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Question
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes `6/5`. Thus, we have
`(2x)/(y-5)=6/5`
`⇒ 10x=6(y-5)`
`⇒ 10x=6y-30`
`⇒ 10x-6y+30 =0`
`⇒ 2(5x-3y+15)=0`
`⇒ 5x - 3y+15=0`
If the denominator is doubled and the numerator is increased by 8, the fraction becomes `2/5`. Thus, we have
`(x+8)/(2y)=2/5`
`⇒ 5(x+8)=4y`
`⇒ 5x+40=4y`
`⇒ 5x-4y+40=0`
So, we have two equations
`5x-3y+15=0`
`5x-4y+40=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-3)xx40-(-4)xx15)=-y/(5xx40-5xx15)=1/(5xx(-4)-5xx(-3))`
`⇒ x/(-120+60)=(-y)/(200-75)=1/(-20+15)`
`⇒x/(-60)=-y/125``=1/-5`
`⇒ x= 60/5,y=125/5`
`⇒ x=12,y=25`
Hence, the fraction is `12/25`
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If 3 is added to the denominator and 2 is subtracted from the numerator, the fraction becomes `1/4`. Thus, we have
`(x-2)/(y+3)=1/4`
`⇒ 4(x-2)=y+3`
`⇒ 4x-8=y+3`
`⇒ 4x-y-11=0`
If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes `2/3`. Thus, we have
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`⇒ 3(x+6)=6y`
`⇒ 3x +18 =6y`
`⇒ 3x-6y+18=0`
`⇒ 3(x-2y+6)=0`
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Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-1)xx6-(-2)xx(-11))=(-y)/(4xx6-1xx(-11))=1/(4xx(-2)-1xx(-1))`
`⇒ x/(-6-22)=-y/(24+11)=1/(-8+1)`
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`⇒ x= 28/7,y=35/7`
`⇒ x= 4,y=5`
Hence, the fraction is`4/5`
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