Question

the numerator of a fraction x/y is 2 less than the denominator. If 3 is added to both the numerator and denominator, then the sum of the new fraction and the original fraction is 53/35 find the quadratic equation.

Answer 1

17y^2 - 89y - 210 = 0

Explanation:

Original fraction: x/y

The numerator is 2 les than the denominator, x = y - 2

Original fraction: (y - 2)/y

New fraction:

Add 3 to the numerator and denominator of the original fraction:

(y - 2 + 3)/(y + 3) = (y + 1)/(y + 3)

Add the new fraction and old fraction and get 53/35.

(y - 2)/y + (y + 1)/(y + 3) = 53/35

35y(y + 3)[(y - 2)/y] + 35y(y + 3)[(y + 1)/(y + 3)] = 35y(y + 3)(53/35)

35(y + 3)(y - 2) + 35y(y + 1) = 53(y^2 + 3y)

35(y^2 + y - 6) + 35(y^2 + y) = 53y^2 + 159y

17y^2 - 89y - 210 = 0

Answer 2

Explanation:

Fraction :x/y

numerator x = y - 2

Now the fraction can be written as: (y-2)/y

3 is added to both denominator and numerator.

After adding 3, the new fraction =
`(y-2 +3 )/(y + 3 ) = (y+ 1 )/(y +3)`

The sum of the new fraction and the original fraction is 53/35

`(y+1)/(y+3) + (y-2)/(y) = (53)/(35)\\\\((y+1)*y)/((y+3)*y)+((y-2)*(y+3))/(y(y+3))=(53)/(35)\\\\(y^(2)+y)/(y^(2)+3y)+(y^(2)+y-6)/(y^(2)+3y)=(53)/(35)\\\\(y^(2)+y+y^(2)+y-6)/(y^(2)+3y)=(53)/(35)\\\\(2y^(2) + 2y - 6)/(y^(2)+3y)=(53)/(35)\\\\`

35*(2y² + 2y - 6) = 53 *(y² + 3y)

35*2y² + 2y * 35 - 6*35 = 53 *y² + 53*3y

70y² + 70y - 210 = 53y² + 159y

70y² + 70y - 210 - 53y² - 159y = 0

70y² - 53y² + 70y - 159y - 210 = 0

17y² - 89y - 210 = 0