Question

How do you solve \[\dfrac{x}{{2x + 1}} = \dfrac{5}{{4 - x}}\] ?

Answer 1

Hint: Here, we need to solve this rational expression by factorizing the numerator and denominator. If there is any common factor, take out commonly then find the value of \[x\] . A rational expression is an expression in the form of ratio of two polynomial which can be written in the form of \[\dfrac{{f(x)}}{{g(x)}}\] where \[f(x)\] and \[g(x)\] are the polynomials.

Complete step by step solution:
Given expression is \[\dfrac{x}{{2x + 1}} = \dfrac{5}{{4 - x}} \to (1)\]
By perform cross multiplication, we get
 \[\dfrac{{x(4 - x)}}{{(2x + 1)(4 - x)}} = \dfrac{{5(2x + 1)}}{{(4 - x)(2x + 1)}}\]
Remove the denominator on both sides are same, we get
 \[x(4 - x) = 5(2x + 1)\]
By simplify the expression, we get
 \[4x - {x^2} = 10x + 5\]
 \[0 = {x^2} - 4x + 10x + 5\]
Now, we have
 \[{x^2} + 6x + 5 = 0\]
By factorize the above equation, we get
 \[(x + 1)(x + 5) = 0\]
Here, \[x = - 1\] and \[x = - 5\]
Let’s check by substitute the value into equation \[(1)\]
By applying the value \[x\] If the two sides are equal, then the value is correct.
Case \[1\] : If \[x = - 1 \Rightarrow \dfrac{{ - 1}}{{2( - 1) + 1}} = \dfrac{5}{{4 - ( - 1)}}\]
 \[\dfrac{{ - 1}}{1} = \dfrac{5}{5}\]
Then, we get
 \[1 = 1\]
So, the value is correct.
Case \[2\] : If \[x = - 5 \Rightarrow \dfrac{{ - 5}}{{2( - 5) + 1}} = \dfrac{5}{{4 - ( - 5)}}\]
 \[\dfrac{{ - 5}}{{ - 9}} = \dfrac{5}{9}\] then, \[\dfrac{5}{9} = \dfrac{5}{9}\]
So, the value is correct.
Therefore, the value of \[x = - 1\] ..
So, the correct answer is “ \[x = - 1\] and \[x = - 5\]”.

Note: Here, we need to solve the rational factor by using factoring method after that we get the value. If you want to check it is correct, apply it into the given solution. An algebraic expression can also be expressed in the form of its factors. An algebraic expression consists of variables, constants and operators. An algebraic expression consists of terms separated by an addition operation.