The value of x is –7.
Solution:
Given expression:
[tex]$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}[/tex]
Let us factor [tex]x^2+4x+3[/tex].
[tex]x^2+4x+3=(x+1)(x+3)[/tex]
Substitute this in the fraction.
[tex]$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}[/tex]
To make the denominator same, multiply and divide the first term by (x +1).
[tex]$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}[/tex]
Denominators are same, you can add the fractions.
[tex]$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}[/tex]
[tex]$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}[/tex]
Cancel the common term in the numerator and denominator.
[tex]$\frac{x+7}{x+1} \cdot \frac{1}{x+1}[/tex]
Multiply the fractions.
[tex]$\frac{x+7}{(x+1)^2}[/tex]
[tex]$\frac{x+7}{x^2+2x+1}[/tex]
The expression is simplified to one rational expression.
Suppose the expression is equal to 0.
[tex]$\frac{x+7}{x^2+2x+1}=0[/tex]
Do cross multiplication.
[tex]${x+7}=0\times (}{x^2+2x+1})[/tex]
Any number or variable multiplied by 0 gives 0.
[tex]${x+7}=0[/tex]
Subtract 7 from both sides of the equation.
[tex]${x+7-7}=0-7[/tex]
x = –7
The value of x is –7.
