Question

answer please friends​

Answer 1

Given Equation

$\sf\to 5x +\dfrac{1}{2} =7\bigg(x-\dfrac{7}{9} \bigg)-2$

To find value x

Now take

$\sf\to 5x +\dfrac{1}{2} =7\bigg(x-\dfrac{7}{9} \bigg)-2$

Take LCM

$\sf\to \dfrac{5x\times 2+1}{2} =7\bigg(\dfrac{9x-7}{9} \bigg)-2$

$\sf\to\dfrac{10x+1}{2} =\dfrac{7(9x-7)-2\times9}{9}$

$\sf\to \dfrac{10x+1}{2} =\dfrac{63x-49-18}{9}$

$\sf\to \dfrac{10x+1}{2} =\dfrac{63x-67}{9}$

Now Using Cross multiplication

$\sf\to 9(10x+1)=2(63x-67)$

$\sf\to 90x+9=126x-134$

$\sf\to 126x-90x=9+134$

$\sf\to 36x=143$

$\sf\to x=\dfrac{143}{36}$

Answer

$\sf\to x=\dfrac{143}{36}$

Answer 2

Answer:

$\large \underline{\sf \pmb{Given}}$

$: \implies \sf{ 5x + \dfrac{1}{2} = 7 \bigg(x - \dfrac{7}{9} \bigg) - 2}$

$\large \underline{\sf \pmb{To \: Find }}$

$: \implies \sf{value \: of \: x}$

$\large \underline{\sf \pmb{Solution}}$

$: \implies \sf{ 5x + \dfrac{1}{2} = 7 \bigg(x - \dfrac{7}{9} \bigg) - 2}$

• ⇒ Take the LCM of denominators.

${: \implies \sf{ \dfrac{(5x \times 2) + 1}{2} = 7 \times \bigg(\dfrac{(x \times 9 )- 7 - (2 \times 9)}{9} \bigg) }}$

${: \implies \sf{ \dfrac{10x+ 1}{2} =\dfrac{( 9x \times 7) - (7 \times 7) - (2 \times 9)}{9} }}$

${: \implies \sf{ \dfrac{10x+ 1}{2} =\dfrac{63x- 49- 18}{9} }}$

${: \implies \sf{ \dfrac{10x+ 1}{2} =\dfrac{63x(- 49- 18)}{9} }}$

${: \implies \sf{ \dfrac{10x+ 1}{2} =\dfrac{63x-67}{9} }}$

• ⇒ By cross multiplication

${: \implies \sf{9 (10x+ 1) = {2(63x-67)} }}$

${: \implies \sf{ (90x+ 9) = {(126x-134)} }}$

$: \implies\sf{126x-90x=9+134}$

$: \implies\sf{36x=143}$

$: \implies\sf{x= \dfrac{143}{36} }$

$: \implies\red{ \boxed{\bf{x= \dfrac{143}{36} }}}$

$\large \underline {\sf \pmb{Verification}}$

${: \implies \sf{5x + \dfrac{1}{2} = 7 \bigg(x - \dfrac{7}{9} \bigg) - 2}}$

• ⇒ Substituting the values of "x".

${: \implies \sf{\dfrac{(5 \times 143)}{36} + \dfrac{1}{2} = 7 \bigg( \dfrac{143}{36} - \dfrac{7}{9} \bigg) - 2}}$

${: \implies \sf \bigg({\dfrac{715}{36} + \dfrac{1}{2} \bigg) = 7 \bigg( \dfrac{143}{36} - \dfrac{7}{9} \bigg) - 2}}$

• ⇒ Taking the LCM of denominators

${: \implies \sf {\dfrac{715 + (1 \times 18)}{36} = 7 \times \bigg (\dfrac{143 -(7 \times 4) }{36} \bigg) -2 }}$

${: \implies \sf {\dfrac{715 + 18}{36} = 7 \times \bigg (\dfrac{143 -28 }{36} \bigg) - 2}}$

${: \implies \sf {\dfrac{733}{36} = 7 \times \bigg (\dfrac{115}{36} \bigg) - 2}}$

${: \implies \sf {\dfrac{733}{36} = \bigg (\dfrac{7 \times 115}{36} \bigg) - 2}}$

${: \implies \sf {\dfrac{733}{36} = \dfrac{805}{36} - 2}}$

• ⇒ Again takin the LCM of denominators

${: \implies \sf {\dfrac{733}{36} = \dfrac{805 - (2 \times 36)}{36} }}$

${: \implies \sf {\dfrac{733}{36} = \dfrac{805 - 72}{36} }}$

${: \implies \sf {\dfrac{733}{36} = \dfrac{773}{36} }}$

${: \implies\red{ \boxed{\bf {LHS=RHS} }}}$

• ⇒ ${\sf \underline {Hence \: Verified} \bf{\checkmark}}$