Question

Solve the following equation and check your result. If $x = \dfrac{4}{5}\left( {x + 10} \right)$ , then $x = $
$A\,.\,12$
$B.\,112$
$C.\,40$
$D.\,\,39$

Answer 1

Hint: In this question, a simple concept of transposing is used while we have generally seen in question the quadratic equation and linear equation. Here a simple linear equation is given which we have to solve by using the above concept. Linear equations are those equations in which the highest power of x is 1 and the other are constants. Both constant as well as variable terms are written in L.H.S as well as R.H.S terms.

Complete step-by-step solution:
Given,
$x = \dfrac{4}{5}\left( {x + 10} \right)$
On cross-multiplication, we get
$5x = 4\left( {x + 10} \right)$
On multiplication, we get
$5x = 4x + 40$
Now, transposing $4x$ to L.H.S
$5x - 4x = 40$
$x = 40$
Hence, the value of x is $40$.
Now, let’s check our result
We have, L.H.S $ = x = 40$
Now,
R.H.S $ = \dfrac{4}{5}\left( {x + 10} \right)$
On putting the value of x, we get
$ = \dfrac{4}{5}\left( {40 + 10} \right)$
$ = \dfrac{4}{5} \times 50$
On calculation, we get
$ = 40$
Therefore, L.H.S $ = $ R.H.S
Hence verified.
Additional information: A linear equation is an equation of a straight line, written in one variable. The only power of the variable is $1$ . Linear equations in one variable may take the form $ax + b = 0$ and are solved using basic algebraic operations. We can classify linear equations in one variable in two types: identity equations, conditional or inconsistent equations. An identity equation is true for all values of the variable. For example, $3x = 2x + x$ . A conditional equation is true for only some values of the variable. For example, $5x + 2 = 3x - 6$ . An inconsistent equation results in a false statement. For example, $5x - 15 = 5\left( {x - 4} \right)$.

Note: Generally, the questions of the linear equations are very easy. The complication of the question starts as the degree of the variable stars increases. Operations like addition or subtraction in variable terms can only be done with other variable terms and the same is true for constant terms as here in this question $5x$ is subtracted from $4x$.