Solve the following system of equation by substitution method :

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194.

Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y-axis :
x + 3y = 6
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 Multiple Choice QuestionsShort Answer Type

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195. Solve the following system of equation by substitution method :

fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2
straight x over 3 plus straight y over 2 equals 13 over 6


The given equation are 

fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2 space comma space space space space space space space straight x over 3 plus straight y over 2 equals 13 over 6

rightwards double arrow space space space space space space 9 x minus 10 y equals negative 2 cross times 6 rightwards double arrow fraction numerator 2 x plus 3 y over denominator 6 end fraction equals 13 over 6
rightwards double arrow space 9 x minus 10 y equals negative 12 space space space space rightwards double arrow 2 x plus 3 y equals 13

Thus, we have,
9x - 10y = -12    ...(i)
2x + 3y = 13    ...(ii)
From (ii), we get

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Substitution the value of eqn. (iii) in (i). we get

9 open parentheses fraction numerator 13 minus 3 straight y over denominator 2 end fraction close parentheses minus 10 straight y equals negative 12
rightwards double arrow space space space space fraction numerator 9 left parenthesis 13 minus 3 straight y right parenthesis minus 20 straight y over denominator 2 end fraction equals negative 12
rightwards double arrow 9 left parenthesis 13 minus 3 straight y right parenthesis minus 20 straight y space equals negative 24
rightwards double arrow space 117 minus 27 straight y minus 20 straight y equals negative 24
rightwards double arrow space 117 minus 47 straight y space equals negative 24
rightwards double arrow space minus 47 straight y space equals negative 24 minus 117
rightwards double arrow space minus 47 straight y space equals negative 141
rightwards double arrow space straight y space equals space 141 over 47 equals 3
Putting the value of ‘y’ in (iii), we get


straight x space equals space fraction numerator 13 minus 3 cross times 3 over denominator 2 end fraction
rightwards double arrow space space space straight x equals fraction numerator 13 minus 9 over denominator 2 end fraction equals 4 over 2 equals 2
Hence, the solution is x = 2, y = 3.
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square root of 2 straight x end root plus square root of 3 straight y end root equals 9 space space space space space space space space space space space
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