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How do you solve $\dfrac{2}{3}x+11=7$?

Answer
VerifiedVerified
550.2k+ views
Hint: In this problem we need to find the solution of the given equation which is $\dfrac{2}{3}x+11=7$. We can say that the given equation is a linear equation in a single variable. So, we need to perform some arithmetical operations to solve the given linear equation in a single variable. We can observe that $11$ is in addition to the given equation. To eliminate $11$, we will subtract the $11$ from both sides of the equation and simplify the obtained equation. Now we can observe that $3$ is in division, so we will multiply the whole equation with $3$ and simplify the equation. After that we can observe $2$ is in multiplication, so we will divide the whole obtained equation with $2$ to get the final result.

Complete step by step solution:
Given equation, $\dfrac{2}{3}x+11=7$.
In the above equation we have $11$ in addition, so subtracting the $11$ on both sides of the above equation, then we will get
$\Rightarrow \dfrac{2}{3}x+11-11=17-11$
We know that $+a-a=0$, from this rule the above equation is modified as
$\Rightarrow \dfrac{2}{3}x=6$
In the above equation we can observe that $3$ in division, so multiplying the $3$ on both sides of the above equation, then we will get
$\Rightarrow 3\times \dfrac{2}{3}x=6\times 3$
We have $a\times \dfrac{1}{a}=1$, then we can write the above equation as
$\Rightarrow 2x=18$
In the above equation we have $2$ in multiplication, so dividing the $2$ on both sides of the equation and simplifying it, then we will have
$\begin{align}
  & \Rightarrow \dfrac{2x}{2}=\dfrac{18}{2} \\
 & \Rightarrow x=9 \\
\end{align}$
Hence the solution for the given equation $\dfrac{2}{3}x+11=7$ is $x=9$.

Note:
In the above solution, we have considered the fraction $\dfrac{2}{3}$ individually i.e., $3$ is in division and $2$ is in multiplication and used an anti arithmetic operation. We can also multiply the inverse of the fraction which is $\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}$ on both sides of the equation to get the result.