Question

ALGEBRA QUESTION PLS HELPS

Answer 1

The value of x is –7.

Solution:

Given expression:

$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}$

Let us factor $x^2+4x+3$.

$x^2+4x+3=(x+1)(x+3)$

Substitute this in the fraction.

$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

To make the denominator same, multiply and divide the first term by (x +1).

$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

Denominators are same, you can add the fractions.

$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}$

$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}$

Cancel the common term in the numerator and denominator.

$\frac{x+7}{x+1} \cdot \frac{1}{x+1}$

Multiply the fractions.

$\frac{x+7}{(x+1)^2}$

$\frac{x+7}{x^2+2x+1}$

The expression is simplified to one rational expression.

Suppose the expression is equal to 0.

$\frac{x+7}{x^2+2x+1}=0$

Do cross multiplication.

${x+7}=0\times (}{x^2+2x+1})$

Any number or variable multiplied by 0 gives 0.

${x+7}=0$

Subtract 7 from both sides of the equation.

${x+7-7}=0-7$

x = –7

The value of x is –7.