Question

Solve the linear equation $\dfrac{{x + 2}}{3} - \dfrac{{x + 1}}{5} = \dfrac{{x + 3}}{4} - 1$

Answer 1

Hint: First we have to define what the terms we need to solve the problem are. Since this question is of linear equation problem, we know some of the properties to solve the given linear equation (the given equation is linear because it’s of degree one).
A pair of linear equations in two variables can be solved using these three methods. Elimination method, substitution method and cross multiplication method. For one variable we can use the substitution method and cross multiplication method.

Complete step-by-step solution:
Since the given equation is linear it depends on one variable.
We further solve $\dfrac{{x + 2}}{3} - \dfrac{{x + 1}}{5} = \dfrac{{x + 3}}{4} - 1$
Let $x$ be the only unknown value and all others are numbers, so first we apply the least common multiple method to solve further.
The Least Common Multiple (LCM) of a group of numbers is the smallest number that is a multiple of all the numbers. For instance, the LCM of \[16\] and \[20\] is \[80\]; \[80\] is the smallest number that is both a multiple of \[16\] and a multiple of \[20\].
Hence from this L.C.M method (L.C.M of $3$ and $5 = 15$ in left hand side and L.C.M $4$ in right hand side)
So cross multiplying we get $\dfrac{{5(x + 2) - 3(x + 1)}}{{15}} = \dfrac{{x + 3 - 4}}{4}$
Simplifying the above equation, we get $\dfrac{{5x + 10 - 3x - 3}}{{15}} = \dfrac{{x - 1}}{4}$
$ \Rightarrow \dfrac{{2x + 7}}{{15}} = \dfrac{{x - 1}}{4}$
Again, cross multiply $15$ and $4$ from left- and right-hand side equations we get
$ \Rightarrow 4(2x + 7) = 15(x - 1)$ (giving multiplication terms to the variables)
$ \Rightarrow 8x + 28 = 15x - 15$
Now turn all the variable elements to the right side and all the numbers to the left side we get
$ \Rightarrow 43 = 7x$
Hence $x = \dfrac{{43}}{7}$ or approximately to $x = 6.14$

Note: To find the linear equations we may have many methods but for one variable we need to apply only substitution method and cross multiplication method.
The LCM of these numbers is \[2{\text{ }} \times {\text{ }}2{\text{ }} \times {\text{ }}3{\text{ }} \times {\text{ }}5{\text{ }} = {\text{ }}60\]. $3$ is the largest number that divides each of these numbers, and hence, is the HCF for these numbers, HCF is also known as Greatest Common Divisor (GCD). To find the HCF of two or more numbers, express each number as a product of prime numbers.