Mathematics
Grade 12
15 min
Solve a system of equations by graphing: word problems (Tutorial only)
Solve a system of equations by graphing: word problems (Tutorial only)
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Introduction & Learning Objectives
Learning Objectives
Translate a real-world scenario into a system of two or more equations.
Identify and define the independent and dependent variables within the context of a word problem.
Accurately graph each equation of a system on the same coordinate plane, respecting domain and range constraints.
Locate the point(s) of intersection on a graph and interpret their meaning as the solution to the word problem.
Verify a graphical solution by substituting the intersection coordinates back into the original system of equations.
Analyze the graph to answer contextual questions, such as determining when one quantity is greater than another.
📈 A startup's costs are high but its revenue is growing fast. When will they finally turn a profit? Let's graph the path to the...
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Key Concepts & Vocabulary
TermDefinitionExample
System of EquationsA set of two or more equations that share the same variables. The solution to the system is the set of variable values that satisfies all equations simultaneously.A company's cost function is `C(x) = 20x + 800` and its revenue function is `R(x) = 60x`. The system is `{y = 20x + 800, y = 60x}`.
Graphical SolutionThe point or points of intersection of the graphs of the equations in a system. Each coordinate `(x, y)` of an intersection point is a solution to the system.If the graphs of `y = -x + 5` and `y = 2x - 1` intersect at `(2, 3)`, then `x=2, y=3` is the solution to the system.
Break-Even PointIn a business or economic context, the point at which total cost and total revenue are equal, resulting in neither a net loss nor a gain. Graphically...
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Core Formulas
Linear Function
y = mx + b
Models situations with a constant rate of change (`m`) and an initial or fixed value (`b`). Ideal for problems involving constant speed, simple interest, or fixed/variable costs.
Quadratic Function
y = ax^2 + bx + c
Models situations where the rate of change is not constant, such as projectile motion under gravity, area calculations, or revenue models where price is dependent on demand.
Condition for Solution
f(x) = g(x)
A graphical solution to a system `y = f(x)` and `y = g(x)` exists at the x-value(s) where the function outputs are equal. This is the algebraic basis for finding the intersection point.
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Challenging
A company's profit is P(x) = R(x) - C(x). The break-even point occurs when P(x) = 0. How can you find the break-even point by graphing a system based on the tutorial's core concepts, without ever graphing the profit function P(x) itself?
A.Graph y = R(x) and y = C(x) on the same plane and find their point of intersection.
B.Graph y = R(x) and find its x-intercept.
C.Graph y = C(x) and find its y-intercept.
D.Graph y = R(x) - C(x) and y = x on the same plane and find their intersection.
Challenging
Consider the flare system: h(t) = -5t^2 + 40t + 5 and h(t) = 65. If the initial velocity of the flare were increased, the equation might change to h(t) = -5t^2 + 50t + 5. Graphically, how would the two intersection points with the line h=65 be affected by this change?
A.Both intersection points would shift to the left (occur earlier).
B.The first intersection point would shift left (earlier), and the second would shift right (later).
C.The vertex of the parabola would shift up and to the right, causing the time interval between the intersections to become wider.
D.The intersection points would not change, but the parabola would become narrower.
Challenging
A word problem comparing a company's linear growth and a competitor's quadratic decay results in a system with two intersection points: (-5, 50) and (10, 200). If the independent variable represents 'years from today', how should the solution to the word problem be interpreted?
A.The companies had the same value 5 years ago and will have the same value again in 10 years.
B.Only the intersection at (10, 200) is a valid solution in the context of the problem, as time cannot be negative.
C.Both points are valid solutions, representing two moments in time.
D.Neither point is a valid solution because one has a negative time coordinate.
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