Solve rational equations and apply to real-world problems. Math 30-1 Alberta.
Lesson 6.7 of Radical & Rational Functions in Math 30-1 — Alberta curriculum lessons.
Definition. A rational equation is an equation where rational functions (fractions with polynomials) appear on one or both sides.
For example: (2x-5)/(x-1) - 2 = (3)/(x+2)
Your goal is to find the x-value(s) that make both sides equal.
You learned two methods in Math 20-1 — graphical and algebraic — and both work here.
The Idea. The graphical method is exactly what you already know from solving radical equations graphically.
You split the equation into two functions and find where they intersect.
If the left side equals the right side at some x-value, then both graphs will have the same y-value at that x — they will cross.
The Steps. Step 1 — Split the equation into two functions:
Left side → y_1, Right side → y_2
Step 2 — Graph both on the same axes
Step 3 — Find the x-coordinate of the intersection point(s)
That x-value is your solution.
The Equation. (2x-5)/(x-1) - 2 = (3)/(x+2)
Split into Two Functions. y_1 = (2x-5)/(x-1) - 2
y_2 = (3)/(x+2)
Graph both functions on the same axes and look for where they cross.
The Intersection. The intersection occurs at x = -0.5
Verify the Solution. Substitute x = -0.5 into both sides of the original equation:
Left side:
(2(-0.5)-5)/(-0.5-1) - 2 = (-1-5)/(-1.5) - 2 = (-6)/(-1.5) - 2 = 4 - 2 = 2
Right side:
(3)/(-0.5+2) = (3)/(1.5) = 2 ✓
Solution: x = -0.5
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