The denominator of a fraction is 4 more than the numerator. if both the numerator and the denominator are increased by 1, the resulting fraction equals 1/2. find the original fraction.

Step-by-step explanation:

Given as :

Let The original fraction =

Where numerator = x

And denominator = y

The denominator of a fraction is 4 more than the numerator.

I.e y = x + 4          

Or, y - x =  4           ........A

Again

If both the numerator and the denominator are increased by 1, the resulting fraction = 1/2

I.e   =  

using cross multiplication

2 × (x + 1) = 1 × (y + 1)

Or, 2 x + 2 = y + 1

Or, 2 x - y = 1 - 2

i.e 2 x - y = - 1          ..........B

Solving eq A and eq B

So, (2 x - y) + (y - x) = - 1 + 4

Or, (2 x - x) + ( - y + y) = 3

Or, x + 0 = 3

∴  x = 3

So, The numerator = x = 3

Now, put the value of x into eq A

∵ y - x =  4

Or, y - 3 = 4

Or, y = 4 + 3

∴   y = 7

So, The denominator = y = 7

So, The original fraction = =

Hence, The original fraction is . Answer

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.