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Compare the given equations with the general equation of linear equations. Check them with the conditions of consistency for linear equations.

“Complete step-by-step answer:”

Let us consider the general linear equation ax + by + c = 0

and another equation mx + ny + d = 0.

ax + by + c = 0

mx + ny + d = 0

Compare both the equation with the conditions of consistency for linear equations;

(i) System of linear equations is consistent with unique solution if $\dfrac{a}{m}\ne \dfrac{b}{n}$

(ii) System of linear equation is consistent with infinitely many solutions if $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$

(iii) System of linear equation is inconsistent i.e., it has no solution if $\dfrac{a}{m}=\dfrac{b}{n}\ne \dfrac{c}{d}$

Let us consider 3x – y = 2, compare it with general equation,

ax + by + c = 0

$\therefore $ a = 3, b = -1, c = -2

Compare ax – 3y = 6 with general equation mx + ny + d = 0.

$\therefore $m = 9, n = -3, d = -6

Now check with all three conditions.

$\begin{align}

& \dfrac{a}{m}\ne \dfrac{b}{n}\Rightarrow \dfrac{3}{9}=\dfrac{a}{m} \\

& \therefore \dfrac{a}{m}=\dfrac{1}{3} \\

& \dfrac{b}{n}=\dfrac{-1}{-3}=\dfrac{1}{3} \\

\end{align}$

Where shows $\dfrac{a}{m}=\dfrac{b}{n}$

$\therefore $Condition not satisfied.

(ii) $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$

$\dfrac{a}{m}=\dfrac{3}{9}=\dfrac{1}{3}$

$\begin{align}

& \dfrac{b}{n}=\dfrac{-1}{-3}=\dfrac{1}{3} \\

& \dfrac{c}{d}=\dfrac{-2}{-6}=\dfrac{1}{3} \\

\end{align}$

$\therefore $This condition is satisfied.

$(iii)\text{ }\dfrac{a}{m}=\dfrac{b}{n}\ne \dfrac{c}{d}$

We got $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$, so condition not satisfied.

So in this case, condition 2 is true i.e., $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$;

Hence, it has an infinite number of solutions.

So m = infinity, hence $\dfrac{1}{m}=0$.

Note: Substitute values of a, b, c, m, n and d on each condition of consistency.

If a system has at least 1 solution, it is consistent.

If a consistent system has exactly 1 solution, it is independent.

If a consistent system has an infinite number of solutions, it is dependent.

“Complete step-by-step answer:”

Let us consider the general linear equation ax + by + c = 0

and another equation mx + ny + d = 0.

ax + by + c = 0

mx + ny + d = 0

Compare both the equation with the conditions of consistency for linear equations;

(i) System of linear equations is consistent with unique solution if $\dfrac{a}{m}\ne \dfrac{b}{n}$

(ii) System of linear equation is consistent with infinitely many solutions if $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$

(iii) System of linear equation is inconsistent i.e., it has no solution if $\dfrac{a}{m}=\dfrac{b}{n}\ne \dfrac{c}{d}$

Let us consider 3x – y = 2, compare it with general equation,

ax + by + c = 0

$\therefore $ a = 3, b = -1, c = -2

Compare ax – 3y = 6 with general equation mx + ny + d = 0.

$\therefore $m = 9, n = -3, d = -6

Now check with all three conditions.

$\begin{align}

& \dfrac{a}{m}\ne \dfrac{b}{n}\Rightarrow \dfrac{3}{9}=\dfrac{a}{m} \\

& \therefore \dfrac{a}{m}=\dfrac{1}{3} \\

& \dfrac{b}{n}=\dfrac{-1}{-3}=\dfrac{1}{3} \\

\end{align}$

Where shows $\dfrac{a}{m}=\dfrac{b}{n}$

$\therefore $Condition not satisfied.

(ii) $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$

$\dfrac{a}{m}=\dfrac{3}{9}=\dfrac{1}{3}$

$\begin{align}

& \dfrac{b}{n}=\dfrac{-1}{-3}=\dfrac{1}{3} \\

& \dfrac{c}{d}=\dfrac{-2}{-6}=\dfrac{1}{3} \\

\end{align}$

$\therefore $This condition is satisfied.

$(iii)\text{ }\dfrac{a}{m}=\dfrac{b}{n}\ne \dfrac{c}{d}$

We got $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$, so condition not satisfied.

So in this case, condition 2 is true i.e., $\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}$;

Hence, it has an infinite number of solutions.

So m = infinity, hence $\dfrac{1}{m}=0$.

Note: Substitute values of a, b, c, m, n and d on each condition of consistency.

If a system has at least 1 solution, it is consistent.

If a consistent system has exactly 1 solution, it is independent.

If a consistent system has an infinite number of solutions, it is dependent.

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