400, choose Option B; above 400, choose Option A.
The break-even is at 400 thousand gallons. When usage is less than that, Option B costs less; when usage is more, Option A's low running cost makes it cheaper. - Answer: x = 450.
Set 2x + 1,200 = 4x + 300. Subtract 2x: 1,200 = 2x + 300. Subtract 300: 900 = 2x, so x = 450 thousand gallons.
Differentiation
Support
Give students a completed graph of Options A and B on the same axes so they can see the intersection near x = 400 before solving algebraically. Provide a filled example of setting two equations equal so they know how to begin problem 3.
Extension
Ask students to find the range of monthly usage, in gallons, over which Option C (from problem 8) is the cheapest of all three options, and to justify the range with the break-even points they calculate.
ELL supports
Pre-teach "fixed cost," "break-even," and "intersection" using a picture of two lines crossing with the crossing point circled. Provide the sentence frame "Below ___ thousand gallons, Option ___ is cheaper because ___."