Question

solve for x
5(x+1)=3(3x+2)-4​

Answer 1

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Answer 2

$\large{ \rm \underline{Given - }}$

$\sf \dashrightarrow \: 5(x+1)=3(3x+2)-4 \:$

$\large{ \rm \underline{To \: find - }}$

$\sf \dashrightarrow \: Value \: of \: x = \: ?$

$\large{ \rm \underline{Solution - }}$

$\sf \dashrightarrow \: 5(x+1)=3(3x+2)-4 \:$

$\sf \dashrightarrow \: 5x +5 =9x+6-4$

$\sf \dashrightarrow \: 5x +5 =9x+2$

$\sf \dashrightarrow \: - 4x + 5 = 2$

$\sf \dashrightarrow \: - 4x = 2 - 5$

$\sf \dashrightarrow \: - 4x = - 3$

$\sf \dashrightarrow \: x = \dfrac{ - 3}{ - 4}$

$\sf \dashrightarrow \: x = \dfrac{ 3}{4}$

$\sf \dashrightarrow\boxed{ \sf Hence \: the \: value \: of \: x \: is \: \dfrac{ 3}{4} }$

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$\large{ \rm \underline{Verification - }}$

$\sf \dashrightarrow \: 5(x+1)=3(3x+2)-4 \:$

$\sf \dashrightarrow \: 5 \left(\dfrac{ 3}{4} +1\right) \bf= \sf3\left(3\left(\dfrac{ 3}{4} \right)+2\right)-4 \:$

$\bf LHS$

$\sf \dashrightarrow \: 5 \left(\dfrac{ 3}{4} +1\right)$

$\sf \dashrightarrow \: 5 \left(\dfrac{ 3}{4} + \dfrac{4}{4} \right)$

$\sf \dashrightarrow \: 5 \left(\dfrac{ 3 + 4}{4} \right)$

$\sf \dashrightarrow \: 5 \times \dfrac{ 7}{4}$

$\sf \dashrightarrow \: \dfrac{ 35}{4}$

$\bf RHS$

$\sf \dashrightarrow \: \sf3\left(3\left(\dfrac{ 3}{4} \right)+2\right)-4 \:$

$\sf \dashrightarrow \: \sf3\left(\dfrac{ 9}{4} +2\right)-4 \:$

$\sf \dashrightarrow \: \sf3\left(\dfrac{ 9}{4} + \dfrac{2 \times 4}{4} \right)-4 \:$

$\sf \dashrightarrow \: \sf3\left(\dfrac{ 9}{4} + \dfrac{8}{4} \right)-4 \:$

$\sf \dashrightarrow \: \sf3\left(\dfrac{ 9 + 8}{4} \right)-4 \:$

$\sf \dashrightarrow \: \sf3\left(\dfrac{ 17}{4} \right)-4 \:$

$\sf \dashrightarrow \: \sf\dfrac{ 51}{4} - 4$

$\sf \dashrightarrow \: \sf\dfrac{ 51}{4} - \dfrac{4 \times 4}{4}$

$\sf \dashrightarrow \: \sf\dfrac{ 51}{4} - \dfrac{16}{4}$

$\sf \dashrightarrow \: \sf\dfrac{ 51 - 16}{4}$

$\sf \dashrightarrow \: \sf\dfrac{ 35}{4}$

$\bf{Hence},$

$\sf \dashrightarrow \: \sf\dfrac{ 35}{4} = \dfrac{ 35}{4}$

$\sf \dashrightarrow \: LHS = RHS$

$\bf\:Hence \: verified \: !!$