Q.Examine the consistency of the following system of equations: $x + 3y = 5$ $2x + 6y = 8$

The key idea is to write the system as a matrix equation $A\mathbf{x} = \mathbf{b}$ and check the rank of $A$ versus the rank of the augmented matrix $[A \mid \mathbf{b}]$.

Step 1: Matrix form

Step 2: Compare ranks

The second row of $A$ is $2$ times the first row, so $\text{rank}(A) = 1$.

The augmented matrix is

Row-reducing: $R_2 \to R_2 - 2R_1$ gives $\begin{pmatrix} 1 & 3 & 5 \\ 0 & 0 & -2 \end{pmatrix}$.

The second row is $(0\;0\;|\; -2)$, so $\text{rank}([A \mid \mathbf{b}]) = 2$.

Step 3: Consistency condition

Since $\text{rank}(A) = 1 \neq 2 = \text{rank}([A \mid \mathbf{b}])$, the system is inconsistent — no solution exists.

Why This Approach Works

When we talk about consistency of a system of linear equations, we are asking: Does there exist at least one pair $(x, y)$ that satisfies all equations simultaneously?

The most intuitive way to check this is to compare the ratios of coefficients. For two linear equations in two variables:

If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct — no intersection, hence inconsistent.

If all three ratios are equal, the lines coincide — infinitely many solutions (consistent).

If the coefficient ratios are unequal, the lines intersect at exactly one point — consistent with a unique solution.

Here, we simply check these ratios.

Step-by-Step Solution

1. Write the system clearly

2. Compare the coefficients of $x$ and $y$

From the first equation: $a_1 = 1$, $b_1 = 3$, $c_1 = 5$

From the second equation: $a_2 = 2$, $b_2 = 6$, $c_2 = 8$

Compute the ratios:

So $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ — the left-hand sides are proportional.

3. Check the constant term ratio

Now $\frac{1}{2} \neq \frac{5}{8}$.

4. Interpret the result

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the two lines are parallel and distinct. They never meet, so no $(x, y)$ can satisfy both equations simultaneously.

5. Conclusion on consistency

The system is inconsistent — it has no solution.

Method: The Coefficient-Ratio Test for Two-Variable Systems

For a system of exactly two linear equations in two unknowns, comparing the ratios of coefficients is a quick alternative to computing a determinant, and it also tells you WHICH kind of inconsistency or dependency you have.

Steps

Step 1: Write the two equations in standard form

Step 2: Compute the three ratios

Step 3: Classify using the ratios

Step 4: Applying — never stop after just the first two ratios

Finding $\frac{a_1}{a_2}=\frac{b_1}{b_2}$ only tells you $\det(A)=0$ — it does NOT by itself tell you whether the system is inconsistent or has infinitely many solutions. You must always go on and compare $\frac{c_1}{c_2}$ against that common value before concluding either way.

Common Mistakes

Mistake 1: Stopping after checking only the $x$- and $y$-coefficient ratios

Why it's wrong: here $\frac{a_1}{a_2}=\frac{1}{2}=\frac{b_1}{b_2}$, which only tells you $\det(A)=0$ — it does NOT by itself mean the system has infinitely many solutions. You must also compare $\frac{c_1}{c_2}=\frac{5}{8}$; since $\frac{1}{2}\neq\frac{5}{8}$, the system is actually inconsistent, not dependent. Correct approach: always compute and compare all three ratios ($a$, $b$, and $c$) before concluding either 'infinitely many' or 'no solution' — two matching ratios alone are not enough.

Mistake 2: Missing the direct contradiction check as a shortcut

Why it's wrong: doubling the first equation gives $2x+6y=10$, but the second equation states $2x+6y=8$ — the same left-hand side is forced to equal two different numbers, an outright contradiction that confirms 'no solution' without needing the ratio test at all. Skipping this check means relying solely on the ratio comparison and missing an easy independent confirmation. Correct approach: whenever one equation is a multiple of another on the left-hand side, immediately test whether the same multiple applies to the constant term too.