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Question

Solve: fraction numerator 5 over denominator x plus y end fraction minus fraction numerator 2 over denominator x minus y end fraction equals negative 1 and fraction numerator 15 over denominator x plus y end fraction plus fraction numerator 7 over denominator x minus y end fraction equals 10

hintHint:

Let fraction numerator 1 over denominator x plus y end fraction equals a and fraction numerator 1 over denominator x minus y end fraction equals b now we get the equation in a and b
Now find values of a and b .After finding a and b substitute them in fraction numerator 1 over denominator x plus y end fraction equals a and fraction numerator 1 over denominator x minus y end fraction equals b and find the values of x and y.

The correct answer is: x = 3 ; y = 2


    Let fraction numerator 1 over denominator x plus y end fraction equals a text  and  end text fraction numerator 1 over denominator x minus y end fraction equals b now we get the equation in a and b

    5 a minus 2 b equals negative 1— eq1

    15 a plus 7 b equals 10 —eq2
    Step 1 :- find value of b by eliminating a
    Do eq2 - 3(eq1) to eliminate a

    15 a plus 7 b minus 3 left parenthesis 5 a minus 2 b right parenthesis equals 10 minus 3 left parenthesis negative 1 right parenthesis

    15 a minus 15 a plus 7 b plus 6 b equals 10 plus 3 not stretchy rightwards double arrow 13 b equals 13

    ∴ b = 1
    Step 2 :- find value of a by substituting b=1 in eq1

    5a - 2b = -1

    not stretchy rightwards double arrow 5 a minus 2 left parenthesis 1 right parenthesis equals negative 1 not stretchy rightwards double arrow 5 a equals 2 minus 1

    not stretchy rightwards double arrow 5 a equals 1

    ∴ a = 1/5
    Step 3 :-  substitute a and b in fraction numerator 1 over denominator x plus y end fraction equals a text  and  end text fraction numerator 1 over denominator x minus y end fraction equals b

    x plus y equals 1 divided by a equals 5 and x minus y equals 1 divided by b equals 1

    Now let x + y = 5 — eq3

    x - y = 1— eq4
    Step 4:- finding x by eliminating y
    Adding eq3 and eq4 we get

    x plus y plus x minus y equals 5 plus 1

    not stretchy rightwards double arrow 2 x equals 6

     therefore space x space equals space 3

    Step 5:- finding y by substituting value of x in eq3

    x plus y equals 5 not stretchy rightwards double arrow 3 plus y equals 5

    not stretchy rightwards double arrow y equals 5 minus 3

     therefore space y space equals space 2
    therefore space x space equals space 3 spaceand y = 2 is the solution to the given pair of equation

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