
Solve the following equation
$\dfrac{1}{2}x - 3 = 5 + \dfrac{1}{3}x$
Answer
483.3k+ views
Hint: Solving the equation is indirectly mentioning to find the value of some unknown. So there are two methods to solve the equation one is the substitution method if two equations are given and the other is by using basic operations such as addition or subtraction if a single equation is given. Check first how many questions are given and follow the method.
Complete step-by-step answer:
Given the equation is $\dfrac{1}{2}x - 3 = 5 + \dfrac{1}{3}x$.
First, observe how many unknown variables are there in the equation. By the above equation given to us, we came to know that the number of variables of the equation is one.
Hence, we can calculate the value of one unknown only that is $x$.
To find the value of $x$, we can use basic operations such as addition or subtraction.
To calculate the value of an unknown variable, first, bring all the like terms on one side and constants on another side.
Let us assume that the given equation as $\dfrac{1}{2}x - 3 = 5 + \dfrac{1}{3}x.....(1)$
After bringing all the like terms on one side and constants on another side, the equation $(1)$becomes
\[\dfrac{1}{2}x - \dfrac{1}{3}x = 5 + 3.....(2)\]
By subtracting the terms in the above equation,
The denominators are not the same, so we need to find the common factor to solve the equation,
The common factor for $2$ and $3$ is $6$.
Let us consider $I = \dfrac{1}{2}x - \dfrac{1}{3}x$
$I = \dfrac{3}{6}x - \dfrac{2}{6}x$
Subtract the values in the numerator,
$I = \dfrac{1}{6}x$
By substitute $I = \dfrac{1}{6}x$in $(2)$.
$\dfrac{1}{6}x = 5 + 3$
Adding the values on the right side of the above equation,
$\dfrac{1}{6}x = 8$
By bringing the denominator to the numerator,
$x = 8 \times 6$
By multiplying the terms in the above equation,
$x = 48$.
Note: First, check how many unknown variables are there in the equation. Then find the method to solve the equations. Find the factors of multiples to make the denominators the same. Hence, there are two methods to solve the equation one is the substitution method if two equations are given and the other is by using basic operations such as addition or subtraction if a single equation is given. Check first how many questions are given and follow the method.
Complete step-by-step answer:
Given the equation is $\dfrac{1}{2}x - 3 = 5 + \dfrac{1}{3}x$.
First, observe how many unknown variables are there in the equation. By the above equation given to us, we came to know that the number of variables of the equation is one.
Hence, we can calculate the value of one unknown only that is $x$.
To find the value of $x$, we can use basic operations such as addition or subtraction.
To calculate the value of an unknown variable, first, bring all the like terms on one side and constants on another side.
Let us assume that the given equation as $\dfrac{1}{2}x - 3 = 5 + \dfrac{1}{3}x.....(1)$
After bringing all the like terms on one side and constants on another side, the equation $(1)$becomes
\[\dfrac{1}{2}x - \dfrac{1}{3}x = 5 + 3.....(2)\]
By subtracting the terms in the above equation,
The denominators are not the same, so we need to find the common factor to solve the equation,
The common factor for $2$ and $3$ is $6$.
Let us consider $I = \dfrac{1}{2}x - \dfrac{1}{3}x$
$I = \dfrac{3}{6}x - \dfrac{2}{6}x$
Subtract the values in the numerator,
$I = \dfrac{1}{6}x$
By substitute $I = \dfrac{1}{6}x$in $(2)$.
$\dfrac{1}{6}x = 5 + 3$
Adding the values on the right side of the above equation,
$\dfrac{1}{6}x = 8$
By bringing the denominator to the numerator,
$x = 8 \times 6$
By multiplying the terms in the above equation,
$x = 48$.
Note: First, check how many unknown variables are there in the equation. Then find the method to solve the equations. Find the factors of multiples to make the denominators the same. Hence, there are two methods to solve the equation one is the substitution method if two equations are given and the other is by using basic operations such as addition or subtraction if a single equation is given. Check first how many questions are given and follow the method.
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