QuestionFind numbers to exclude from the domain of these rational expressions: 1. $\frac{7}{x-3}$ , 2. $\frac{13}{x+9}$ , 3. $\frac{x+5}{x^{2}-25}$ , 4. $\frac{x+7}{x^{2}-49}$ , 5. $\frac{x-1}{x^{2}+11 x+10}$ , 6. $\frac{x-3}{x^{2}+4 x-45}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with rational expressions. . A rational expression is undefined when the denominator is zero.
3. We need to find the values of x that make the denominator zero, as these values must be excluded from the domain.

STEP 2

For the first rational expression $\frac{7}{x-}$ , we set the denominator equal to zero and solve for x.
$x - =0$

STEP 3

olve the equation to find the value of x that makes the denominator zero.
$x =3$ So,3 must be excluded from the domain of the first rational expression.

STEP 4

For the second rational expression $\frac{13}{x+9}$ , we set the denominator equal to zero and solve for x.
$x +9 =0$

STEP 5

olve the equation to find the value of x that makes the denominator zero.
$x = -9$ So, -9 must be excluded from the domain of the second rational expression.

STEP 6

For the third rational expression $\frac{x+5}{x^{2}-25}$ , we set the denominator equal to zero and solve for x.
$x^{2} -25 =0$

STEP 7

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +5) =0$

STEP 8

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +5 =0 \Rightarrow x = -5$ So,5 and -5 must be excluded from the domain of the third rational expression.

STEP 9

For the fourth rational expression $\frac{x+7}{x^{2}-49}$ , we set the denominator equal to zero and solve for x.
$x^{2} -49 =$

STEP 10

This is a quadratic equation, so we can factor it to find the values of x.
$(x -7)(x +7) =0$

STEP 11

Setting each factor equal to zero gives the solutions.
$x -7 =0 \Rightarrow x =7$ $x +7 =0 \Rightarrow x = -7$ So,7 and -7 must be excluded from the domain of the fourth rational expression.

STEP 12

For the fifth rational expression $\frac{x-}{x^{2}+11x+10}$ , we set the denominator equal to zero and solve for x.
$x^{2} +11x +10 =0$

STEP 13

This is a quadratic equation, so we can factor it to find the values of x.
$(x +)(x +10) =0$

STEP 14

Setting each factor equal to zero gives the solutions.
$x + =0 \Rightarrow x = -$ $x +10 =0 \Rightarrow x = -10$ So, - and -10 must be excluded from the domain of the fifth rational expression.

STEP 15

For the sixth rational expression $\frac{x-3}{x^{2}+4x-45}$ , we set the denominator equal to zero and solve for x.
$x^{2} +4x -45 =0$

STEP 16

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +9) =0$

STEP 17

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +9 =0 \Rightarrow x = -9$ So,5 and -9 must be excluded from the domain of the sixth rational expression.

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QuestionFind numbers to exclude from the domain of these rational expressions: 1. $\frac{7}{x-3}$ , 2. $\frac{13}{x+9}$ , 3. $\frac{x+5}{x^{2}-25}$ , 4. $\frac{x+7}{x^{2}-49}$ , 5. $\frac{x-1}{x^{2}+11 x+10}$ , 6. $\frac{x-3}{x^{2}+4 x-45}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with rational expressions. . A rational expression is undefined when the denominator is zero.
3. We need to find the values of x that make the denominator zero, as these values must be excluded from the domain.

STEP 2

For the first rational expression $\frac{7}{x-}$ , we set the denominator equal to zero and solve for x.
$x - =0$

STEP 3

olve the equation to find the value of x that makes the denominator zero.
$x =3$ So,3 must be excluded from the domain of the first rational expression.

STEP 4

For the second rational expression $\frac{13}{x+9}$ , we set the denominator equal to zero and solve for x.
$x +9 =0$

STEP 5

olve the equation to find the value of x that makes the denominator zero.
$x = -9$ So, -9 must be excluded from the domain of the second rational expression.

