Given below are three equations, Two of them have infinite solutions and two have no solution.
State the two pairs.
$2 x - 3 y = 4, 4 x - 6 y = 7, 6 x - 9 y = 12$

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Solution

Step 1:Applying and Analysing
The given system of equations is of the forms
$2 x - 3 y - 4 = 0 \Rightarrow \Rightarrow 1 4 x - 6 y - 7 = 0 \Rightarrow \Rightarrow 2 6 x - 9 y - 12 = 0 \Rightarrow \Rightarrow 3 T h e g i v e n s y s t e m o f e q u a t i o n s i s o f t h e f o r m s a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
Step 2: Pairing equation 1 & 2
$w h e r e, a_{1} = 2, b_{1} = - 3, c_{1} = - 4 a_{2} = 4, b_{2} = - 6, c_{2} = - 7 \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2} \frac{b_{1}}{b_{2}} = \frac{- 3}{- 6} = \frac{1}{2} \frac{c_{1}}{c_{2}} = \frac{- 4}{- 7} = \frac{4}{7} C l e a r l y, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \Rightarrow \frac{1}{2} = \frac{1}{2} \neq \frac{4}{7} T h e r f o r e, w e c a n s a y t h a t t h e g i v e n s y s t e m o f e q u a t i o n h a s n o s o l u t i o n .$
Step3: pairing, equation 2 & 3
$w h e r e, a_{1} = 4, b_{1} = - 6, c_{1} = - 7 a_{2} = 6, b_{2} = - 9, c_{2} = - 12 t h e r f o r e, \frac{a_{1}}{a_{2}} = \frac{4}{6} = \frac{2}{3} \frac{b_{1}}{b_{2}} = \frac{- 6}{- 9} = \frac{2}{3} \frac{c_{1}}{c_{2}} = \frac{- 7}{- 12} = \frac{7}{12} C l e a r l y, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \Rightarrow \frac{2}{3} = \frac{2}{3} \neq \frac{7}{12}$
Therefore we can say that the given system of equation has no solution.
Step 4: Pairing equation 1 & 3,
$w h e r e, a_{1} = 2, b_{1} = - 3, c_{1} = - 4 a_{2} = 6, b_{2} = - 9, c_{2} = - 12 \frac{a_{1}}{a_{2}} = \frac{2}{6} = \frac{1}{3} \frac{b_{1}}{b_{2}} = \frac{- 3}{- 9} = \frac{1}{3} \frac{c_{1}}{c_{2}} = \frac{- 4}{- 12} = \frac{1}{3} C l e a r l y, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \Rightarrow \frac{1}{3} = \frac{1}{3} = \frac{1}{3}$
Step 5: Result
Therefore, We can say that the given system of equation have infinitely many solutions.
Hence, If pairing the given equation $2 x - 3 y = 4$ , $4 x - 6 y = 7$ the system of equation have no solutions. If pairing $4 x - 6 y = 7$ and $6 x - 9 y = 12$ the system of equation have no solution. If pairing 2x-3y=4 and 6x-9y=12.
Therefore,the system of equation have infinitely many solutions.

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Given below are three equations. Two of them have infinite solutions and two have a unique solutions. state the two pairs.

$3 x - 2 y = 4, 6 x + 2 y = 4, 9 x - 6 y = 12$

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