Question

Find the value of a, if:Modulus(2a-3)=3a+2​

Answer 1

$\textbf{Concept:}$

$\text{If } |x|=u, \text{ then} \;x=\pm\;u$

$\textbf{Given:}$

$|2a-3|=3a+2$

$\text{Using the given concept,}$

$2a-3=\pm(3a+2)$

$\text{case(i):}$

$2a-3=3a+2$

$-2-3=3a-2a$

$\implies\bf\;a=-5$

$\text{case(ii):}$

$2a-3=-(3a+2)$

$2a-3=-3a-2$

$2a+3a=3-2$

$5a=1$

$\implies\bf\;a=\frac{1}{5}$

$\text{But a=-5 does not satisfy the given equation }$

$\therefore\text{The solution is }\bf\;a=\frac{1}{5}$

Answer 2

Answer:

\textbf{Concept:}Concept:

\text{If } |x|=u, \text{ then} \;x=\pm\;uIf ∣x∣=u, thenx=±u

\textbf{Given:}Given:

|2a-3|=3a+2∣2a−3∣=3a+2

\text{Using the given concept,}Using the given concept,

2a-3=\pm(3a+2)2a−3=±(3a+2)

\text{case(i):}case(i):

2a-3=3a+22a−3=3a+2

-2-3=3a-2a−2−3=3a−2a

\implies\bf\;a=-5⟹a=−5

\text{case(ii):}case(ii):

2a-3=-(3a+2)2a−3=−(3a+2)

2a-3=-3a-22a−3=−3a−2

2a+3a=3-22a+3a=3−2

5a=15a=1

\implies\bf\;a=\frac{1}{5}⟹a=51

\text{But a=-5 does not satisfy the given equation }But a=-5 does not satisfy the given equation 

\therefore\text{The solution is }\bf\;a=\frac{1}{5}∴The solution is a=51