Question

solve the following
equation​

Answer 1

Answer:

2p+6/7=1/3-p

2p+1p=1/3 -6/7

3p=7-18/21

3p=-11/21

p=-11/21 ×1/3

p=-11/63

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Answer 2

Question

$\sf \large 2p + \dfrac{6}{7} = \dfrac{1}{3} - p$

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \longmapsto2p + \dfrac{6}{7} = \dfrac{1}{3} - p$

$\sf \longmapsto \dfrac{14p + 6}{7} = \dfrac{1 - 3p}{3}$

Now,cross multiply to both sides:

$\sf \longmapsto3(14p + 6) = 7(1 - 3p)$

=>expanding the terms and removing brackets

=>3×14p+3×6=7×1+7×-3p

$\sf \longmapsto42p + 21p = 7 - 18$

$\sf \longmapsto63p = - 11$

$\sf \longmapsto \: p = - \dfrac{11}{63}$

Check:

$\sf \longmapsto2p + \dfrac{6}{7} = \dfrac{1}{3} - p$

$\sf \large\longmapsto2 \times - \frac{11}{63} + \frac{6}{7} = \frac{1}{3} - ( - \frac{11}{63} )$

$\sf \longmapsto - \dfrac{22}{63} + \dfrac{6}{7} = \dfrac{1}{3} + \dfrac{11}{63}$

$\sf \longmapsto \dfrac{ - 22 + 54}{63} = \dfrac{21 + 11}{63}$

$\sf \longmapsto \dfrac{32}{63} = \dfrac{32}{63}$

Since,LHS =RHS ( verified)