Question

Solve the algebraic equation: \[x - 5 = - \left( {\dfrac{3}{2}x - \dfrac{5}{2}} \right)\]

Answer 1

Hint:The given equation is a linear equation which has the same one variable on both sides of the separating ‘equals to’ sign. So we apply the BODMAS rule and combine the like terms together then separate the constants on the other side. This way we would get the value of the variable.

Complete solution step by step:
Firstly, we write down the equation provided in the question
\[x - 5 = - \left( {\dfrac{3}{2}x - \dfrac{5}{2}} \right)\]

As we can see the highest power of the variable present in the question is one. This means the equation is linear in nature with one variable on both sides of it.

Now we apply the BODMAS rule on the right side of the equation.
BODMAS Rule: BODMAS is an acronym which stands for “Bracket-Order (Powers & Square Roots)-

Division-Multiplication-Addition-Subtraction”. It provides us the order to operate on mathematical expressions and the method to open all three kinds of brackets i.e.

 \[\left( {} \right),\left\{ {}
\right\},\left[ {} \right]\] in a way that is a bigger bracket first then middle one and finally the small bracket.

Given that the expression is a linear equation and has only one bracket, we solve the bracket first then move our like variables to one side of the equation like this
$
x - 5 = - \left( {\dfrac{3}{2}x - \dfrac{5}{2}} \right) = - \dfrac{3}{2}x + \dfrac{5}{2} \\
\Rightarrow x + \dfrac{3}{2}x = \dfrac{5}{2} + 5 \\
$
Now it is easier for us to solve this with like terms together on each side of the equation.

Now we take ‘x’ common on LHS and take LCM on both sides to add the fractional values which gives us
$
x\left( {1 + \dfrac{3}{2}} \right) = \dfrac{5}{2} + 5 \\
\Rightarrow x\left( {\dfrac{{2 + 3}}{2}} \right) = \dfrac{{5 + 10}}{2} \\
\Rightarrow \dfrac{{5x}}{2} = \dfrac{{15}}{2} \\
$

Now we cancel the denominator in each side and shift ‘5’ to the RHS and solve the equation
$
\Rightarrow x = \dfrac{{15}}{5} = 3 \\
\Rightarrow x = 3 \\
$
So our solution for the equation is solved in this way.

Note: Knowing the order of operations here in the question helps us to evaluate the correct answer otherwise a glitch may have occurred. While solving any algebraic equation, containing variables and numbers, our goal is to put terms together and then find the value of the variable.