
How do you combine $\dfrac{2}{{{x}^{2}}+8x+15}+\dfrac{1}{{{x}^{2}}+11x+30}$ ?
Answer
476.1k+ views
Hint: The problem that we are given here consists of two fractions that must be added to each other in order to get the solution. Both the fractions present in the given expression have a quadratic expression in their denominators, so we factorize both the quadratic equations one by one. After doing that so we do the summation of the two fractions by finding the LCM of the factors present in the denominator and adding them accordingly which leads us to the final result of the problem.
Complete step by step solution:
The given expression is
$\dfrac{2}{{{x}^{2}}+8x+15}+\dfrac{1}{{{x}^{2}}+11x+30}$
In the above expression the two factors consist of a quadratic equation each at their denominators.
Hence, we rewrite those quadratic equations in terms of their corresponding factors.
The first fraction has the quadratic equation
$\Rightarrow {{x}^{2}}+8x+15$
As the number $15$ has the factors $5$ and $3$ we can write $5+3$ in place of $8$ as a coefficient of $x$ as shown below
$\Rightarrow {{x}^{2}}+\left( 5+3 \right)x+15$
Now, applying the distributive law we get
$\Rightarrow {{x}^{2}}+5x+3x+15$
$\Rightarrow x\left( x+5 \right)+3\left( x+5 \right)$
$\Rightarrow \left( x+5 \right)\left( x+3 \right)$
The second fraction has the quadratic equation
$\Rightarrow {{x}^{2}}+11x+30$
As the number $30$ has the factors $5$ and $6$ we can write $5+6$ in place of $11$ as a coefficient of $x$ as shown below
$\Rightarrow {{x}^{2}}+\left( 5+6 \right)x+30$
Now, applying the distributive law we get
$\Rightarrow {{x}^{2}}+5x+6x+30$
$\Rightarrow x\left( x+5 \right)+6\left( x+5 \right)$
$\Rightarrow \left( x+5 \right)\left( x+6 \right)$
Therefore, we can rewrite the given expression as
$\Rightarrow \dfrac{2}{\left( x+5 \right)\left( x+3 \right)}+\dfrac{1}{\left( x+5 \right)\left( x+6 \right)}$
Hence, we get the LCM of the denominator as $\left( x+6 \right)\left( x+5 \right)\left( x+3 \right)$ and write it as the common denominator as
$\Rightarrow \dfrac{2\left( x+6 \right)+1\cdot \left( x+3 \right)}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{2\cdot x+2\cdot 6+1\cdot x+1\cdot 3}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{3x+15}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{3\left( x+5 \right)}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
Cancelling the factor $\left( x+5 \right)$ from both the numerator and denominator we get
$\Rightarrow \dfrac{3}{\left( x+6 \right)\left( x+3 \right)}$
Therefore, the simplified form is $\dfrac{3}{\left( x+6 \right)\left( x+3 \right)}$
Note: While calculating the LCM of the factors in the denominator we have to be careful, so that errors can be avoided. Also, we must properly cancel the common terms from the numerator and denominator to get the result in the simplified form.
Complete step by step solution:
The given expression is
$\dfrac{2}{{{x}^{2}}+8x+15}+\dfrac{1}{{{x}^{2}}+11x+30}$
In the above expression the two factors consist of a quadratic equation each at their denominators.
Hence, we rewrite those quadratic equations in terms of their corresponding factors.
The first fraction has the quadratic equation
$\Rightarrow {{x}^{2}}+8x+15$
As the number $15$ has the factors $5$ and $3$ we can write $5+3$ in place of $8$ as a coefficient of $x$ as shown below
$\Rightarrow {{x}^{2}}+\left( 5+3 \right)x+15$
Now, applying the distributive law we get
$\Rightarrow {{x}^{2}}+5x+3x+15$
$\Rightarrow x\left( x+5 \right)+3\left( x+5 \right)$
$\Rightarrow \left( x+5 \right)\left( x+3 \right)$
The second fraction has the quadratic equation
$\Rightarrow {{x}^{2}}+11x+30$
As the number $30$ has the factors $5$ and $6$ we can write $5+6$ in place of $11$ as a coefficient of $x$ as shown below
$\Rightarrow {{x}^{2}}+\left( 5+6 \right)x+30$
Now, applying the distributive law we get
$\Rightarrow {{x}^{2}}+5x+6x+30$
$\Rightarrow x\left( x+5 \right)+6\left( x+5 \right)$
$\Rightarrow \left( x+5 \right)\left( x+6 \right)$
Therefore, we can rewrite the given expression as
$\Rightarrow \dfrac{2}{\left( x+5 \right)\left( x+3 \right)}+\dfrac{1}{\left( x+5 \right)\left( x+6 \right)}$
Hence, we get the LCM of the denominator as $\left( x+6 \right)\left( x+5 \right)\left( x+3 \right)$ and write it as the common denominator as
$\Rightarrow \dfrac{2\left( x+6 \right)+1\cdot \left( x+3 \right)}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{2\cdot x+2\cdot 6+1\cdot x+1\cdot 3}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{3x+15}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
$\Rightarrow \dfrac{3\left( x+5 \right)}{\left( x+5 \right)\left( x+6 \right)\left( x+3 \right)}$
Cancelling the factor $\left( x+5 \right)$ from both the numerator and denominator we get
$\Rightarrow \dfrac{3}{\left( x+6 \right)\left( x+3 \right)}$
Therefore, the simplified form is $\dfrac{3}{\left( x+6 \right)\left( x+3 \right)}$
Note: While calculating the LCM of the factors in the denominator we have to be careful, so that errors can be avoided. Also, we must properly cancel the common terms from the numerator and denominator to get the result in the simplified form.
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