
Solve the algebraic equation: \[5x+\dfrac{7}{2}=\dfrac{3}{2}x-14\].
Answer
588.3k+ views
Hint: Rearrange the given expression, apply the basic algebraic operations addition and subtraction and solve the equation and get the value of x.
Complete step-by-step answer:
The given expression is an algebraic expression. It is built up from integer constants, variables and the algebraic operations like addition, subtraction, multiplication, division etc.
We have been given an algebraic expression \[5x+\dfrac{7}{2}=\dfrac{3}{2}x-14\].
The only variable that has been used is x. Thus we need to solve this equation and find the value of x.
\[5x+\dfrac{7}{2}=\dfrac{3}{2}x-14\]
Let us first rearrange the above expression.
\[5x+\dfrac{3}{2}x=\dfrac{-7}{2}-14\]
Take LCM on both sides, i.e. take LCM of LHS and RHS and simply the expression obtained.
\[\begin{align}
& \dfrac{\left( 2\times 5 \right)x-3x}{2}=\dfrac{-7-14\times 2}{2} \\
& \dfrac{10x-3x}{2}=\dfrac{-7-28}{2} \\
& \therefore \dfrac{7x}{2}=\dfrac{-35}{2} \\
\end{align}\]
Cancel out 2 in both the denominators of LHS and RHS.
\[\begin{align}
& 7x=-35 \\
& \therefore x=\dfrac{-35}{7}=-5 \\
\end{align}\]
Thus we got the value of x = -5.
Note: For solving an expression with one variable only one equation is needed. If there were 2 variables namely x and y, then we require 2 equations to get the x and y. Similarly if there were 3 variables then we will need 3 equations to solve the same.
Complete step-by-step answer:
The given expression is an algebraic expression. It is built up from integer constants, variables and the algebraic operations like addition, subtraction, multiplication, division etc.
We have been given an algebraic expression \[5x+\dfrac{7}{2}=\dfrac{3}{2}x-14\].
The only variable that has been used is x. Thus we need to solve this equation and find the value of x.
\[5x+\dfrac{7}{2}=\dfrac{3}{2}x-14\]
Let us first rearrange the above expression.
\[5x+\dfrac{3}{2}x=\dfrac{-7}{2}-14\]
Take LCM on both sides, i.e. take LCM of LHS and RHS and simply the expression obtained.
\[\begin{align}
& \dfrac{\left( 2\times 5 \right)x-3x}{2}=\dfrac{-7-14\times 2}{2} \\
& \dfrac{10x-3x}{2}=\dfrac{-7-28}{2} \\
& \therefore \dfrac{7x}{2}=\dfrac{-35}{2} \\
\end{align}\]
Cancel out 2 in both the denominators of LHS and RHS.
\[\begin{align}
& 7x=-35 \\
& \therefore x=\dfrac{-35}{7}=-5 \\
\end{align}\]
Thus we got the value of x = -5.
Note: For solving an expression with one variable only one equation is needed. If there were 2 variables namely x and y, then we require 2 equations to get the x and y. Similarly if there were 3 variables then we will need 3 equations to solve the same.
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