Question

Solve this equation \[x = \dfrac{4}{5}(x + 10)\].

Answer 1

Hint:Here, we need to find the value of x, which is an unknown variable.The given equation is not a linear equation. Any term of an equation may be taken from one side to other with the change in its sign, this does not affect the equality of the statement and this process is called transposition. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS).

Complete step by step answer:
Given the equation as below,
\[x = \dfrac{4}{5}(x + 10)\]
Solve this equation in the following way,
\[ \Rightarrow x = \dfrac{{4(x + 10)}}{5}\]
Multiplying by \[5\]on both the sides, we get,
\[ \Rightarrow x(5) = 5 \times \dfrac{{4(x + 10)}}{5}\]
\[ \Rightarrow x(5) = 4(x + 10)\]
Removing the brackets, we get,
\[ \Rightarrow 5x = 4x + 40\]
By using transposition and moving \[4x\]to LHS, we get,
\[ \Rightarrow 5x - 4x = 40\]
Simplify this above equation, we get,
\[ \Rightarrow x = 40\]

Let us check the answer by substituting the value of x in the given equation.Let's see if the answer is correct or not. First we will calculate the LHS part. So,
\[LHS = x\]
Substituting the value of \[x\], we get,
\[\therefore LHS = 40\]
Next, we will calculate the RHS part.
\[ RHS= \dfrac{4}{5}(x + 10)\]
Substituting the value of \[x\], we get,
\[ RHS= \dfrac{4}{5}(40 + 10)\]
Solving the bracket first, we get,
\[RHS = \dfrac{4}{5}(50)\]
Simplify this above expression, we get,
\[ RHS= \dfrac{{4(50)}}{5}\]
Dividing by \[5\]we get,
\[RHS = 4(10)\]
\[\therefore RHS = 40\]

Hence, LHS = RHS and so the value of x is correct. Thus, it is verified.

Note: The standard form of a linear equation with one variable is of the form $ax + b = 0$, where $x$ is a variable, and ‘$a$’ and ‘$b$’ are constants. A linear equation with two variables is of the form $ax + by = c$, where $x$ and $y$ are variables and $a, b$ and $c$ are constants. We can check if the answer is correct or not by substituting the value of $x$ in the given equation.