Maths-
General
Easy

Question

fraction numerator x minus 2 over denominator 3 end fraction plus fraction numerator y minus 1 over denominator 4 end fraction equals 13 over 12 and fraction numerator 3 plus y over denominator 3 end fraction plus fraction numerator x minus 2 over denominator 2 end fraction equals 11 over 6 Solve by using the elimination method.

Hint:

simplify the given equation and solve them using the elimination method.

The correct answer is: x=-15 ; y=28


    Ans :- x equals negative 15 space semicolon space y equals 28
    Explanation :-
    fraction numerator x minus 2 over denominator 3 end fraction plus fraction numerator y minus 1 over denominator 4 end fraction equals 13 over 12
    not stretchy rightwards double arrow x over 3 plus y over 4 equals 13 over 12 plus 2 over 3 plus 1 fourth
    not stretchy rightwards double arrow x over 3 plus y over 4 equals fraction numerator 13 plus 8 plus 3 over denominator 12 end fraction
    x over 3 plus y over 4 equals 2 —eq1
    fraction numerator 3 plus y over denominator 3 end fraction plus fraction numerator x minus 2 over denominator 2 end fraction equals 11 over 6
    not stretchy rightwards double arrow 1 plus y over 3 plus x over 2 minus 1 equals 11 over 6
    not stretchy rightwards double arrow y over 3 plus x over 2 equals 11 over 6 — eq2
    Step 1 :-Eliminate x to find y
    Doing 2(eq2) -3(eq1) to eliminate x
    2 open parentheses y over 3 plus x over 2 close parentheses minus 3 open parentheses x over 3 plus y over 4 close parentheses equals 2 open parentheses 11 over 6 close parentheses minus 3 left parenthesis 2 right parenthesis
    not stretchy rightwards double arrow fraction numerator 2 y over denominator 3 end fraction minus fraction numerator 3 y over denominator 4 end fraction plus x minus x equals 11 over 3 minus 6
    not stretchy rightwards double arrow negative y over 12 equals negative 7 over 3 not stretchy rightwards double arrow y equals 12 space cross times space 7 over 3 equals 28
    ∴ y = 28
    Step 2:- substitute the value of y = 28 in eq1
    x over 3 plus y over 4 equals 2 not stretchy rightwards double arrow x over 3 plus 28 over 4 equals 2 not stretchy rightwards double arrow x over 3 plus 7 equals 2
    not stretchy rightwards double arrow x over 3 equals 2 minus 7 not stretchy rightwards double arrow x over 3 equals negative 5
    ∴ x = -15
    therefore x = -15 and y = 28 is the solution to the given pair of equations.

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