Question

solve the following equations​

Answer 1

Answer:

value of x is -2........

Answer 2

Required Solution:

$\sf \blue { \dfrac{3x}{10(2-x)-3(x+2)}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{3x}{10(2-x)-3(x+2)}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{3x}{20-10x-3x-6}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{3x}{20-6-10x-3x}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{3x}{14-13x}=\dfrac{-6}{40} }$

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$\rm {⇢ 3x(40) = -6(14-13x) }$

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$\rm {⇢120x= -84+78x }$

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$\rm {⇢-84= 120x-78x }$

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$\rm {⇢-84= 42x }$

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$\rm {⇢x = \dfrac{-84}{42} }$

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$\boxed { \rm {⇢x = -2} }$

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Verification!

Let's verify it also by substituting the value of x in the equation.

$\rm {⇢ \dfrac{3x}{10(2-x)-3(x+2)}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{3(-2)}{10[2-(-2)]-3(-2+2)}=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{-6}{10(4) \cancel{+6-6} }=\dfrac{-6}{40} }$

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$\rm {⇢ \dfrac{-6}{40}=\dfrac{-6}{40} }$

Therefore,LHS = RHS

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