Question

CHAPTER → LINEAR EQUATION IN ONE VARIABLE

CLASS → 6TH

BOOK → DR.RAJESH KUMAR THAKUR

SUB→MATHS


I will give him a BRAINILAGIST MARK​

Answer 1

3x + 3(x + 1) + 3(x + 2) = 999

By taking 3 common from all terms in LHS,

⇒ 3[(x) + (x + 1) + (x + 2)] = 999

⇒ x + x + 1 + x + 2 = 999/3

⇒ 3x + 3 = 333

⇒ 3x = 333 - 3

⇒ 3x = 330

⇒ x = 330/3

⇒ x = 110 (ans.)

{Verification:-

RHS: 999

LHS:

3x + 3(x + 1) + 3(x + 2)

= 3(110) + 3(110 + 1) + 3(110 + 2)

= 330 + 333 + 336

= 999 = RHS.

Hence, verified.}

Answer 2

$\LARGE{\underline{\purple{\sf{Explanation}}}}$

3x + 3(x + 1) + 3(x + 2) = 999

3x + 3x + 3 + 3x + 6 = 999

3x + 3x + 3x + 3 + 6 = 999

9x + 9 = 999

9x = 999 – 9

$x \:=\:\frac{990}{9}$

$\fbox\purple{x \:= \:110}$

.

$\fbox\orange{Let's \:verify}$

x = 110

3x + 3(x + 1) + 3(x + 2) = 999

3(110) + 3(110 + 1) + 3(110 + 2) = 999

330 + 330 + 3 + 330 + 6 = 999

999 = 999

LHS = RHS

Hence, verified

Therefore, x = 110

_________________

$\LARGE{\underline{\purple{\sf{Answer}}}}$

$\fbox\purple{x \:= \:110}$

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