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Rajan@2021
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Solve the following systems of equations by using the method of cross multiplication: x/6+y/15=4, x/3-y/12=19/4

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Sir please give me a detailed solution of this question as it was already asked in various important examinations in which we have given two equations x/6+y/15=4, x/3-y/12=19/4 and we have to find the value of variables x and y by using the method of cross multiplication

RS Aggarwal, Class 10, chapter 3C, question no 8

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  1. Simplify given equations:
    x/6+y/15=4
    (5x+2y)/30=4
    5x+2y−120=0
    And x/3−y/12=19/4
    4x−y−57=0
    New set of equation is:
    5x+2y−120=0....(1)
    4x−y−57=0....(2)
    From equation (1):a1​=5,b1​=2 and c1​=−120
    From equation (2):a2​=4,b2​=−1 and c2​=−57

    Using cross multiplication
    x/(−114−120)=y/(−480+285​)=1/(−5−8)​

    x/−234=y/−195​=1/−13​

    x/−234​=1/−13⇒x=−234/-13​

    ⇒x=18
    And
    y/−195​=1/−13​
    Or y=15
    Answer x=18 and y=15

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