Ninth Grade Mathematics: Concept of Solutions & Ordered Pairs
A linear equation in one variable, such as 2*x + 6 = 0, possesses exactly one unique solution (x = -3). By contrast, introducing a second variable creates a relationship of co-dependence between x and y, yielding an infinite continuum of solutions. In this chapter, you will explore the nature of ordered pairs, learn how to calculate solutions systematically, and understand why linear equations in two variables have infinitely many solutions.
The Anatomy of an Ordered Pair Solution
A solution to a linear equation in two variables, ax + by + c = 0, is an ordered pair of real numbers (x_0, y_0) such that substituting x = x_0 and y = y_0 into the equation makes the left-hand side equal to the right-hand side.
Testing an Ordered Pair:
Equation: 2*x + 3*y = 12
Test Point (3, 2):
Left-Hand Side: 2*(3) + 3*(2) = 6 + 6 = 12
Right-Hand Side: 12
Since LHS = RHS, (3, 2) IS a solution.
Test Point (1, 4):
Left-Hand Side: 2*(1) + 3*(4) = 2 + 12 = 14
Right-Hand Side: 12
Since LHS != RHS, (1, 4) is NOT a solution.
The order in an ordered pair is essential: the first coordinate always represents the independent variable x, while the second coordinate represents the dependent variable y. The ordered pair (3, 2) is entirely different from (2, 3).
Why There Are Infinitely Many Solutions
A single equation with two unknowns does not fix a single value for either variable. Instead, for every arbitrary value you choose for x, there exists a corresponding unique value of y that satisfies the equation.
To find solutions to a*x + b*y + c = 0:
1. Isolate one variable (say y) in terms of the other:
-a * x - c
y = ------------
b
2. Substitute any real value for x.
3. Compute the resulting value for y.
Because there are infinitely many real numbers that can be substituted for x, there are infinitely many corresponding values of y. Hence, every linear equation in two variables has infinitely many solutions.
Generating Solutions for x + 2*y = 6:
Isolate y: y = (6 - x) / 2
Choose x = 0 ===> y = (6 - 0) / 2 = 3 ===> Solution: (0, 3)
Choose x = 2 ===> y = (6 - 2) / 2 = 2 ===> Solution: (2, 2)
Choose x = 4 ===> y = (6 - 4) / 2 = 1 ===> Solution: (4, 1)
Choose x = 6 ===> y = (6 - 6) / 2 = 0 ===> Solution: (6, 0)
Choose x = -2 ===> y = (6 - (-2))/2 = 4 ===> Solution: (-2, 4)
Geometric Identity: Points on the Line
There is an absolute geometric equivalence between the algebraic solutions of an equation and the points on its graph:
Algebraic Truth Geometric Reality
-------------------------------------------------------------------------
(x_0, y_0) is a solution of equation <===> Point (x_0, y_0) lies on the line
(x_1, y_1) is NOT a solution <===> Point (x_1, y_1) lies off the line
Every single point that lies on the straight line is a valid solution to the equation, and every solution to the equation corresponds to a point lying on that line.
Chapter Exercises
Exercise 1: Find four distinct ordered pair solutions for the linear equation 3x + 4y = 24. Show the step-by-step substitution for each chosen value.
Exercise 2: Check whether each of the following ordered pairs is a solution to the equation 2x - 5y = 10: (0, -2), (5, 0), (1, 1), and (-5, -4).
Exercise 3: If the point (2k - 1, k) lies on the line 3x - 2y + 5 = 0, find the value of the unknown parameter k.
Solutions and Explanations
Solution 1: Isolate y in the equation 3x + 4y = 24: 4y = 24 - 3x, so y = (24 - 3x) / 4. For the first solution, choose x = 0: y = (24 - 0) / 4 = 6, giving the ordered pair (0, 6). For the second solution, choose x = 4: y = (24 - 12) / 4 = 12 / 4 = 3, giving (4, 3). For the third solution, choose x = 8: y = (24 - 24) / 4 = 0, giving (8, 0). For the fourth solution, choose x = -4: y = (24 - 3(-4)) / 4 = (24 + 12) / 4 = 36 / 4 = 9, giving (-4, 9). These four ordered pairs—(0, 6), (4, 3), (8, 0), and (-4, 9)—are all valid solutions.
Solution 2: Substitute each ordered pair into the left-hand side LHS = 2x - 5y. For (0, -2): LHS = 2(0) - 5(-2) = 0 + 10 = 10, which equals RHS, so (0, -2) is a solution. For (5, 0): LHS = 2(5) - 5(0) = 10 - 0 = 10, which equals RHS, so (5, 0) is a solution. For (1, 1): LHS = 2(1) - 5(1) = 2 - 5 = -3 != 10, so (1, 1) is not a solution. For (-5, -4): LHS = 2(-5) - 5(-4) = -10 + 20 = 10, which equals RHS, so (-5, -4) is a solution.
Solution 3: If the point (2k - 1, k) lies on the line 3x - 2y + 5 = 0, the coordinates x = 2k - 1 and y = k must satisfy the equation. Substitute these expressions into the equation: 3(2k - 1) - 2(k) + 5 = 0. Distribute the multiplication: 6k - 3 - 2k + 5 = 0. Combine like terms: 4k + 2 = 0. Subtract 2 from both sides: 4k = -2. Divide by 4: k = -2/4 = -1/2. Thus, the value of k is -1/2.