pl some one answer these questions.
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bhargu:
i have to submit this by 9 june
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i) x + y =3.3 and (0.6/3x-2y)-2y= -1
=>x=3.3-y------------(1) and 0.6/{3(3.3-y)}-2y=-1
=>0.6/(9.9-3y-2y)=-1----------------From (1)
=>0.6/(9.9-5y)=-1 =>0.6=-1(9.9-5y) =>0.6=-9.9+5y =>10.5=5y =>y=2.1
ii)1/2x-1/y=-1 and 1/x+1/2y=8
=>y-2x/2xy=-1 =>y-2x=-2xy =>y+2xy=2x =>y(1+2x)=2x =>y=2x/(1+2x)
Therefore,1/x+1/2y=8 =>1/x+1/2[2x/(1+2x)]=8 =>1/x+2x/(2+4x)=8 =>2+4x+2x²/[x(2+4x)]=8
=>2x²+4x+2/(2x+4x²)=8 =>2x²+4x+2=8(2x+4x²) =>2x²+4x+2=16x+32x²
=>32x²-2x²+16x-4x-2=0 =>30x²+12x-2=0 =>2(15x²+6x-1)=0 =>15x²+6x-1=0 =>...........question is may be wrong.
iii) 43x+67y=- 24 and 67x+ 43y= 24
=>
=>x=3.3-y------------(1) and 0.6/{3(3.3-y)}-2y=-1
=>0.6/(9.9-3y-2y)=-1----------------From (1)
=>0.6/(9.9-5y)=-1 =>0.6=-1(9.9-5y) =>0.6=-9.9+5y =>10.5=5y =>y=2.1
ii)1/2x-1/y=-1 and 1/x+1/2y=8
=>y-2x/2xy=-1 =>y-2x=-2xy =>y+2xy=2x =>y(1+2x)=2x =>y=2x/(1+2x)
Therefore,1/x+1/2y=8 =>1/x+1/2[2x/(1+2x)]=8 =>1/x+2x/(2+4x)=8 =>2+4x+2x²/[x(2+4x)]=8
=>2x²+4x+2/(2x+4x²)=8 =>2x²+4x+2=8(2x+4x²) =>2x²+4x+2=16x+32x²
=>32x²-2x²+16x-4x-2=0 =>30x²+12x-2=0 =>2(15x²+6x-1)=0 =>15x²+6x-1=0 =>...........question is may be wrong.
iii) 43x+67y=- 24 and 67x+ 43y= 24
=>
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