Solved on Dec 05, 2023

Find the product and the domain of $\frac{5a}{5a+5} \cdot (10a+10)$ and $\frac{x^2-5x}{x^2-3x} \cdot \frac{x+3}{x-5}$ .

STEP 1

Assumptions
1. We are asked to find the product of two expressions and their domain.
2. The first expression is $\frac{5 a}{5 a+5} \cdot(10 a+10)$ .
3. The second expression is $\frac{x^{2}-5 x}{x^{2}-3 x} \cdot \frac{x+3}{x-5}$ .
4. The domain of an expression is the set of all possible values of the variable for which the expression is defined.

STEP 2

First, we will simplify the first expression $\frac{5 a}{5 a+5} \cdot(10 a+10)$ .

STEP 3

Factor out the common factor of 5 in the denominator and the common factor of 10 in the second term.
$\frac{5 a}{5(a+1)} \cdot 10(a+1)$

STEP 4

The factor of $5$ in the numerator and the factor of $5$ in the denominator cancel each other out, as do the $(a+1)$ terms.
$\frac{5 a}{5(a+1)} \cdot 10(a+1) = a \cdot 10 = 10a$

STEP 5

The product of the first expression is $10a$ .

STEP 6

Now, let's find the domain of the first expression. The domain is all the values of $a$ for which the expression is defined.

STEP 7

The original expression has a denominator of $5a+5$ . To find the domain, we set the denominator not equal to zero.
$5a+5 \neq 0$

STEP 8

Solve for $a$ .
$a \neq -1$

STEP 9

The domain of the first expression is all real numbers except $a \neq -1$ .

STEP 10

Now, let's simplify the second expression $\frac{x^{2}-5 x}{x^{2}-3 x} \cdot \frac{x+3}{x-5}$ .

STEP 11

Factor out $x$ from the numerator and the denominator of the first fraction.
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5}$

STEP 12

Cancel out the common factors of $x$ and $(x-5)$ .
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5} = \frac{x}{x-3} \cdot \frac{x+3}{1}$

STEP 13

Now, multiply the remaining factors.
$\frac{x}{x-3} \cdot (x+3)$

STEP 14

The product of the second expression is $\frac{x(x+3)}{x-3}$ .

STEP 15

Next, let's find the domain of the second expression. The domain is all the values of $x$ for which the expression is defined.

STEP 16

The original expression has denominators of $x^{2}-3 x$ and $x-5$ . To find the domain, we set each denominator not equal to zero.
$x^{2}-3 x \neq 0 \quad \text{and} \quad x-5 \neq 0$

STEP 17

Solve for $x$ in each inequality.
$x(x-3) \neq 0 \quad \text{implies} \quad x \neq 0 \quad \text{and} \quad x \neq 3$ $x-5 \neq 0 \quad \text{implies} \quad x \neq 5$

STEP 18

The domain of the second expression is all real numbers except $x \neq 0$ , $x \neq 3$ , and $x \neq 5$ .

STEP 19

Summarize the results.
The product of the first expression is $10a$ , and its domain is all real numbers except $a \neq -1$ .
The product of the second expression is $\frac{x(x+3)}{x-3}$ , and its domain is all real numbers except $x \neq 0$ , $x \neq 3$ , and $x \neq 5$ .

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Find the product and the domain of $\frac{5a}{5a+5} \cdot (10a+10)$ and $\frac{x^2-5x}{x^2-3x} \cdot \frac{x+3}{x-5}$ .

STEP 1

STEP 2

First, we will simplify the first expression $\frac{5 a}{5 a+5} \cdot(10 a+10)$ .

STEP 3

Factor out the common factor of 5 in the denominator and the common factor of 10 in the second term.
$\frac{5 a}{5(a+1)} \cdot 10(a+1)$

STEP 4

The factor of $5$ in the numerator and the factor of $5$ in the denominator cancel each other out, as do the $(a+1)$ terms.
$\frac{5 a}{5(a+1)} \cdot 10(a+1) = a \cdot 10 = 10a$

STEP 5

The product of the first expression is $10a$ .

