Question

How do you solve \[\dfrac{1}{{x + 3}} + \dfrac{1}{{x + 5}} = 1\]?

Answer 1

Hint: Given a rational expression. We have to find the solution of the equation. First, we will solve the rational expression on the left hand side of the equation. Then, cross multiply the terms. Then, multiply the linear polynomial on the right hand side of the expression. Then, we will rewrite the expression in the standard form of quadratic equation. Then, find the solution of the quadratic equation, by applying the quadratic formula.

Formula used:
The quadratic formula to solve the quadratic equation of the form \[a{x^2} + bx + c = 0\]
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:
We are given the equation, \[\dfrac{1}{{x + 3}} + \dfrac{1}{{x + 5}} = 1\]

Rewrite the expression by representing each term over the common denominator.

\[ \Rightarrow \dfrac{1}{{x + 3}} \cdot \dfrac{{x + 5}}{{x + 5}} + \dfrac{1}{{x + 5}} \cdot \dfrac{{x + 3}}{{x + 3}} = 1\]

Multiply the terms we get:

\[ \Rightarrow \dfrac{{x + 5}}{{\left( {x + 3} \right)\left( {x + 5} \right)}} + \dfrac{{x + 3}}{{\left( {x + 3} \right)\left( {x + 5} \right)}} = 1\]

Add the numerator over the common denominator.

\[ \Rightarrow \dfrac{{x + 5 + x + 3}}{{\left( {x + 3} \right)\left( {x + 5} \right)}} = 1\]

On combining like terms, we get:

\[ \Rightarrow \dfrac{{2x + 8}}{{\left( {x + 3} \right)\left( {x + 5} \right)}} = 1\]

Multiply each side of the equation by \[\left( {x + 3} \right)\left( {x + 5} \right)\]

\[ \Rightarrow \dfrac{{2x + 8}}{{\left( {x + 3} \right)\left( {x + 5} \right)}} \times \left( {x + 3} \right)\left( {x + 5} \right) = 1 \times \left( {x + 3} \right)\left( {x + 5} \right)\]

On combining like terms, we get:

\[ \Rightarrow 2x + 8 = {x^2} + 3x + 5x + 15\]

Simplify the equation, we get:

\[ \Rightarrow 2x + 8 = {x^2} + 8x + 15\]

Subtract both sides of equation by \[2x\]

 \[ \Rightarrow 2x + 8 - 2x = {x^2} + 8x - 2x + 15\]

On combining like terms, we get:

\[ \Rightarrow 8 = {x^2} + 6x + 15\]

Subtract 8 from both sides of the equation.

\[ \Rightarrow 8 - 8 = {x^2} + 6x + 15 - 8\]

On combining like terms, we get:

\[ \Rightarrow 0 = {x^2} + 6x + 7\]

Now, apply the quadratic formula, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to solve the equation.

Here, \[a = 1,b = 6,c = 7\]. Substitute the values into the quadratic formula.

Thus, the solution is \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4 \times 1 \times 7} }}{{2 \times 1}}\]

\[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 28} }}{2}\]

On simplifying the expression further, we get:

\[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt 8 }}{2}\]

On simplifying the expression, we get:

\[ \Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt 2 }}{2}\]

Cancel out the common terms from numerator and denominator.

\[ \Rightarrow x = - 3 \pm \sqrt 2 \]

Final answer: Hence the solution of the equation are \[x = - 3 + \sqrt 2 \] and \[x = - 3 - \sqrt 2 \].
Note: Please note that in such types of questions students can check the answer by substituting back the final answer into the given equation and determine whether LHS is equal to RHS.
Here, substitute \[x = - 3 + \sqrt 2 \] into the given equation.

\[\dfrac{1}{{ - 3 + \sqrt 2 + 3}} + \dfrac{1}{{ - 3 + \sqrt 2 + 5}} = 1\]
\[\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{2 + \sqrt 2 }} = 1\]
\[\dfrac{{2 + \sqrt 2 + \sqrt 2 }}{{\sqrt 2 \left( {2 + \sqrt 2 } \right)}} = 1\]
\[\dfrac{{2 + 2\sqrt 2 }}{{2\sqrt 2 + 2}} = 1\]
\[1 = 1\]
Therefore, LHS is equal to RHS.
Similarly, substitute \[x = - 3 - \sqrt 2 \] into the given equation, we get:
\[\dfrac{1}{{ - 3 - \sqrt 2 + 3}} + \dfrac{1}{{ - 3 - \sqrt 2 + 5}} = 1\]
\[\dfrac{1}{{ - \sqrt 2 }} + \dfrac{1}{{2 - \sqrt 2 }} = 1\]
\[\dfrac{{2 - \sqrt 2 - \sqrt 2 }}{{ - \sqrt 2 \left( {2 - \sqrt 2 } \right)}} = 1\]
\[\dfrac{{2 - 2\sqrt 2 }}{{ - 2\sqrt 2 + 2}} = 1\]
 \[1 = 1\]
Therefore, LHS is equal to RHS.