Tenth Grade Mathematics: Conditions for Consistency, Inconsistency, and Dependency

While drawing graphs provides a clear geometric visual of how two linear equations interact, constructing detailed plots for every system is time-consuming and prone to plotting errors. Algebra provides a direct, elegant analytical alternative: by comparing the ratios of the coefficients of x, the coefficients of y, and the constant terms, one can immediately predict whether a system has a unique solution, no solution, or infinitely many solutions without drawing a single line.

Terminology of System Consistency

In mathematical terminology, the solvability of any system of equations is classified into two primary categories:

1. Consistent System:
   A system of linear equations is called consistent if it possesses at least
   one valid solution.
   - Independent Consistent: Exactly one unique solution (Intersecting lines).
   - Dependent Consistent:   Infinitely many solutions (Coincident lines).

2. Inconsistent System:
   A system of linear equations is called inconsistent if it possesses no
   solution whatsoever (Parallel lines).

The Three Fundamental Ratio Tests

Consider a pair of linear equations written in standard general form:

a1*x + b1*y + c1 = 0
a2*x + b2*y + c2 = 0

The relationship between the coefficient ratios governs the geometric nature of the system completely:

+-----------------------------+--------------------+------------------------+------------------+
| Ratio Comparison            | Graphical Behavior | Algebraic Solution     | Classification   |
+-----------------------------+--------------------+------------------------+------------------+
| (a1 / a2) != (b1 / b2)      | Intersecting lines | Exactly one (unique)   | Consistent       |
|                             | at one point       | solution               | (Independent)    |
+-----------------------------+--------------------+------------------------+------------------+
| (a1 / a2) == (b1 / b2)      | Coincident lines   | Infinitely many        | Consistent       |
|           == (c1 / c2)      | (identical line)   | solutions              | (Dependent)      |
+-----------------------------+--------------------+------------------------+------------------+
| (a1 / a2) == (b1 / b2)      | Parallel lines     | No solution            | Inconsistent     |
|           != (c1 / c2)      | (never meet)       |                        |                  |
+-----------------------------+--------------------+------------------------+------------------+

Derivation from Slope-Intercept Form

To understand the origin of these ratio rules, convert both equations into slope-intercept form y = m*x + c:

Equation 1:  y = (-a1 / b1)*x + (-c1 / b1)  ===>  Slope m1 = -a1/b1,  Intercept k1 = -c1/b1
Equation 2:  y = (-a2 / b2)*x + (-c2 / b2)  ===>  Slope m2 = -a2/b2,  Intercept k2 = -c2/b2

For two lines to intersect at a single point, their slopes must be different: m1 != m2. -a1 / b1 != -a2 / b2 ===> a1 / a2 != b1 / b2. This gives the first condition.

If their slopes are identical (m1 = m2, meaning a1 / a2 = b1 / b2), they are either the exact same line or separate parallel lines: If their y-intercepts are also identical (k1 = k2, meaning -c1 / b1 = -c2 / b2, which gives c1 / c2 = b1 / b2), the lines coincide completely, yielding infinitely many solutions. If their y-intercepts are different (k1 != k2, meaning c1 / c2 != b1 / b2), the lines are parallel and never intersect, yielding no solution.

Finding Unknown Parameters for Solvability Conditions

A central application of these ratio conditions is determining the value of unknown parameters (often denoted by k or p) so that a system satisfies a prescribed solvability requirement.

Sample Problem:
Find the value of k for which the system of equations:
  k*x + 3*y = k - 3
  12*x + k*y = k
has infinitely many solutions.

Step 1: Write in standard form:
        k*x + 3*y - (k - 3) = 0   ===>  a1 = k,   b1 = 3,   c1 = -(k - 3)
        12*x + k*y - k = 0       ===>  a2 = 12,  b2 = k,   c2 = -k

Step 2: Apply condition for infinitely many solutions:
        a1 / a2 = b1 / b2 = c1 / c2
        k / 12 = 3 / k = (k - 3) / k

Step 3: Solve the first equality:
        k / 12 = 3 / k  ===>  k^2 = 36  ===>  k = +6 or k = -6

Step 4: Verify with the second equality 3 / k = (k - 3) / k:
        If k = 6:   3 / 6 = (6 - 3) / 6 = 3 / 6 = 1/2.  (Consistent!)
        If k = -6:  3 / (-6) = -1/2, while (-6 - 3) / (-6) = -9 / (-6) = 3/2.
                    Since -1/2 != 3/2, k = -6 does not satisfy all ratios.

Conclusion:
The unique value of k for which the system has infinitely many solutions is k = 6.

Chapter Exercises

Exercise 1: On comparing the ratios a1/a2, b1/b2, and c1/c2, determine whether the following systems are consistent or inconsistent: (a) 3x + 2y = 5 and 2x - 3y = 7; (b) 2x - 3y = 8 and 4x - 6y = 9; (c) (4/3)x + 2y = 8 and 2x + 3y = 12.

Exercise 2: For what value of p will the following pair of linear equations have a unique solution: 4x + py + 8 = 0 and 2x + 2y + 2 = 0?

Exercise 3: For what values of a and b does the following pair of linear equations have an infinite number of solutions: 2x + 3y = 7 and (a - b)x + (a + b)y = 3*a + b - 2?

Solutions and Explanations

Solution 1: In system (a), a1 = 3, b1 = 2, c1 = -5 and a2 = 2, b2 = -3, c2 = -7. Comparing ratios: a1/a2 = 3/2 and b1/b2 = 2/(-3) = -2/3. Since 3/2 != -2/3, a1/a2 != b1/b2. The lines intersect at a unique point, and the system is consistent. In system (b), a1 = 2, b1 = -3, c1 = -8 and a2 = 4, b2 = -6, c2 = -9. a1/a2 = 2/4 = 1/2; b1/b2 = -3/(-6) = 1/2; c1/c2 = -8/(-9) = 8/9. Since a1/a2 = b1/b2 != c1/c2, the lines are parallel, and the system is inconsistent. In system (c), a1 = 4/3, b1 = 2, c1 = -8 and a2 = 2, b2 = 3, c2 = -12. a1/a2 = (4/3)/2 = 2/3; b1/b2 = 2/3; c1/c2 = -8/(-12) = 2/3. Since a1/a2 = b1/b2 = c1/c2, the lines are coincident, and the system is dependent and consistent.

Solution 2: For the system to have a unique solution, the ratio of the coefficients of x must not equal the ratio of the coefficients of y: a1/a2 != b1/b2. Here, a1 = 4, a2 = 2, b1 = p, and b2 = 2. Substituting these into the condition: 4/2 != p/2. Simplifying gives 2 != p/2, which yields p != 4. Therefore, the system will have a unique solution for all real values of p except p = 4.

Solution 3: For infinitely many solutions, the ratio condition is a1/a2 = b1/b2 = c1/c2. Here, a1 = 2, b1 = 3, c1 = -7, and a2 = a - b, b2 = a + b, c2 = -(3a + b - 2). Thus: 2/(a - b) = 3/(a + b) = 7/(3a + b - 2). From the first equality: 2(a + b) = 3(a - b) ==> 2a + 2b = 3a - 3b ==> a - 5b = 0, so a = 5b. From the second equality: 3(3a + b - 2) = 7(a + b) ==> 9a + 3b - 6 = 7a + 7b ==> 2a - 4b = 6 ==> a - 2b = 3. Substitute a = 5b: 5b - 2b = 3 ==> 3b = 3 ==> b = 1. Then a = 5*(1) = 5. Thus, the system has infinitely many solutions when a = 5 and b = 1.