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If the number of solutions of $3x-y=2$ and $9x-3y=6$ equations are m, then find $\dfrac{1}{m}$.
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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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