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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 1 and fraction numerator 15 over denominator x plus y end fraction minus 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 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
Now find values of a and b .After finding a and b substitute them in and and find the values of x and y .

The correct answer is: x=-24/91 ; y=-11/91


    Let  and now we get the equation in a and b

    5 a minus 2 b equals 1 —- eq1

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

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

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

    ∴ b = -7
    Step 2 :- find value of a by substituting b= -7 in eq1

    5 a minus 2 b equals 1

    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 5 a minus 2 left parenthesis negative 7 right parenthesis equals 1 end cell row cell not stretchy rightwards double arrow 5 a plus 14 equals 1 end cell row cell not stretchy rightwards double arrow 5 a equals negative 13 end cell end table

    ∴ a = -13/5
    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 negative 5 divided by 13 and x minus y equals 1 divided by b equals negative 1 divided by 7

    Now let x plus y equals negative 5 divided by 13 — eq3

    x minus y equals negative 1 divided by 7 —- eq4
    Step 4:- finding x by eliminating y
    Adding eq3 and eq4 we get

    x plus y plus x minus y equals negative 5 divided by 13 minus 1 divided by 7

    not stretchy rightwards double arrow 2 x equals negative 48 divided by 91

     therefore space x space equals space fraction numerator negative 24 over denominator 91 end fraction space left parenthesis o r right parenthesis space minus space 0.2637362637363
    Step 5:- finding y by substituting value of x in eq 3

    x plus y equals negative 5 divided by 13 not stretchy rightwards double arrow negative 24 divided by 91 plus y equals negative 5 divided by 13

    not stretchy rightwards double arrow y equals 24 divided by 91 minus 5 divided by 13 not stretchy rightwards double arrow y equals left parenthesis 24 minus 35 right parenthesis divided by 91

     therefore space y space equals space fraction numerator negative 11 over denominator 91 end fraction space left parenthesis o r right parenthesis space minus 0.1208791208791
    therefore space x space equals space fraction numerator negative 24 over denominator 91 end fraction y space equals space fraction numerator negative 11 over denominator 91 end fraction is the solution to the given pair of equations.


     
     

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