Question

show that 1.83 can be expressed in the form of p/q where p and q are integers and q ne 0​

Answer 1

$\sf Let x = 1.8\bar3$

• Multiple both sides by 10.

$\\\implies \sf 10x = 18.\bar3$

$\implies \sf 10x = 18 + 0.\bar3$

$\implies \sf 10x = 18 + \dfrac{3}{9}$

$\implies \sf 10x = \dfrac{18 \times 9 + 3}{9}$

$\implies \sf 10x = \dfrac{162 + 3}{9}$

$\implies \sf 10x = \dfrac{165}{9}$

$\implies \sf x = \dfrac{ \cancel{165}}{ \cancel9} \times \dfrac{1}{10}$

$\implies \sf x = \dfrac{ 55}{3} \times \dfrac{1}{10}$

$\implies \sf x = \dfrac{ 55}{30}$

$\implies \sf x = \dfrac{11}{6}$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{1.8\bar3 = \bf \dfrac{11}{6} }$

Answer 2

$\Rsh\small\tt\purple{ Solution}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple\Rsh\small\tt{Let \: x = 1.8333}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\Rsh\small\tt\purple{ Multiply \: 10 \: both \: sides}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple\Rsh\small\tt{ 10x =10 \times 1.8333}$

$\purple\Rsh\small\tt{ 10x =18.333 - - - - - - - eq(1)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\Rsh\small\tt\purple{ Multiply \: 100 \: both \: sides}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple\Rsh\small\tt{ 100x =100 \times 1.8333}$

$\purple\Rsh\small\tt{ 100x = 183.333 - - - - - - eq(2)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\Rsh\small\tt\purple{ Subtract \: equation \: 1 \: from \: equation \: 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple\Rsh\small\tt{ 100x - 10x = 183.333 - 18.333}$

$\purple\Rsh\small\tt{ 90x = 183.333 - 18.333}$

$\purple\Rsh\small\tt{ 90x = 165}$

$\purple\Rsh\tt{ x = \cancel\frac{165}{90} }$

$\small \boxed{\purple\Rsh\tt{ x = \cancel\frac{33}{18} }}$