
What is the answer for this: Solve for $x$: $x - \dfrac{2}{3} = \dfrac{{2x}}{7}$?
Answer
524.1k+ views
Hint: To solve the above linear equation for $x$, go step by step solving the terms and cancelling the constant terms on one side, so that we are left with only the variable $x$. Initiate the process by adding both the sides by $\dfrac{2}{3}$ and then subtracting both the sides by $\dfrac{{2x}}{7}$.
Complete step-by-step answer:
We are given with a linear equation $x - \dfrac{2}{3} = \dfrac{{2x}}{7}$.
To solve for $x$, we start by adding both the sides by $\dfrac{2}{3}$, and we get:
$
x - \dfrac{2}{3} + \dfrac{2}{3} = \dfrac{{2x}}{7} + \dfrac{2}{3} \\
x = \dfrac{{2x}}{7} + \dfrac{2}{3} \\
$
Subtracting both the sides by $\dfrac{{2x}}{7}$, so that we have constant on one side and values with variables on one side and we get:
$
x - \dfrac{{2x}}{7} = \dfrac{{2x}}{3} + \dfrac{2}{3} - \dfrac{{2x}}{3} \\
x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
$
Since, we can see that there are two values having the variable $x$, so to have only one variable, we are multiplying and divide the first operand $x$ with $7$ to make the common denominator as the other operand, and then subtract the other operand through it and we get:
$
x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
\dfrac{7}{7}x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
\dfrac{{7x - 2x}}{7} = \dfrac{2}{3} \\
\dfrac{{5x}}{7} = \dfrac{2}{3} \\
$
Multiplying both the sides by $7$, to remove the denominator from the left-hand side, and we get:
$
\dfrac{{5x}}{7} \times 7 = \dfrac{2}{3} \times 7 \\
5x = \dfrac{{14}}{3} \\
$
Lastly, dividing both the sides by $5$, so that we are left only with the variable $x$, whose value we had to find, and we get:
$
5x = \dfrac{{14}}{3} \\
\dfrac{{5x}}{5} = \dfrac{{14}}{3} \times \dfrac{1}{5} \\
x = \dfrac{{14}}{{15}} \\
$
Therefore, the value of $x$ for the equation $x - \dfrac{2}{3} = \dfrac{{2x}}{7}$ is $\dfrac{{14}}{{15}}$.
Note: We can check the answer whether it is correct on not, by substituting the value of $x = \dfrac{{14}}{{15}}$ on both the sides of the equation and solving it. If both the sides that is the Left and Right-hand side values are equal, then the result obtained is absolutely correct.
Always solve the equations step by step rather than solving at one step, otherwise it may lead to problems sometimes for larger values.
Complete step-by-step answer:
We are given with a linear equation $x - \dfrac{2}{3} = \dfrac{{2x}}{7}$.
To solve for $x$, we start by adding both the sides by $\dfrac{2}{3}$, and we get:
$
x - \dfrac{2}{3} + \dfrac{2}{3} = \dfrac{{2x}}{7} + \dfrac{2}{3} \\
x = \dfrac{{2x}}{7} + \dfrac{2}{3} \\
$
Subtracting both the sides by $\dfrac{{2x}}{7}$, so that we have constant on one side and values with variables on one side and we get:
$
x - \dfrac{{2x}}{7} = \dfrac{{2x}}{3} + \dfrac{2}{3} - \dfrac{{2x}}{3} \\
x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
$
Since, we can see that there are two values having the variable $x$, so to have only one variable, we are multiplying and divide the first operand $x$ with $7$ to make the common denominator as the other operand, and then subtract the other operand through it and we get:
$
x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
\dfrac{7}{7}x - \dfrac{{2x}}{7} = \dfrac{2}{3} \\
\dfrac{{7x - 2x}}{7} = \dfrac{2}{3} \\
\dfrac{{5x}}{7} = \dfrac{2}{3} \\
$
Multiplying both the sides by $7$, to remove the denominator from the left-hand side, and we get:
$
\dfrac{{5x}}{7} \times 7 = \dfrac{2}{3} \times 7 \\
5x = \dfrac{{14}}{3} \\
$
Lastly, dividing both the sides by $5$, so that we are left only with the variable $x$, whose value we had to find, and we get:
$
5x = \dfrac{{14}}{3} \\
\dfrac{{5x}}{5} = \dfrac{{14}}{3} \times \dfrac{1}{5} \\
x = \dfrac{{14}}{{15}} \\
$
Therefore, the value of $x$ for the equation $x - \dfrac{2}{3} = \dfrac{{2x}}{7}$ is $\dfrac{{14}}{{15}}$.
Note: We can check the answer whether it is correct on not, by substituting the value of $x = \dfrac{{14}}{{15}}$ on both the sides of the equation and solving it. If both the sides that is the Left and Right-hand side values are equal, then the result obtained is absolutely correct.
Always solve the equations step by step rather than solving at one step, otherwise it may lead to problems sometimes for larger values.
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