How do you solve #(1 - y)/(1 + y) = 2/3#?
1 Answer
May 30, 2016
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#