Question

solve the equation and check your answer:
$\frac{2x}{3} +1=\frac{7x}{15} +3$

Answer 1

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✺Answer:

♦️ To solve:

• $\Large{ \rm{ \frac{2x}{3} + 1 = \frac{7x}{15} + 3}}$

We have to find the value of x

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✺Explanation of Q.

The above question is a equation in one variable, Here we can see that the variable is x. Next, we can observe that this is a linear equation i.e. Highest degree = 1

How? Here in LHS and RHS, the coefficient of x are 2/3 and 7/15 which are not equal to each other, thus they don't cancel each other. Thus, we will eventually get a equation i.e. a linear equation in one variable.

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Hence, we can solve this by just equating LHS and RHS and finding the value of x.

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✺Solution:

Solving the equation,

$\large{ \rm{ \rightarrow \: \frac{2x}{3} + 1 = \frac{7x}{15} + 3}}$

Shifting 7x/15 to the left hand side and 1 to the Right hand side,

$\large{ \rm{ \rightarrow \: \frac{2x}{3} - \frac{7x}{15} = 3 - 1}} \\ \\ \large{ \rm{ \rightarrow \: \frac{10x - 7x}{15} = 2}} \\ \\\large{ \rm{ \rightarrow \: \frac{3x}{15} = 2}} \\ \\\large{ \rm{ \rightarrow \: x = \cancel{ \frac{2 \times 15}{3}}}} \\ \\ \large{ \rm{ \rightarrow \: x = \boxed{ \red{ \rm{10}}}}}$

$\large{ \rm{ \therefore{ \underline{ \purple{ \: The \: required \: value \: of \: x = 10}}}}}$

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✺Verification:

Putting the value of x in LHS and RHS,

$\large{ \rm{ \rightarrow \: \frac{2(10)}{3} + 1 = \frac{7(10)}{15} + 3}} \\ \\\large{ \rm{ \rightarrow \: \frac{20}{3} + 1 = \frac{70}{15} + 3}} \\ \\ \large{ \rm{ \rightarrow \: \frac{20 + 3}{3} = \frac{70 + 45}{15} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{23}{3} = \cancel{\frac{115}{15}} \: \: \: \frac{23}{3} }} \\ \\ \large{ \rm{ \rightarrow \: \pink{LHS = RHS}}} \\ \\ \large{ \therefore{ \underline{ \rm{ \purple{Hence \: verified....!!}}}}}$

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Answer 2

$\huge\bf{ \color{pink}{{Answer}}}$

TO SOLVE :

$\frac{2x}{3} +1=\frac{7x}{15} +3$

We have to find the value of x.

EXPLANATION  OF QUESTION :

The above question is a equation in one variable, Here we can see that the variable is x.

Next, we can observe that this is a linear equation i.e. Highest degree = 1

HOW ?

Here in LHS and RHS, the coefficient of x are 2/3 and 7/15 which are not equal to each other, thus they don't cancel each other.

Thus, we will eventually get a equation i.e. a linear equation in one variable.

Hence, we can solve this by just equating LHS and RHS and finding the value of x.

SOLUTION :

SOLVING THE EQUATION :

$\implies\frac{2x}{3} +1=\frac{7x}{15} +3$

Shifting 7x/15 to the left hand side and 1 to the Right hand side.

$\implies\frac{2x}{3} - \frac{7x}{15} = 3 - 1$

$\implies\frac{10x - 7x}{15} = 2$

$\implies \frac{3x}{15} = 2$

$\implies \: x = \cancel \frac{ 2 \times 15}{3} </p><p></p><p>$

$\implies x = \fbox{10}$

The required value for x = 10

VERIFICATION :

Putting the value of x in LHS and RHS.

$\implies \frac{2(10)}{3} + 1 = \frac{7(10)}{15} + 3$

$\implies \frac{20}{3} + 1 = \frac{70}{15} + 3$

$\implies \frac{20 + 3}{3} = \frac{70 + 45}{15}$

$\implies \frac{23}{3} = \cancel \frac{115}{15} \: \: \frac{23}{3}$

$\implies \bf{\color{gold}{LHS = RHS}}$

Hence Verified !