STEP 6

For the third rational expression $\frac{x+5}{x^{2}-25}$ , we set the denominator equal to zero and solve for x.
$x^{2} -25 =0$

STEP 7

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +5) =0$

STEP 8

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +5 =0 \Rightarrow x = -5$ So,5 and -5 must be excluded from the domain of the third rational expression.

STEP 9

For the fourth rational expression $\frac{x+7}{x^{2}-49}$ , we set the denominator equal to zero and solve for x.
$x^{2} -49 =$

STEP 10

This is a quadratic equation, so we can factor it to find the values of x.
$(x -7)(x +7) =0$

STEP 11

Setting each factor equal to zero gives the solutions.
$x -7 =0 \Rightarrow x =7$ $x +7 =0 \Rightarrow x = -7$ So,7 and -7 must be excluded from the domain of the fourth rational expression.

STEP 12

For the fifth rational expression $\frac{x-}{x^{2}+11x+10}$ , we set the denominator equal to zero and solve for x.
$x^{2} +11x +10 =0$

STEP 13

This is a quadratic equation, so we can factor it to find the values of x.
$(x +)(x +10) =0$

STEP 14

Setting each factor equal to zero gives the solutions.
$x + =0 \Rightarrow x = -$ $x +10 =0 \Rightarrow x = -10$ So, - and -10 must be excluded from the domain of the fifth rational expression.

STEP 15

For the sixth rational expression $\frac{x-3}{x^{2}+4x-45}$ , we set the denominator equal to zero and solve for x.
$x^{2} +4x -45 =0$

STEP 16

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +9) =0$

STEP 17

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +9 =0 \Rightarrow x = -9$ So,5 and -9 must be excluded from the domain of the sixth rational expression.

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Studdy Solution

STEP 1

Assumptions1. We are dealing with rational expressions. . A rational expression is undefined when the denominator is zero.3. We need to find the values of x that make the denominator zero, as these values must be excluded from the domain.

STEP 2

For the first rational expression 7x−\frac{7}{x-}x−7​, we set the denominator equal to zero and solve for x.x−=0x - =0x−=0

STEP 3

olve the equation to find the value of x that makes the denominator zero.x=3x =3x=3So,3 must be excluded from the domain of the first rational expression.

STEP 4

For the second rational expression 13x+9\frac{13}{x+9}x+913​, we set the denominator equal to zero and solve for x.x+9=0x +9 =0x+9=0

STEP 5

olve the equation to find the value of x that makes the denominator zero.x=−9x = -9x=−9So, -9 must be excluded from the domain of the second rational expression.

STEP 6

For the third rational expression x+5x2−25\frac{x+5}{x^{2}-25}x2−25x+5​, we set the denominator equal to zero and solve for x.x2−25=0x^{2} -25 =0x2−25=0

STEP 7

This is a quadratic equation, so we can factor it to find the values of x.(x−5)(x+5)=0(x -5)(x +5) =0(x−5)(x+5)=0

STEP 8

Setting each factor equal to zero gives the solutions.x−5=0⇒x=5x -5 =0 \Rightarrow x =5x−5=0⇒x=5x+5=0⇒x=−5x +5 =0 \Rightarrow x = -5x+5=0⇒x=−5So,5 and -5 must be excluded from the domain of the third rational expression.

STEP 9

For the fourth rational expression x+7x2−49\frac{x+7}{x^{2}-49}x2−49x+7​, we set the denominator equal to zero and solve for x.x2−49=x^{2} -49 =x2−49=

STEP 10

This is a quadratic equation, so we can factor it to find the values of x.(x−7)(x+7)=0(x -7)(x +7) =0(x−7)(x+7)=0

STEP 11

Setting each factor equal to zero gives the solutions.x−7=0⇒x=7x -7 =0 \Rightarrow x =7x−7=0⇒x=7x+7=0⇒x=−7x +7 =0 \Rightarrow x = -7x+7=0⇒x=−7So,7 and -7 must be excluded from the domain of the fourth rational expression.

