Question

How do you solve \[\dfrac{1}{4}x-\dfrac{5}{2}=-2\]?

Answer 1

Hint: This type of problem is based on the concept of linear equations with one variable. First, we have to consider the left-hand side of the equation. Take LCM 4 and simplify the left-hand side of the equation. Then, equate it with the right-hand side. Multiply the whole equation by 4 and cancel the common terms from the left-hand side and right-hand side of the equation. And then, add 10 on both the sides of the equation to find the value of x.

Complete step by step solution:
According to the question, we are asked to solve the equation \[\dfrac{1}{4}x-\dfrac{5}{2}=-2\].
We have been given the inequality is \[\dfrac{1}{4}x-\dfrac{5}{2}=-2\]. -----(1)
We first have to consider the left-hand side of the equation (1), that is \[\dfrac{1}{4}x-\dfrac{5}{2}\].
Now, take LCM and simplify the left-hand side.
\[\Rightarrow \dfrac{1}{4}x-\dfrac{5}{2}=\dfrac{x-5\times 2}{4}\]
On further simplification, we get
\[\dfrac{1}{4}x-\dfrac{5}{2}=\dfrac{x-10}{4}\]
Now, we have to equate the simplified left-hand side with the right-hand side of the equation (1).
\[\Rightarrow \dfrac{x-10}{4}=-2\]
Multiply the whole equation by 4. We get
\[\dfrac{x-10}{4}\times 4=-2\times 4\]
We find that 4 are common in both the numerator and denominator of the LHS.
On cancelling 4, we get
\[x-10=-2\times 4\]
\[\Rightarrow x-10=-8\]
Let us now add 10 on both the sides of the equation.
\[\Rightarrow x-10+10=-8+10\]
We know that terms with same magnitude and opposite signs cancel out.
Therefore, on cancelling 10, we get
\[x=-8+10\]
On further simplification, we get
\[x=2\]

Hence, the value of x in the equation \[\dfrac{1}{4}x-\dfrac{5}{2}=-2\] is 2.

Note: Whenever you get this type of problems, we should always try to make the necessary changes in the given equation to get the variable x in the left-hand side of the equation. We should avoid calculation mistakes based on sign conventions.
We can check whether the answer obtained is correct or not.
Consider the LHS of the equation
LHS=\[\dfrac{1}{4}x-\dfrac{5}{2}\]
But, we know that x=2. On substituting the value of x, we get
LHS=\[\dfrac{1}{4}\times 2-\dfrac{5}{2}\]
On further simplification, we get
LHS=\[\dfrac{1}{2}-\dfrac{5}{2}\]
On taking the LCM, we get
LHS=\[\dfrac{1-5}{2}\]
That is, LHS=\[\dfrac{1-5}{2}\].
LHS=\[\dfrac{-4}{2}\]
Therefore, LHS=-2
Now, consider the RHS.
RHS=-2.
Here, we find that LHS=RHS.