Q.Examine the consistency of the following system of equations: $2x - y = 5$ $x + y = 4$

The key idea is to write the system as a matrix equation $A\mathbf{x} = \mathbf{b}$ and check whether $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$.

Step 1 — Write the augmented matrix.

Step 2 — Row reduce.

Swap rows:

Replace $R_2$ with $R_2 - 2R_1$:

Step 3 — Compare ranks.

Both $A$ and $[A|\mathbf{b}]$ have 2 non-zero rows, so $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = 2$, which equals the number of variables. The system is consistent and has a unique solution.

Why This Problem Matters

When you hear "consistency" in a system of linear equations, you're really asking: Can these equations all be true at the same time? If yes, the system is consistent; if no, it's inconsistent. For two equations in two variables, consistency means the lines either intersect (one solution) or coincide (infinitely many solutions). Parallel lines? That's inconsistency — no point satisfies both.

Here, we have two simple linear equations. Let's check whether they play nicely together.

Step-by-Step Solution

1. Write the system clearly

We have:

2. Choose a method — elimination is cleanest here

Notice the $y$ terms: one is $-y$, the other is $+y$. If we add the two equations, $y$ cancels out immediately. That's the fastest path.

3. Add the equations

Add left-hand sides and right-hand sides separately:

Simplify:

4. Solve for $x$

Divide both sides by 3:

5. Substitute back to find $y$

Use the second equation $x + y = 4$ (it's simpler):

6. Verify with the first equation

Plug $(3, 1)$ into $2x - y = 5$:

It checks out perfectly.

7. Interpret the result

We found exactly one solution: $(x, y) = (3, 1)$. This means the two lines intersect at a single point. The system is consistent (has at least one solution) and independent (exactly one solution).

Why This Approach Works

Elimination is powerful because it reduces the system to a single equation in one variable. Here, the coefficients of $y$ were opposites ($-1$ and $+1$), so adding eliminated $y$ instantly. If they weren't opposites, we'd multiply one equation to make them so — but that wasn't needed.

Alternatively, you could solve by substitution (from $x + y = 4$, get $y = 4 - x$, then plug into $2x - (4 - x) = 5$). You'd get the same answer. The method doesn't matter; the logic of consistency does.

Final Answer

Method: Consistency by the Determinant Test, Confirmed by Elimination

Use this method whenever you are asked to decide if a pair of linear equations has a solution, without necessarily being asked for the solution itself.

Steps

Step 1: Identify the coefficient matrix

For a system $a_1x+b_1y=c_1,\ a_2x+b_2y=c_2$, write

ignoring the constants for now — consistency for a non-zero determinant depends only on $A$.

Step 2: Compute $\det(A)=a_1b_2-a_2b_1$

A non-zero value means $A$ is invertible, so a unique $(x,y)$ exists — the system is consistent by that fact alone, with no need to look at the constants.

Step 3: State the consistency conclusion

Step 4 — Applying to this problem: confirm with elimination

Once you know a unique solution exists, actually finding it is a good habit (and often asked for anyway). Pick whichever variable has opposite-sign coefficients across the two equations and add the equations directly to eliminate it in one step, rather than first hunting for a common multiple — the fastest path when the signs already cooperate.

Common Mistakes

Mistake 1: Assuming 'consistent' means 'infinitely many solutions'

Why it's wrong: consistent only means at least one solution exists — a single unique solution, like the $(3,1)$ found here, is just as consistent as an infinite family. 'Infinitely many' is one particular kind of consistent system (dependent equations), not a requirement for consistency. Correct approach: read off consistency from $\det(A)\neq0$ alone; the count of solutions (one vs. infinite) is a separate question that only arises once $\det(A)=0$.

Mistake 2: A sign slip when adding the equations to eliminate $y$

Why it's wrong: here $2x-y=5$ and $x+y=4$ have opposite-sign $y$ coefficients ($-1$ and $+1$) specifically so that ADDING the equations cancels $y$; subtracting them instead would double $y$ (giving $x-2y=1$) rather than eliminate it, sending the whole elimination off track. Correct approach: before eliminating, check the sign of the coefficient in each equation — add when the signs are already opposite, subtract only when they are the same.