Solve the equation for x: \[\dfrac{{\left( {3x + 2} \right)}}{3} = 2x + 1\]
Answer
637.2k+ views
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.
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.
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