STEP 6

Now, let's find the domain of the first expression. The domain is all the values of $a$ for which the expression is defined.

STEP 7

The original expression has a denominator of $5a+5$ . To find the domain, we set the denominator not equal to zero.
$5a+5 \neq 0$

STEP 8

Solve for $a$ .
$a \neq -1$

STEP 9

The domain of the first expression is all real numbers except $a \neq -1$ .

STEP 10

Now, let's simplify the second expression $\frac{x^{2}-5 x}{x^{2}-3 x} \cdot \frac{x+3}{x-5}$ .

STEP 11

Factor out $x$ from the numerator and the denominator of the first fraction.
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5}$

STEP 12

Cancel out the common factors of $x$ and $(x-5)$ .
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5} = \frac{x}{x-3} \cdot \frac{x+3}{1}$

STEP 13

Now, multiply the remaining factors.
$\frac{x}{x-3} \cdot (x+3)$

STEP 14

The product of the second expression is $\frac{x(x+3)}{x-3}$ .

STEP 15

Next, let's find the domain of the second expression. The domain is all the values of $x$ for which the expression is defined.

STEP 16

The original expression has denominators of $x^{2}-3 x$ and $x-5$ . To find the domain, we set each denominator not equal to zero.
$x^{2}-3 x \neq 0 \quad \text{and} \quad x-5 \neq 0$

STEP 17

Solve for $x$ in each inequality.
$x(x-3) \neq 0 \quad \text{implies} \quad x \neq 0 \quad \text{and} \quad x \neq 3$ $x-5 \neq 0 \quad \text{implies} \quad x \neq 5$

STEP 18

The domain of the second expression is all real numbers except $x \neq 0$ , $x \neq 3$ , and $x \neq 5$ .

STEP 19

Summarize the results.
The product of the first expression is $10a$ , and its domain is all real numbers except $a \neq -1$ .
The product of the second expression is $\frac{x(x+3)}{x-3}$ , and its domain is all real numbers except $x \neq 0$ , $x \neq 3$ , and $x \neq 5$ .

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Find the product and the domain of 5a5a+5⋅(10a+10)\frac{5a}{5a+5} \cdot (10a+10)5a+55a​⋅(10a+10) and x2−5xx2−3x⋅x+3x−5\frac{x^2-5x}{x^2-3x} \cdot \frac{x+3}{x-5}x2−3xx2−5x​⋅x−5x+3​.

STEP 1

STEP 2

First, we will simplify the first expression 5a5a+5⋅(10a+10)\frac{5 a}{5 a+5} \cdot(10 a+10)5a+55a​⋅(10a+10).

STEP 3

Factor out the common factor of 5 in the denominator and the common factor of 10 in the second term.5a5(a+1)⋅10(a+1)\frac{5 a}{5(a+1)} \cdot 10(a+1)5(a+1)5a​⋅10(a+1)

STEP 4

The factor of 555 in the numerator and the factor of 555 in the denominator cancel each other out, as do the (a+1)(a+1)(a+1) terms.5a5(a+1)⋅10(a+1)=a⋅10=10a\frac{5 a}{5(a+1)} \cdot 10(a+1) = a \cdot 10 = 10a5(a+1)5a​⋅10(a+1)=a⋅10=10a

STEP 5

The product of the first expression is 10a10a10a.

STEP 6

Now, let's find the domain of the first expression. The domain is all the values of aaa for which the expression is defined.

STEP 7

The original expression has a denominator of 5a+55a+55a+5. To find the domain, we set the denominator not equal to zero.5a+5≠05a+5 \neq 05a+5=0

STEP 8

Solve for aaa.a≠−1a \neq -1a=−1

STEP 9

The domain of the first expression is all real numbers except a≠−1a \neq -1a=−1.

STEP 10

Now, let's simplify the second expression x2−5xx2−3x⋅x+3x−5\frac{x^{2}-5 x}{x^{2}-3 x} \cdot \frac{x+3}{x-5}x2−3xx2−5x​⋅x−5x+3​.

