Consistency & Inconsistency of Linear Equations

In algebra, when solving systems of linear equations, we determine if a solution exists (consistent system) or if no solution exists (inconsistent system). We can determine this without actually solving the equations by analyzing the coefficients of the variables and the constant terms. This analysis is done by comparing the ratios of the coefficients. For a system of two linear equations in two variables, $x$ and $y$, given by:

$a_1x + b_1y = c_1$

$a_2x + b_2y = c_2$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are constants.

By comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$, we can determine the nature of the solution.

Formulae

For a system of two linear equations:

1. Consistent System (Unique Solution): If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, then the system has a unique solution (intersecting lines).

2. Consistent System (Infinitely Many Solutions): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, then the system has infinitely many solutions (coincident lines).

3. Inconsistent System (No Solution): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, then the system has no solution (parallel lines).

Examples

Example-1: Determine if the following system is consistent or inconsistent.

$2x + 3y = 7$

$4x + 6y = 14$

Here, $a_1 = 2$, $b_1 = 3$, $c_1 = 7$, $a_2 = 4$, $b_2 = 6$, $c_2 = 14$.

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system is consistent with infinitely many solutions.

Example-2: Determine if the following system is consistent or inconsistent.

$x – y = 5$

$2x – 2y = 12$

Here, $a_1 = 1$, $b_1 = -1$, $c_1 = 5$, $a_2 = 2$, $b_2 = -2$, $c_2 = 12$.

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{5}{12}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system is inconsistent (no solution).

Theorem with Proof

Theorem: For a system of two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system is consistent with a unique solution.

2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system is consistent with infinitely many solutions.

3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system is inconsistent (no solution).

Proof:

1. Case 1: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Multiply the first equation by $b_2$ and the second by $b_1$:

$a_1b_2x + b_1b_2y = c_1b_2$

$a_2b_1x + b_1b_2y = c_2b_1$

Subtract the second equation from the first:

$(a_1b_2 – a_2b_1)x = c_1b_2 – c_2b_1$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, then $a_1b_2 \neq a_2b_1$, so $a_1b_2 – a_2b_1 \neq 0$. Thus, we can solve for $x$:

$x = \frac{c_1b_2 – c_2b_1}{a_1b_2 – a_2b_1}$

Substituting this value of x back into the first equation gives a unique value for y. Hence a unique solution exists.

2. Case 2: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k$ (where k is a constant)

This implies $a_1 = ka_2$, $b_1 = kb_2$, and $c_1 = kc_2$. Substituting into the first equation:

$ka_2x + kb_2y = kc_2$

Dividing by $k$ (assuming $k \neq 0$):

$a_2x + b_2y = c_2$

This is the same as the second equation. The two equations are equivalent, representing the same line. Thus, there are infinitely many solutions.

3. Case 3: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Let $\frac{a_1}{a_2} = \frac{b_1}{b_2} = k$. Then $a_1 = ka_2$ and $b_1 = kb_2$. Substituting into the first equation:

$ka_2x + kb_2y = c_1$

Dividing by k:

$a_2x + b_2y = \frac{c_1}{k}$

Since $\frac{c_1}{c_2} \neq k$, $\frac{c_1}{k} \neq c_2$. Thus, we have:

$a_2x + b_2y = \frac{c_1}{k}$

and

$a_2x + b_2y = c_2$

which is a contradiction. This means that no solution exists, and the system is inconsistent (parallel lines).

Common mistakes by students

• Incorrectly calculating ratios: Students often make arithmetic errors when calculating the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.
• Forgetting to check all ratios: They may only check $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$ and fail to check the ratio with the constant terms.
• Misinterpreting the results: Incorrectly concluding that a system is consistent when it’s inconsistent, or vice versa, based on the ratio comparisons.
• Not simplifying ratios: Not simplifying the ratios fully, leading to incorrect comparisons. For example, using 2/4 instead of 1/2, leading to incorrect conclusions.

Real Life Application

This concept is used in various real-life scenarios:

• Business and Economics: Analyzing supply and demand curves. If two supply and demand equations lead to intersecting lines (a unique solution), a market equilibrium exists. If the lines are parallel (no solution), there’s no equilibrium.
• Computer Graphics: Determining if two lines representing objects in a scene intersect.
• Chemistry: Balancing chemical equations can sometimes lead to systems of linear equations. Determining if a unique solution for the coefficients exists is related to consistency.
• Engineering: Analyzing electrical circuits, where the voltages and currents can be represented by linear equations. Consistency determines if a solution for the currents exists.

Fun Fact

The concept of consistency and inconsistency extends to systems of linear equations with more than two variables. The concepts are similar, relying on the properties of the coefficient matrix of the system.

Recommended YouTube Videos for Deeper Understanding

Q.1 Consider the following system of linear equations: $2x + 3y = 5$ $4x + 6y = k$ For what value of $k$ is the system consistent with infinitely many solutions?

Check Solution

Ans: A

For infinitely many solutions, $a_1/a_2 = b_1/b_2 = c_1/c_2$. Here, $a_1 = 2, a_2 = 4, b_1 = 3, b_2 = 6, c_1 = 5, c_2 = k$. We have $2/4 = 3/6 = 5/k$, which simplifies to $1/2 = 1/2 = 5/k$. Therefore, $k = 10$.

Q.2 The following system of equations is given: $x – 2y = 3$ $3x + ky = 9$ For what value of $k$ will the system be inconsistent?

Check Solution

Ans: A

For an inconsistent system, $a_1/a_2 = b_1/b_2 \neq c_1/c_2$. Here, $a_1 = 1, a_2 = 3, b_1 = -2, b_2 = k, c_1 = 3, c_2 = 9$. We need $1/3 = -2/k \neq 3/9$. Solving $1/3 = -2/k$, we get $k = -6$. Also, $3/9 = 1/3$, so we must have $-2/k \neq 1/3$, hence $k \neq -6$. Since $3/9=1/3$, so $k=-6$ makes it inconsistent.

Q.3 Determine the value of $p$ for which the following system of equations has a unique solution: $px + 2y = 5$ $3x + y = 1$

Check Solution

Ans: B

For a unique solution, $a_1/a_2 \neq b_1/b_2$. Here, $a_1 = p, a_2 = 3, b_1 = 2, b_2 = 1$. We need $p/3 \neq 2/1$, which simplifies to $p \neq 6$.

Q.4 For what value of $m$ are the following equations consistent? $mx + y = 3$ $2x + y = 6$

Check Solution

Ans: B

The system will be consistent for all $m$ except when $m/2 = 1/1 \ne 3/6$. So, $m \ne 2$.

Q.5 Consider the following system of linear equations: $x + 2y = 5$ $2x + 4y = k$ The system has no solution when $k$ is:

Check Solution

Ans: D

For no solution(inconsistent), $a_1/a_2 = b_1/b_2 \neq c_1/c_2$. Here, $a_1=1, a_2=2, b_1=2, b_2=4, c_1=5, c_2=k$. We have $1/2 = 2/4 \neq 5/k$, which simplifies to $1/2 \neq 5/k$. Therefore, $k \neq 10$. The options A,B,C give us cases when the equation is possible with $k=10$, which would give infinite solutions.

Next Topic: Methods of Solving Linear Equations: Substitution & Elimination