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

If the number of solutions of 3x−y=23x-y=2 and 9x−3y=69x-3y=6 equations are m, then find 1m\dfrac{1}{m}.
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Explanation

Solution

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 am≠bn\dfrac{a}{m}\ne \dfrac{b}{n}
(ii) System of linear equation is consistent with infinitely many solutions if am=bn=cd\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}
(iii) System of linear equation is inconsistent i.e., it has no solution if am=bn≠cd\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.
am≠bn⇒39=am ∴am=13 bn=−1−3=13 \begin{aligned} & \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{aligned}
Where shows am=bn\dfrac{a}{m}=\dfrac{b}{n}
∴\therefore Condition not satisfied.
(ii) am=bn=cd\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}
am=39=13\dfrac{a}{m}=\dfrac{3}{9}=\dfrac{1}{3}
bn=−1−3=13 cd=−2−6=13 \begin{aligned} & \dfrac{b}{n}=\dfrac{-1}{-3}=\dfrac{1}{3} \\\ & \dfrac{c}{d}=\dfrac{-2}{-6}=\dfrac{1}{3} \\\ \end{aligned}
∴\therefore This condition is satisfied.
(iii) am=bn≠cd(iii)\text{ }\dfrac{a}{m}=\dfrac{b}{n}\ne \dfrac{c}{d}
We got am=bn=cd\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d}, so condition not satisfied.
So in this case, condition 2 is true i.e., am=bn=cd\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{d};
Hence, it has an infinite number of solutions.
So m = infinity, hence 1m=0\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.