How do you solve abs((5y+2)/(2))= 6?

1 Answer
Apr 2, 2015

The answer: {-14/5,2}

Absolute value operation always return non-negative.

So,

if (5y+2)/2>=0 ->abs((5y+2)/2) = (5y+2)/2

if (5y+2)/2<0 ->abs((5y+2)/2) = (-1) * ((5y+2)/2)

Since we don't know the value of y, we need to check all these possibilities.

First

Assume that (5y+2)/2 >= 0

We need to find the range of y

2 * ((5y+2)/2) >= 2 * 0

5y +2 >= 0

A: y>=-2/5

We need to remember this.

Now, try to solve the given equation.

(5y+2)/2 = 6

5y+2=12
5y=10
y = 2

Remember expression A, y=2 satisfies the inequality in expression A. So 2 is in the solution set.

Second

Assume that (5y+2)/2<0

B: y < -2/5

Now try to solve the given equation again, but this time the absolute value operation will return (-1) times of the input since the input is assumed negative.

(-1) * (5y+2)/2 = 6

(5y+2)/2 = -6

5y+2 = -12

5y = -14

y = -14/5

y=-14/5 satisfies the inequality in the expression B so -14/5 is also in our solution set.

Result: {-14/5,2}