Modulus(2a-3) = 3a+2, which can also be written as |2a-3| = 3a+2.
Modulus function f(x) = |x| is defined as
|x| = x, for x>0,
|x| = -x, for x<0
So, we get two cases here,
Case (1):
2a-3>0
$$\Rightarrow a>\frac{3}{2}$$........................(1)
|2a-3| = 2a-3 = 3a+2
$$\Rightarrow a = -5$$
But from equation (1), The above value of a = -5 is not a possible case.
Case (2):
2a-3<0
$$\Rightarrow a<\frac{3}{2}$$....................(2)
|2a-3| = -(2a-3) = 3a+2
$$\Rightarrow a = \frac{1}{5}$$
This is a possible case, hence the value of $$a = \frac{1}{5}$$
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