Question

(x/2)=(x/3)+1 solve the equation​

Answer 1

$\tt\ \: \:Answer:$

$\tt\ \: \: x = 6$

$\tt\ \: \:Step-by-Step\:\: Explanation:$

$\tt\ \: \: \bigg(\frac{x}{2}\bigg)=\bigg(\frac{x}{3}\bigg )+1$

$\tt\ \: \:\bigg(\frac{x}{2}\bigg)-\bigg(\frac{x}{3}\bigg )=1\bigg[\:Transposing\: \bigg(\frac{x}{3}\bigg)\: to\: LHS\: \:\bigg]$

$\tt\ \: \:\frac{3x-2x}{6} =1$

$\tt\ \: \:\frac{x}{6} = 1$

$\tt\ \: \:x=1 * 6$

$\tt\ \: \:x=6$

$\tt\ \: \:Hope\:\: it\:\: helps\: \:you!!$(^◕.◕^)

Answer 2

Solve the following equation :-

$Q) \: \sf \bigg( \dfrac{x}{2} \bigg) = \bigg(\dfrac{x}{3} \bigg) + 1$

Removing the brackets,

$\longmapsto \sf\dfrac{x}{2} = \dfrac{x}{3} + 1$

Let's add x/3 and 1 first.

$\longmapsto\sf\dfrac{x}{2} = \dfrac{x}{3} + \dfrac{1}{1}$

The LCM of 3 and 1 is 3, so multiplying the fractions using their denominators,

$\longmapsto\sf\dfrac{x}{2} = \dfrac{x \times 1 + 1 \times 3}{3}$

On simplifying,

$\longmapsto\sf\dfrac{x}{2} = \dfrac{x + 3}{3}$

Transposing 3 from RHS to LHS, changing it's sign,

$\longmapsto\sf\dfrac{x}{2} \times 3 = x + 3$

Multiplying x/2 with 3,

$\longmapsto\sf\dfrac{3x}{2} = x + 3$

Transposing 3 from RHS to LHS, changing it's sign,

$\longmapsto\sf\dfrac{3x}{2} - 3 = x$

Now let's subtract 3 from 3x/2.

$\longmapsto\sf \dfrac{3x}{2} - \dfrac{3}{1} = x$

The LCM of 2 and 1 is 2, so multiplying the fractions using their denominators,

$\longmapsto\sf\dfrac{3x \times 1 - 3 \times 2}{2} = x$

On simplifying,

$\longmapsto\sf\dfrac{3x - 6}{2} = x$

Transposing 2 from LHS to RHS, changing it's sign,

$\longmapsto\sf3x - 6 = x \times 2$

Multiplying x with 2,

$\longmapsto\sf3x - 6 = 2x$

Transposing 6 from LHS to RHS, changing it's sign,

$\longmapsto\sf3x = 2x + 6$

Transposing 2x from RHS to LHS, changing it's sign,

$\longmapsto\sf3x - 2x = 6$

Subtracting 2x from 3x,

$\longmapsto\sf x = 6$

• The value of x is 6.

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$\underline{\rm Verification :-}$

To check our answer, let's put 6 in the place of x and see whether LHS = RHS.

$\underline{ \underline{ \mathfrak{Substituting \: the \: value \: of \: x,}}}$

LHS

$\Rightarrow \bigg(\bf\dfrac{6}{2} \bigg)$

Removing the brackets,

$\Rightarrow \bf \dfrac{6}{2}$

Reducing the fraction,

$\Rightarrow \bf \dfrac{3}{1}$

RHS

$\Rightarrow \bigg(\bf\dfrac{6}{3} \bigg) + 1$

Removing the brackets,

$\Rightarrow \bf \dfrac{6}{3} + 1$

Let's add 6/3 and 1.

$\Rightarrow \bf\dfrac{6}{3} + \dfrac{1}{1}$

The LCM of 3 and 1 is 3, so multiplying the fractions using their denominators,

$\Rightarrow \bf \dfrac{6 \times 1 + 1 \times 3}{3}$

On simplifying,

$\Rightarrow \bf\dfrac{6 + 3}{3}$

Adding 3 to 6,

$\Rightarrow\bf\dfrac{9}{3}$

Reducing the fraction,

$\Rightarrow\bf\dfrac{3}{1}$

Since LHS = RHS,

Hence verified!!