Linear Equations in Two Variables

[@ManochaAcademy]

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[@ambikasivakumar338]

3:31 we can also do it in this method x+y=10 (1) x -y= 2 (2) Adding eq(1)&(2) x + y=10 x - y=2 2x = 12 x = 6 Subtituting value of x in eq (1) 6+y =10 y = 10-6 y = 4 Solution (6,4)

[@MHARecordsOfficial]

ANS#1: 2x + y =7 (i) 4x - 3y + 1 = 0 (ii) (i)2x = 7 - y substituting 2x in equa..(ii) we make 4x as 2(2x) in equation (ii) 2(2x) - 3y + 1= 0 2(7 - y) - 3y + 1 = 0 14 - 2y -3y = -1 -5y = -15 y = -15\-5 y = 3 substituting value of "y" in equation (i) (i) 2x + y = 7 2x + 3 =7 2x = 7 - 3 2x = 4 x = 4\2 x = 2 SOLUTIION SET ( 2 , 3 )

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1. x=2 , y=3 2. x=5 , y=0 3. 78 Manocha sir , your classes are amazing . Understood the concepts very well

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But i got 1st answer as x =1 and y = 5

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Is this correct that aome equation has many solution

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No you are wrong , if we substitute 1 and 5 for x and y the second equation won't hold true

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Ans 1 -> x=2 , y=3 Ans 2 -> solution is (5,0) Ans 2 -> 87 sir you are awesom ✔✔

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[@vishnusai3918]

SIR, PLEASE ANSWER THIS DOUBT. We have the standard form of a linear equation as: ax+b=0 But these are also linear equations: 1. 25x+25=100 (Here the RHS is 100, but the standard form states a linear equation has 0 as RHS.) 2. 25x-25=100 (Here the minus or subtraction(-) sign has been used, but the standard form states that we use the plus or addition sign(+).) SPECIFY SUITABLE PERFECT REASONS FOR THIS.

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Q1 x=2 & y=3 Q2 x=5 & y=0 vertices are: 20,12 5,0 15,6 A=4.32cm2 Q3 78

[@ManochaAcademy]

Very good! Q1 and Q3 answers are correct! Q2, one of the vertex (5, 0) is correct. Other two vertices are not right. You can plot it on the graph and see where these lines are intersecting the y-axis. These will give you the other 2 points. Let me know what answer you get or if there are any doubts!

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1) x=2, y=3 2) x=5, y=0 3)87

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Good luck!

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Welcome