Q.Examine the consistency of the following system of equations: $x + y + z = 1$ $2x + 3y + 2z = 2$ $ax + ay + 2az = 4$

Solve the system $x+y+z=1,\ 2x+3y+2z=2,\ ax+ay+2az=4$ (the third equation carries a parameter $a$).

Eliminate: $(\text{eq2})-2(\text{eq1})$ gives $y=0$. Then eq1 gives $x+z=1$, i.e. $x=1-z$.

Third equation: $a(x+y+2z)=4$. With $y=0$ and $x=1-z$, $x+2z = 1+z$, so $a(1+z)=4$.

• If $a\neq0$: $z=\dfrac{4}{a}-1$, $x=2-\dfrac{4}{a}$, $y=0$ — a unique solution (and $x+y+z=1$). …

This is a system of three linear equations in $x,y,z$, with a parameter $a$ in the third. We reduce it step by step and see how the answer depends on $a$.

1. Remove $y$ using the first two equations

Compute $(\text{eq2})-2(\text{eq1})$:

With $y=0$, eq1 becomes $x+z=1$, i.e. $x=1-z$.

2. Bring in the third equation

Factor the third equation: $a(x+y+2z)=4$. Substituting $y=0$ and $x=1-z$,

so the third equation reads

3. Split on the parameter $a$ …

Method: Consistency of a System with a Parameter

When one equation of a 3-variable system carries a letter parameter (here $a$) instead of a fixed number, the technique is to eliminate the parameter-free variables first, then examine how the parameter controls the remaining equation.

Steps

Step 1: Eliminate variables using only the parameter-free equations

Combine the equations that do NOT contain the parameter (here, equations 1 and 2) using standard elimination, to pin down as many variables as possible before the parameter ever enters the picture.

Step 2: Express the remaining variable(s) in terms of one free variable

Use the parameter-free relation from Step 1 (typically of the form $x+z=\text{constant}$ or similar) to write one variable in terms of another, ready to substitute into the parameter equation.

Step 3: Substitute into the parameter equation and isolate the parameter's role

After substitution, the third equation collapses to the parameter multiplying a single expression in one remaining variable.

Step 4: Split into cases on whether the parameter is zero …

Common Mistakes

Mistake 1: Dividing by $a$ without first considering $a=0$ separately

Why it's wrong: the third equation reduces to $a(1+z)=4$; solving this as $z=\frac{4}{a}-1$ silently assumes $a\neq0$. If $a=0$, this division is undefined, and the equation instead becomes $0=4$, an outright contradiction — a completely different (inconsistent) case that gets missed entirely if you divide first without checking. Correct approach: whenever a parameter multiplies an expression that must be isolated, explicitly test the parameter's zero value as its own case before dividing by it.

Mistake 2: Not verifying the $a\neq0$ solution against the untouched first equation …