Question

plzzzz Solve..,...... ​

Answer 1

$\boxed{\rm{\red{Question:}}}$

$\dfrac{-2x+1}{3}=5-\dfrac{3(x+4)}{2}$

$\boxed{\sf{\blue{Solution:}}}$

$\implies \dfrac{-2x+1}{3}=\dfrac{\red{10} - [3(x+4)]}{2}$

[By taking LCM = 2 in RHS]

$\implies \dfrac{-2x+1}{3}=\dfrac{\red{10} - 3x-12}{2}$

[In RHS 10 - 12 = 22]

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

[By cross multiplication]

⇒ 2(-2x + 1) = 3( -2 - 3x)

⇒ -4x + 2 = -6 - 9x

[By taking 2 to RHS and (-9x) to LHS]

⇒ 9x - 4x = -6 - 2

⇒ 5x = -8

[By transporting 5 to RHS]

⇒ x = $\red{\frac{-8}{5}}$

$\green{\therefore}$ x = $\frac{-8}{5}$

Hence,

The value of x is $\frac{-8}{5}$

---------

Verification:-

LHS

$= \red{\dfrac{-2x+1}{3}}$

[Puuting the value of x]

$= \dfrac{-2 \times \frac{-8}{5} + 1}{3}$

$= \dfrac{ \frac{16}{5} + 1}{3}$

$= \dfrac{ (\frac{16+5}{5})}{3}$

$=(\frac{21}{5}) \div 3$

$=(\frac{21}{5}) \times \frac{1}{3}$

$\boxed{= \frac{7}{5}}$

Now,

RHS

$= \purple{5-\frac{3(x+4)}{2}}$

[Putting the value of x]

$= \purple{5-\dfrac{3(\frac{-8}{5}+4)}{2}}$

$= \purple{5-\dfrac{3(\frac{-8+20}{5})}{2}}$

$= \purple{5-\dfrac{3(\frac{12}{5})}{2}}$

$= \purple{5-\dfrac{(\frac{36}{5})}{2}}$

$= \purple{\dfrac{(10-\frac{36}{5})}{2}}$

$= \purple{\dfrac{(\frac{50-36}{5})}{2}}$

$= \purple{(\frac{14}{5}) \div 2}$

$= \purple{(\frac{14}{5}) \times \frac{1}{2}}$

$\boxed{=\frac{7}{5}}$

Hence,

LHS = RHS

Answer 2

Answer:

$\boxed{\rm{\red{Question:}}}$

$\dfrac{-2x+1}{3}=5-\dfrac{3(x+4)}{2}$

$\boxed{\sf{\blue{Solution:}}}$

$\implies \dfrac{-2x+1}{3}=\dfrac{\red{10} - [3(x+4)]}{2}$

[By taking LCM = 2 in RHS]

$\implies \dfrac{-2x+1}{3}=\dfrac{\red{10} - 3x-12}{2}$

[In RHS 10 - 12 = 22]

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

[By cross multiplication]

⇒ 2(-2x + 1) = 3( -2 - 3x)

⇒ -4x + 2 = -6 - 9x

[By taking 2 to RHS and (-9x) to LHS]

⇒ 9x - 4x = -6 - 2

⇒ 5x = -8

[By transporting 5 to RHS]

⇒ x = $\red{\frac{-8}{5}}$

$\green{\therefore}$ x = $\frac{-8}{5}$

Hence,

The value of x is $\frac{-8}{5}$

---------

Verification:-

LHS

$= \red{\dfrac{-2x+1}{3}}$

[Puuting the value of x]

$= \dfrac{-2 \times \frac{-8}{5} + 1}{3}$

$= \dfrac{ \frac{16}{5} + 1}{3}$

$= \dfrac{ (\frac{16+5}{5})}{3}$

$=(\frac{21}{5}) \div 3$

$=(\frac{21}{5}) \times \frac{1}{3}$

$\boxed{= \frac{7}{5}}$

Now,

RHS

$= \purple{5-\frac{3(x+4)}{2}}$

[Putting the value of x]

$= \purple{5-\dfrac{3(\frac{-8}{5}+4)}{2}}$

$= \purple{5-\dfrac{3(\frac{-8+20}{5})}{2}}$

$= \purple{5-\dfrac{3(\frac{12}{5})}{2}}$

$= \purple{5-\dfrac{(\frac{36}{5})}{2}}$

$= \purple{\dfrac{(10-\frac{36}{5})}{2}}$

$= \purple{\dfrac{(\frac{50-36}{5})}{2}}$

$= \purple{(\frac{14}{5}) \div 2}$

$= \purple{(\frac{14}{5}) \times \frac{1}{2}}$

$\boxed{=\frac{7}{5}}$

Hence,

LHS = RHS