Question

The ratio of the numerator to the denominator of a fraction is 2 to 3. If both the numerator and the denominator are increased by 2, the fraction becomes 3/4. What is the original fraction?

Which of the following systems of equations can be used to solve the problem?
n + 4 = 3 and d + 5 = 4
3n - 2d = 0 and 4n + 2 = 3d + 2
3n = 2d and 4n + 8 = 3d + 6

Answer 1

Let the numerator be x, and the denominator be y.

The ratio of the numerator to the denominator of a fraction is 2 to 3.

>> x/y = 2/3

If both the numerator and the denominator are increased by 2, the fraction becomes 3/4.

>> (x + 2)/(y + 2) = 3/4

From x/y = 2/3, x = (2/3)y

From (x + 2)/(y + 2) = 3/4, cross multiply and we have

4(x + 2) = 3(y + 2)

Substitute x = (2/3)y into the equation above.

4[(2/3)y + 2] = 3(y + 2)

Distribute the lefh hand side and solve for y.

(8/3)y + 8 = 3y + 6

(8/3)y - 3y = 6 - 8

-(1/3)y = -2

y = 6

x = (2/3)y = (2/3)(6) = 4

Hence the numerator is 4, and the denominator is 6.

Hope this helps!

Answer 2

3n = 2d and 4n + 8 = 3d + 6

Explanation:

The ratio of the numerator to the denominator of a fraction is 2 to 3. If both the numerator and the denominator are increased by 2, the fraction becomes 3/4.

Let n be the numerator and d be the denominator

The ratio of the numerator to the denominator of a fraction is 2 to 3

`(n)/(d) =(2)/(3)`

Corss multiply it , the equation becomes

3n = 2d

If both the numerator and the denominator are increased by 2, the fraction becomes 3/4.

add 2 with n and d

`(n+2)/(d+2) =(3)/(4)`

Cross multiply it

4(n+2) = 3(d+2)

4n+8 = 3d+6