STEP 11

Factor out xxx from the numerator and the denominator of the first fraction.x(x−5)x(x−3)⋅x+3x−5\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5}x(x−3)x(x−5)​⋅x−5x+3​

STEP 12

Cancel out the common factors of xxx and (x−5)(x-5)(x−5).x(x−5)x(x−3)⋅x+3x−5=xx−3⋅x+31\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5} = \frac{x}{x-3} \cdot \frac{x+3}{1}x(x−3)x(x−5)​⋅x−5x+3​=x−3x​⋅1x+3​

STEP 13

Now, multiply the remaining factors.xx−3⋅(x+3)\frac{x}{x-3} \cdot (x+3)x−3x​⋅(x+3)

STEP 14

The product of the second expression is x(x+3)x−3\frac{x(x+3)}{x-3}x−3x(x+3)​.

STEP 15

Next, let's find the domain of the second expression. The domain is all the values of xxx for which the expression is defined.

STEP 16

The original expression has denominators of x2−3xx^{2}-3 xx2−3x and x−5x-5x−5. To find the domain, we set each denominator not equal to zero.x2−3x≠0andx−5≠0x^{2}-3 x \neq 0 \quad \text{and} \quad x-5 \neq 0x2−3x=0andx−5=0

STEP 17

Solve for xxx in each inequality.x(x−3)≠0impliesx≠0andx≠3x(x-3) \neq 0 \quad \text{implies} \quad x \neq 0 \quad \text{and} \quad x \neq 3x(x−3)=0impliesx=0andx=3 x−5≠0impliesx≠5x-5 \neq 0 \quad \text{implies} \quad x \neq 5x−5=0impliesx=5

STEP 18

The domain of the second expression is all real numbers except x≠0x \neq 0x=0, x≠3x \neq 3x=3, and x≠5x \neq 5x=5.

STEP 19

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Find the product and the domain of $\frac{5a}{5a+5} \cdot (10a+10)$ and $\frac{x^2-5x}{x^2-3x} \cdot \frac{x+3}{x-5}$ .

First, we will simplify the first expression $\frac{5 a}{5 a+5} \cdot(10 a+10)$ .

Factor out the common factor of 5 in the denominator and the common factor of 10 in the second term.
$\frac{5 a}{5(a+1)} \cdot 10(a+1)$

The factor of $5$ in the numerator and the factor of $5$ in the denominator cancel each other out, as do the $(a+1)$ terms.
$\frac{5 a}{5(a+1)} \cdot 10(a+1) = a \cdot 10 = 10a$

The product of the first expression is $10a$ .

Now, let's find the domain of the first expression. The domain is all the values of $a$ for which the expression is defined.

The original expression has a denominator of $5a+5$ . To find the domain, we set the denominator not equal to zero.
$5a+5 \neq 0$

Solve for $a$ .
$a \neq -1$

The domain of the first expression is all real numbers except $a \neq -1$ .

Now, let's simplify the second expression $\frac{x^{2}-5 x}{x^{2}-3 x} \cdot \frac{x+3}{x-5}$ .

Factor out $x$ from the numerator and the denominator of the first fraction.
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5}$

Cancel out the common factors of $x$ and $(x-5)$ .
$\frac{x(x-5)}{x(x-3)} \cdot \frac{x+3}{x-5} = \frac{x}{x-3} \cdot \frac{x+3}{1}$

Now, multiply the remaining factors.
$\frac{x}{x-3} \cdot (x+3)$

The product of the second expression is $\frac{x(x+3)}{x-3}$ .

Next, let's find the domain of the second expression. The domain is all the values of $x$ for which the expression is defined.

The original expression has denominators of $x^{2}-3 x$ and $x-5$ . To find the domain, we set each denominator not equal to zero.
$x^{2}-3 x \neq 0 \quad \text{and} \quad x-5 \neq 0$

Solve for $x$ in each inequality.
$x(x-3) \neq 0 \quad \text{implies} \quad x \neq 0 \quad \text{and} \quad x \neq 3$ $x-5 \neq 0 \quad \text{implies} \quad x \neq 5$

The domain of the second expression is all real numbers except $x \neq 0$ , $x \neq 3$ , and $x \neq 5$ .