STEP 12

For the fifth rational expression x−x2+11x+10\frac{x-}{x^{2}+11x+10}x2+11x+10x−​, we set the denominator equal to zero and solve for x.x2+11x+10=0x^{2} +11x +10 =0x2+11x+10=0

STEP 13

This is a quadratic equation, so we can factor it to find the values of x.(x+)(x+10)=0(x +)(x +10) =0(x+)(x+10)=0

STEP 14

Setting each factor equal to zero gives the solutions.x+=0⇒x=−x + =0 \Rightarrow x = -x+=0⇒x=−x+10=0⇒x=−10x +10 =0 \Rightarrow x = -10x+10=0⇒x=−10So, - and -10 must be excluded from the domain of the fifth rational expression.

STEP 15

For the sixth rational expression x−3x2+4x−45\frac{x-3}{x^{2}+4x-45}x2+4x−45x−3​, we set the denominator equal to zero and solve for x.x2+4x−45=0x^{2} +4x -45 =0x2+4x−45=0

STEP 16

This is a quadratic equation, so we can factor it to find the values of x.(x−5)(x+9)=0(x -5)(x +9) =0(x−5)(x+9)=0

STEP 17

Setting each factor equal to zero gives the solutions.x−5=0⇒x=5x -5 =0 \Rightarrow x =5x−5=0⇒x=5x+9=0⇒x=−9x +9 =0 \Rightarrow x = -9x+9=0⇒x=−9So,5 and -9 must be excluded from the domain of the sixth rational expression.

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Assumptions1. We are dealing with rational expressions. . A rational expression is undefined when the denominator is zero.
3. We need to find the values of x that make the denominator zero, as these values must be excluded from the domain.

For the first rational expression $\frac{7}{x-}$ , we set the denominator equal to zero and solve for x.
$x - =0$

olve the equation to find the value of x that makes the denominator zero.
$x =3$ So,3 must be excluded from the domain of the first rational expression.

For the second rational expression $\frac{13}{x+9}$ , we set the denominator equal to zero and solve for x.
$x +9 =0$

olve the equation to find the value of x that makes the denominator zero.
$x = -9$ So, -9 must be excluded from the domain of the second rational expression.

For the third rational expression $\frac{x+5}{x^{2}-25}$ , we set the denominator equal to zero and solve for x.
$x^{2} -25 =0$

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +5) =0$

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +5 =0 \Rightarrow x = -5$ So,5 and -5 must be excluded from the domain of the third rational expression.

For the fourth rational expression $\frac{x+7}{x^{2}-49}$ , we set the denominator equal to zero and solve for x.
$x^{2} -49 =$

This is a quadratic equation, so we can factor it to find the values of x.
$(x -7)(x +7) =0$

Setting each factor equal to zero gives the solutions.
$x -7 =0 \Rightarrow x =7$ $x +7 =0 \Rightarrow x = -7$ So,7 and -7 must be excluded from the domain of the fourth rational expression.

For the fifth rational expression $\frac{x-}{x^{2}+11x+10}$ , we set the denominator equal to zero and solve for x.
$x^{2} +11x +10 =0$

This is a quadratic equation, so we can factor it to find the values of x.
$(x +)(x +10) =0$

Setting each factor equal to zero gives the solutions.
$x + =0 \Rightarrow x = -$ $x +10 =0 \Rightarrow x = -10$ So, - and -10 must be excluded from the domain of the fifth rational expression.

For the sixth rational expression $\frac{x-3}{x^{2}+4x-45}$ , we set the denominator equal to zero and solve for x.
$x^{2} +4x -45 =0$

This is a quadratic equation, so we can factor it to find the values of x.
$(x -5)(x +9) =0$

Setting each factor equal to zero gives the solutions.
$x -5 =0 \Rightarrow x =5$ $x +9 =0 \Rightarrow x = -9$ So,5 and -9 must be excluded from the domain of the sixth rational expression.