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CBSE Class 10 Answered

Solve the following pairs of equations by reducing them to a pair of linear equations.
1) 6x+3y=6xy; 2x+4y=5xy
2) 1/(3x+y)+1/(3x-y)=3/4; 1/2(3x+y)-1/2(3x-y)=-1/8
Asked by mkmishra101 | 18 Jun, 2016, 05:30: PM
answered-by-expert Expert Answer
begin mathsize 12px style 1 right parenthesis space 6 straight x plus 3 straight y equals 6 xy semicolon space 2 straight x plus 4 straight y equals 5 xy divide space the space first space equation 6 straight x plus 3 straight y equals 6 xy space by space 3 comma space we space get 2 straight x plus straight y space equals space 2 xy... left parenthesis straight i right parenthesis Let space space 2 straight x plus 4 straight y equals 5 xy space be space equation space left parenthesis ii right parenthesis Divide space equations left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space throughout space by space xy comma space we space get 2 over straight y plus 1 over straight x equals 2... left parenthesis iii right parenthesis space and space 2 over straight y plus 4 over straight x equals 5... left parenthesis iv right parenthesis Observe space that space the space equations space are space not space linear Let space 1 over straight x equals straight p space and space 1 over straight y equals straight q So comma space the space equations space become space straight p plus 2 straight q equals 2 space and space 4 straight p plus 2 straight q equals 5 Substitute space straight p equals 2 minus 2 straight q space in space equation space 4 straight p plus 2 straight q equals 5. So comma space 4 straight p plus 2 straight q equals 5 rightwards double arrow 4 left parenthesis 2 minus 2 straight q right parenthesis plus 2 straight q equals 5 rightwards double arrow 8 minus 8 straight q plus 2 straight q equals 5 rightwards double arrow negative 6 straight q equals negative 3 rightwards double arrow straight q equals 1 half straight p equals 2 minus 2 straight q equals 2 minus 2 open parentheses 1 half close parentheses equals 2 minus 1 equals 1 Thus comma space 1 over straight x equals straight p equals 1 rightwards double arrow straight x equals 1 space and space 1 over straight y equals straight q equals 1 half rightwards double arrow straight y equals 2 Hence comma space the space solution space is space straight x space equals space 1 space and space straight y space equals space 2. end style
2 right parenthesis space fraction numerator 1 over denominator left parenthesis 3 straight x plus straight y right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 3 straight x minus straight y right parenthesis end fraction equals 3 over 4 space and space fraction numerator 1 over denominator 2 left parenthesis 3 straight x plus straight y right parenthesis end fraction minus fraction numerator 1 over denominator 2 left parenthesis 3 straight x minus straight y right parenthesis end fraction equals negative 1 over 8 Multiply space the space second space equation space by space 8 comma space we space get fraction numerator 1 over denominator 2 left parenthesis 3 straight x plus straight y right parenthesis end fraction minus fraction numerator 1 over denominator 2 left parenthesis 3 straight x minus straight y right parenthesis end fraction equals negative 1 over 8 rightwards double arrow fraction numerator 4 over denominator left parenthesis 3 straight x plus straight y right parenthesis end fraction minus fraction numerator 4 over denominator left parenthesis 3 straight x minus straight y right parenthesis end fraction equals negative 1 Let space fraction numerator 1 over denominator left parenthesis 3 straight x plus straight y right parenthesis end fraction equals straight p space and space fraction numerator 1 over denominator left parenthesis 3 straight x minus straight y right parenthesis end fraction equals straight q So comma space the space equations space become space straight p plus straight q equals 3 over 4 space and space 4 straight p minus 4 straight q equals negative 1... left parenthesis 1 right parenthesis Multiply space the space first space equation space be space 4 comma space we space get space 4 straight p plus 4 straight q equals 3... left parenthesis 2 right parenthesis Adding space space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get space 4 straight p minus 4 straight q plus 4 straight p plus 4 straight q equals negative 1 plus 3 8 straight p equals 2 rightwards double arrow straight p equals 1 fourth Substitute space the space value space of space straight p space in space equation space left parenthesis 1 right parenthesis comma space we space get space 4 straight p minus 4 straight q equals negative 1 rightwards double arrow 4 left parenthesis 1 fourth right parenthesis minus 4 straight q equals negative 1 rightwards double arrow 1 minus 4 straight q equals negative 1 rightwards double arrow 2 equals 4 straight q rightwards double arrow straight q equals 1 half Thus comma space straight p equals 1 fourth rightwards double arrow fraction numerator 1 over denominator left parenthesis 3 straight x plus straight y right parenthesis end fraction equals 1 fourth space and space straight q equals fraction numerator 1 over denominator left parenthesis 3 straight x minus straight y right parenthesis end fraction equals 1 half So comma space 3 straight x plus straight y equals 4 space and space 3 straight x minus straight y equals 2 Adding space the space equations space comma space we space get space 6 straight x equals 6 rightwards double arrow straight x equals 1 Substitute space straight x space equals space 1 space in space 3 straight x plus straight y equals 4 rightwards double arrow 3 left parenthesis 1 right parenthesis plus straight y equals 4 rightwards double arrow straight y equals 4 minus 3 equals 1 Hence space the space solution space is space straight x space equals space 1 space and space straight y space equals space 1.
Answered by Rebecca Fernandes | 18 Jun, 2016, 07:04: PM

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