Question

Solve the equation:
$\dfrac{{2x - 7}}{5} - \dfrac{{3x + 11}}{2} = \dfrac{{2x + 8}}{3} - 5$

Answer 1

Hint: Start with taking the LCM of the fractions on both right hand side and left hand side. Then cross-multiply the resulting expression to form a linear equation. Solve this equation to find the value of $x$.

Complete step-by-step solution:
According to the question, the given equation is $\dfrac{{2x - 7}}{5} - \dfrac{{3x + 11}}{2} = \dfrac{{2x + 8}}{3} - 5$. We have to solve it to find the value of $x$.
We will take the LCM on both the left hand side and right hand side of the equation. This will give us:
$ \Rightarrow \dfrac{{2\left( {2x - 7} \right) - 5\left( {3x + 11} \right)}}{{5 \times 2}} = \dfrac{{2x + 8 - 5 \times 3}}{3}$
Simplifying it further, we’ll get:
$
   \Rightarrow \dfrac{{4x - 14 - 15x - 55}}{{10}} = \dfrac{{2x + 8 - 15}}{3} \\
   \Rightarrow \dfrac{{ - 11x - 69}}{{10}} = \dfrac{{2x - 7}}{3}
 $
On cross multiplication, this will give us:
$ \Rightarrow 3\left( { - 11x - 69} \right) = 10\left( {2x - 7} \right)$
On further simplification, we’ll get:
$
   \Rightarrow - 33x - 207 = 20x - 70 \\
   \Rightarrow 53x = - 137 \\
   \Rightarrow x = - \dfrac{{137}}{{53}}
 $

Thus the value of $x$ in the above equation is $ - \dfrac{{137}}{{53}}$.

Note: Since this is a linear equation in $x$, it has only one real root. Similarly a quadratic equation has maximum two real roots and a cubic equation has maximum three real roots. If we generalize it, we can say that an $n$ degree equation has maximum $n$ real roots.
If a linear equation is having only one variable, it can be solved directly to get the value of the variable. If it was a two variable equation, we couldn’t have solved it. To determine the values of two different variables, we need two different equations in those variables. Similarly if there are $n$ different variables then we require $n$ different equations to find their values.