Question

Chapter : Linear Equations

Solve the following equations

please solve this

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Answer 1

$\sf{Solution}$

Step by step explanation :-

Given :-

$\sf\dfrac{5k+4}{2}$ - $\sf\dfrac{3(k-4)}{3}$ =

$2 \frac{1}{6}$

In RHS The given fraction is in mixed fraction So, convert into improper fraction

$2 \frac{1}{6} = \frac{2 \times 6 + 1}{6}$

$2 \frac{1}{6} = \frac{13}{6}$

So,

$\sf\dfrac{5k+4}{2}$ - $\sf\dfrac{3(k-4)}{3}$ = $\sf\dfrac{13}{6}$

Take LCM to the denominators in LHS

LCM Of 2 , 3 is 6

$\sf\dfrac{(5k+4)3- 3(k-4) (2)}{6}$ = $\sf\dfrac{13}{6}$

(5k + 4 ) 3 - 3 (k - 4 ) (2) = 13

Simplify the LHS

15k + 12 - 6 (k-4) = 13

15k + 12 - 6k + 24 = 13

9k + 36 = 13

9k = 13 - 36

9k = -23

k = $\sf\dfrac{-23}{9}$

Know more :-

Transportations:-

If we transpose + after transpose it becomes -

If we tranpose - after tranpose it becomes +

If we tranpose × after tranpose it becomes ÷

If we tranpose ÷ after tranpose it becomes ×

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Multiplication of signs:-

(+) ( -) = -

(-) ( +) = -

( + ) ( + ) = +

(-) (-) = +

Answer 2

What to do:

$\to$We have to solve this

Given equation:

$\to \mathsf{\frac{5k + 4}{2} \ - \ \frac{3(k-4)}{3} =2\frac{1}{6} }$

Solution:

$\\\to \mathsf{\frac{5k + 4}{2} \ - \ \frac{3(k-4)}{3} =2\frac{1}{6} }\\\to \mathsf{or, \frac{5k+4}{2} -\frac{3k-12}{3}=\frac{13}{6}} \\\\\to \mathsf{or,\frac{3(5k+4)-2(3k-12)}{6} =\frac{13}{6} }\\\\\to \mathsf{or,\frac{15k+12-6k+24}{6} =\frac{13}{6} }\\\\\to \mathsf{or,6(15k+12-6k+24)=13 \ * \ 6}\\\\\to \mathsf{or,90k+72-36k+144=78}\\\\\to \mathsf{or,90k - 36k=78-144-72}\\\\\to \mathsf{or,54k=-138}\\\\\to \mathsf{or,k=\frac{-138}{54} }\\\\\to \mathsf{or,k=\frac{-23}{9} }$