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olchik [2.2K]
3 years ago
9

Given the two sets, which statement is true?

Mathematics
2 answers:
gogolik [260]3 years ago
8 0

what is the answer fghrtuer rrrrrrrrrr


Dmitrij [34]3 years ago
4 0
Use number 4 option im sure that is correct

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The value of x is –7.

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Given expression:

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

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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}

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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.

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