Question

please someone solve this linear equation with explaination.​

Answer 1

Step-by-step explanation:

hope this is what you are looking for.

Answer 2

Question

$\boxed{\bf{4p - \frac{p-1}{3}=1 - \frac{p-2}{2}}}$

SOLUTION

$\implies \dfrac{\red{12}p - (p-1)}{3} = \dfrac{\red{2-(p-2)}}{2}$

[By taking LCM = 3 in LHS, and taking LCM = 2 in RHS]

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

$\implies \dfrac{11p +1}{3} = \dfrac{4-p}{2}$

[(In LHS, 12p - p = 11p) and (in RHS the numerator, 2+2 = 4)]

⇒ 2(11p - 1) = 3(4 - p)

[By doing cross multiplication]

⇒ 22p + 2 = 12 - 3p

⇒ 22p + 3p = 12 - 2

[By transposting (-3p) to LHS, and (2) to RHS]

⇒ 25p = 10

⇒ p = 10/25

[by dividing both the numerator and denominator by 5, we get]

⇒ p = 2/5

Thus,

• The value of p = $\frac{2}{5}$

---------

Verification:

LHS

$\boxed{\bf{4p - \frac{p-1}{3}}}$

$=4p - \frac{p-1}{3}$

Now putting the value of p = $\frac{2}{5}$

$= 4 * \frac{2}{5} - (\dfrac{\frac{2}{5}-1}{3})$

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

$= \frac{8}{5} - ({\frac{-3}{5}\div 3)$

$= \frac{8}{5} - ({\frac{-\not{3}}{5}\times \frac{1}{\not{3}})$

$= \frac{8}{5} - ({\frac{-1}{5})$

$= \frac{8}{5}+ {\frac{1}{5}$

$= \frac{8 + 1}{5}$

$\boxed{=\frac{9}{5}}$

And,

RHS

$\boxed{\bf{1 - \frac{p-2}{2}}}$

Putting the value of p = $\frac{2}{5}$

$= 1 - \dfrac{(\frac{2}{5})}{2}$

$= 1 - (\frac{2}{5}\div 2)$

$= 1 - (\frac{\not{2}}{5}\times \frac{1}{\not{2}})$

$= 1 - (\frac{1}{5})$

$= \frac{5 -1}{5}$

$\boxed{= \frac{4}{5}}$