2. Form the pair of linear equatioรกs i the tulu(if they exist) by

1.	2. Form the pair of linear equatioรกs i the tulu(if they exist) by the elimination methodwe add I to the numerator and substract I from the denominator, a fraction reduces5 if we only add 1 to the denominator What is the fraction?later Nuri will be twice as
Answer» The fraction : x/y Add 1 to numerator and subtract 1 from denominator,gives us 1: (x + 1) / (y - 1) =1 x+1=y-1 x+2=y [i] Add 1 to denominator,gives us 1/2: x/(y+1)=1/2 2x=y+1 [ii] Substitute [i] in [ii]: 2x=(x+2)+1 2x=x+3 x=3. Substitute x=3 in [i]: 3+2=y y=5. The original fraction : 3/5. Like my answer if you find it useful!

2. Form the pair of linear equatioรกs i the tulu(if they exist) by the elimination methodwe add I to the numerator and substract I from the denominator, a fraction reduces5 if we only add 1 to the denominator What is the fraction?later Nuri will be twice as

Answer»

The fraction : x/y

Add 1 to numerator and subtract 1 from denominator,gives us 1:

(x + 1) / (y - 1) =1

x+1=y-1

x+2=y [i]

Add 1 to denominator,gives us 1/2:

x/(y+1)=1/2

2x=y+1 [ii]

Substitute [i] in [ii]:

2x=(x+2)+1

2x=x+3

x=3.

Substitute x=3 in [i]:

3+2=y

y=5.

The original fraction : 3/5.

Like my answer if you find it useful!