Explanation
⇒3x−23y+7=5x−15y+16⇒15xy−10y+48x−32=15xy+35x−3y−7⇒13x−7y=25.........eq1⇒3x−15x−9=6y−52y+3⇒6xy−30y+9x−45=6xy−54y−5x+45⇒14x+24y=90.......eq2Multiply eq1 by 24 and eq2 by 7⇒312x−168y=600.......eq3⇒98x+168y=630.......eq4Add eq3 and eq4⇒410x=1230⇒x=3Put x=3 in eq 1⇒13×3−7y=25⇒−7y=−14⇒y=2Hence, x=3 and y=2
Let xy be the fraction, where x and y are positive integers.
⇒x+2y+2=911 ..(1)⇒x+3y+3=56 ..(2)
From (1) we get ⇒11x+22=9y+18⇒11x−9y=−4 ...(3)
From (2) we get ⇒6x+18=5y+15⇒6x−5y=−3 ...(4)
From (4) will gety=3+6x5
Now substituting y in (3),
⇒11x−9(3+6x)5=−4
⇒55x−27−54x=−20
⇒55x−54x=−20+27
⇒x=7
Now substituting x in (4)
⇒6×7−5y=−3
⇒42−5y=−3
⇒−5y=−3−42
⇒−5y=−45
⇒y=9
Hence, x=7 and y=9.
And fraction xy=79
The given equations are
\dfrac { 1 }{ x-1 } +\dfrac { 1 }{ y-2 } =2 and
\dfrac { 6 }{ x-1 } -\dfrac { 2 }{ y-2 } =1
Let \dfrac { 1 }{ x-1 } =u and \dfrac { 1 }{ y-2 } =v
u+v=2 .......(i) and
6u-2v=1 ......(ii)
Multiply (i) by 6
\Rightarrow 6u+6v=12 ......(iii)
Subtract (ii) from (iii)
\Rightarrow 8v=11
\Rightarrow v=\dfrac { 11 }{ 8 }
Therefore \dfrac { 1 }{ y-2 } =\dfrac { 11 }{ 8 }
\Rightarrow y-2=\dfrac { 8 }{ 11 }
\Rightarrow y=\dfrac { 30 }{ 11 }
Putting v=\dfrac { 11 }{ 8 } in (i), we get
u+\dfrac { 11 }{ 8 } =2
\Rightarrow u=\dfrac { 5 }{ 8 }
\Rightarrow \dfrac { 1 }{ x-1 } =\dfrac { 5 }{ 8 }
\Rightarrow x-1=\dfrac { 8 }{ 5 }
\Rightarrow x=\dfrac { 13 }{ 5 }
Therefore, \left( x,y \right) =\left( \dfrac { 13 }{ 5 } ,\dfrac { 30 }{ 11 } \right)
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