Question

Solve the equation for x: \[\dfrac{{\left( {3x + 2} \right)}}{3} = 2x + 1\]

Answer 1

Hint:
Here, we need to solve the equation and check our results. We will use the operations of addition, subtraction, multiplication, and division to find the value of \[x\]. Then, we will substitute the value of \[x\] in the given equation to check our results.

Complete step by step solution:
We will use the basic mathematical operations to find the value of \[x\].
Multiplying both sides by 3, we get
\[\begin{array}{l} \Rightarrow \dfrac{{\left( {3x + 2} \right)}}{3} \times 3 = \left( {2x + 1} \right) \times 3\\ \Rightarrow 3x + 2 = 3\left( {2x + 1} \right)\end{array}\]
Multiplying 3 by \[\left( {2x + 1} \right)\] using the distributive law of multiplication, we get
\[\begin{array}{l} \Rightarrow 3x + 2 = 3 \cdot 2x + 3 \cdot 1\\ \Rightarrow 3x + 2 = 6x + 3\end{array}\]
Subtracting 3 from both sides, we get
\[\begin{array}{l} \Rightarrow 3x + 2 - 3 = 6x + 3 - 3\\ \Rightarrow 3x - 1 = 6x\end{array}\]
Subtracting \[3x\] from both the sides, we get
\[\begin{array}{l} \Rightarrow 3x - 1 - 3x = 6x - 3x\\ \Rightarrow - 1 = 3x\end{array}\]
Finally, dividing both sides by 3, we get
\[ \Rightarrow \dfrac{{ - 1}}{3} = \dfrac{{3x}}{3}\]
Therefore, we get
\[ \Rightarrow x = - \dfrac{1}{3}\]

Thus, we get the value of \[x\] as \[ - \dfrac{1}{3}\].

Note:
We have used the distributive law of multiplication in the solution to multiply 3 by \[\left( {2x + 1} \right)\]. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
We can check our answer by using the given equation.
If the left hand side is equal to the right hand side, then our answer is correct.
Substituting \[x = - \dfrac{1}{3}\] in the left hand side (L.H.S.) of the given equation \[\dfrac{{\left( {3x + 2} \right)}}{3} = 2x + 1\], we get
\[ \Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = \dfrac{{3 \times \left( { - \dfrac{1}{3}} \right) + 2}}{3}\]
Multiplying the terms, we get
\[ \Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = \dfrac{{ - 1 + 2}}{3}\]
Simplifying the numerator, we get
\[ \Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = \dfrac{1}{3}\]
Substituting \[x = - \dfrac{1}{3}\] in the right hand side (L.H.S.) of the given equation \[\dfrac{{\left( {3x + 2} \right)}}{3} = 2x + 1\], we get
\[ \Rightarrow {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}} = 2 \times \left( { - \dfrac{1}{3}} \right) + 1\]
Multiplying the terms, we get
\[ \Rightarrow {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}} = - \dfrac{2}{3} + 1\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}} = - \dfrac{2}{3} + \dfrac{3}{3}\\ \Rightarrow {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}} = \dfrac{1}{3}\end{array}\]
Therefore, we can observe that
\[ \Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}}\]
Thus, the value \[x = - \dfrac{1}{3}\] satisfies the given equation.
Hence, we have verified our answer.