Question

3/2x+1=3 solve as directed​

Answer 1

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• $\sf \dfrac{3}{2x \: + \: 1} \: = \: 3. \: Solve \: as \: directed.$

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\sf \dfrac{3}{2x \: + \: 1} \: = \: 3$

• $\sf 3 \: = \: 3(2x \: + \: 1)$

• $\sf 3 \: = \: 6x \: + \: 3$

• $\sf 6x \: = \: 3 \: - \: 3$

• $\sf 6x \: = \: 0$

• $\sf x \: = \: \dfrac{0}{6}$

• $\boxed{\sf x \: = \: 0}$

$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\boxed{\therefore{\sf x \: = \: 0.}}$

$\huge{\bf{\purple{\mathfrak{Verification:-}}}}$

• $\sf \dfrac{3}{2x \: + \: 1} \: = \: 3$

• $\textsf{Substituting x = 0,}$

• $\sf \dfrac{3}{2 \: * \: 0 \: + \: 1} \: = \: 3$

• $\sf \dfrac{3}{1} \: = \: 3$

• $\sf 3 \: = \: 3$

• $\textsf{LHS = RHS}$

• $\textsf{Hence, verified.}$

Answer 2

${\large{\pmb{\sf{\underline{\maltese \: \: Required \; Solution...}}}}}$

Solve as directed -

${\small{\underline{\boxed{\sf{\dfrac{3}{2x+1} \: = 3}}}}}$

According to the question it is cleared that we have to the value of “x” Let's find the value of “x”. Below are some rules to solve this type of questions let's see them.

• (- -) = (+)
• (+ +) = (+)
• (+ -) = (-)
• (- +) = (-)
• (×) = (÷)
• (÷) = (×)

Let's carry on to our given question!

${\small{\underline{\boxed{\sf{\dfrac{3}{2x+1} \: = 3}}}}}$

${\sf{:\implies \dfrac{3}{2x+1} \: = 3}}$

${\sf{:\implies 3 = 3(2x+1)}}$

${\sf{:\implies 3 = 6x + 3}}$

${\sf{:\implies 3 - 3 = 6x}}$

${\sf{:\implies 0 = 6x}}$

${\sf{:\implies \dfrac{0}{6} = x}}$

${\sf{:\implies 0 = x}}$

${\sf{:\implies x = 0}}$

${\small{\underline{\boxed{\sf{\dfrac{3}{2x+1} \: = 3 \: \longrightarrow \bf 0}}}}}$

${\pmb{\sf{\dag \: \: Solution \: is \: 0 \: here}}}$

Let's verify this answer too! To verify we just have to put the value of x as 0.

${\small{\underline{\boxed{\sf{\dfrac{3}{2x+1} \: = 3}}}}}$

${\sf{:\implies \dfrac{3}{2x+1} \: = 3}}$

${\sf{:\implies \dfrac{3}{2(0)+1} \: = 3}}$

${\sf{:\implies \dfrac{3}{0+1} \: = 3}}$

${\sf{:\implies \dfrac{3}{1} \: = 3}}$

${\sf{:\implies 3 = 3}}$

${\sf{:\implies RHS \: = LHS}}$

${\pmb{\sf{\dag \: \: Henceforth, \: verified}}}$