Question

\[Solve{\text{ the linear equation: }}\dfrac{{x - 4}}{7} - x = \dfrac{{5 - x}}{3} + 1\]

Answer 1

Hint: As the equation given is linear it can be simplified to the form $ax + b = 0$ , where a and b are constants. And after that the value of the variable can be found out. In this question the RHS can be subtracted on both sides and then we just need to do the simple mathematical calculation. Lastly we just need to separate the variables and the constant and we will get our required answer.

Complete step-by-step answer:
The equation given is,
\[\dfrac{{x - 4}}{7} - x = \dfrac{{5 - x}}{3} + 1\]
Taking LCM on LHS and RHS,
\[\dfrac{{x - 4 - 7x}}{7} = \dfrac{{5 - x + 3}}{3}\]
\[ \Rightarrow \dfrac{{ - 4 - 6x}}{7} = \dfrac{{8 - x}}{3}\]
Subtracting \[\dfrac{{8 - x}}{3}\] on both sides,
\[ \Rightarrow \dfrac{{ - 4 - 6x}}{7} - \dfrac{{8 - x}}{3} = 0\]
On taking LCM,
\[ \Rightarrow \dfrac{{3\left( { - 4 - 6x} \right) - 7\left( {8 - x} \right)}}{{21}} = 0\]
On simplifying further,
\[\dfrac{{ - 11x - 68}}{{21}} = 0\]
Multiplying 21 both sides,
\[ \Rightarrow - 11x - 68 = 0\]
Subtracting both sides by 68,
\[ \Rightarrow - 11x = 68\]
Finally dividing both sides by -11
\[ \Rightarrow x = - \dfrac{{68}}{{11}}\]
$\therefore {\text{ x = }} - \dfrac{{68}}{{11}}{\text{ is the solution of the given equation}}$
$\therefore {\text{ the final answer is x = }} - \dfrac{{68}}{{11}}$

Note: In this type of questions the equations are simplified as far as possible and then it is solved for the variable in the equation. Calculations should be done carefully to avoid any mistake. Multiply the coefficients of the variable carefully because if the coefficients are wrong then the final answer would be changed. If the final answer comes in fraction then it may be changed into decimal in rounded form. This is a linear equation but if the equation given in question is in quadratic form i.e.
$a{x^2} + bx + c = 0$ then it can solved using quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , where a, b and c are the constants in the general quadratic equation of the form $a{x^2} + bx + c = 0$ .The values of a, band c can be found out be comparing it with the simplified equation. Always try to solve step by step. After the final answer is found out it can be checked whether it satisfies the original equation given in the question by simply substituting its value in the equation.