How do you solve #(1 - y)/(1 + y) = 2/3#?

1 Answer
May 30, 2016

Answer:

#y=1/5#

Explanation:

When we have 2 fractions equal to each other then the way to eliminate the fractions is to #color(blue)"cross-multiply"#

#rArra/b=c/drArr ad=bc#

here we obtain

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

multiply out brackets.

hence : 2 + 2y = 3 - 3y

now collect like terms , y to the left and numbers on right. Remember to change the sign of a term when it is moved from one side to the other.

thus : 2y + 3y = 3 - 2 → 5y = 1

divide both sides by 5

#rArr(cancel(5)^1 y)/cancel(5)^1=1/5rArry=1/5#