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प्रश्न
Solve.
`( 5y^2 + 40y - 12)/( 5y + 10y^2 - 4) = ( y + 8)/( 1 + 2y)`
उत्तर
`( 5y^2 + 40y - 12)/( 5y + 10y^2 - 4) = ( y + 8)/( 1 + 2y)`
If y = 0, then `[5 xx 0 + 40 xx 0 - 12]/[ 5 xx 0 + 10 xx 0 - 4 ] = (0+8)/( 1 + 2 xx 0 ) ⇒ { -12}/{ - 4 } = 8/1`
So, y = 0 is not a solution of the given equation.
Now,
⇒ `( 5y^2 + 40y - 12)/( 5y + 10y^2 - 4) = [5y( y + 8)]/[ 5y( 1 +2y)]` ...[Multiplying numerator and denominator of second ratio by 5y]
⇒ `[ 5y^2 + 40y - 12 - 5y^2 - 40y ]/[ 5y + 10y^2 - 4 - 5y - 10y^2 ]` ...(Theorem of equal ratios)
⇒ `[-12]/[-4]`
`therefore ( y + 8)/( 1 +2y) = [-12]/[-4] = 3/1`
⇒ y + 8 = 3(1 + 2y)
⇒ y + 8 = 3 + 6y
⇒ 6y − y = 8 − 3
⇒ 5y = 5
⇒ y = 1
Thus, the solution of the given equation is y = 1.
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