Solved on Sep 27, 2023

Solve the rational equation $\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{2}-25}{x^{2}-9x+18}$ and simplify the solution set.

STEP 1

Assumptions1. The equation to solve is $\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{}-25}{x^{}-9 x+18}$ . We are looking for the set of real number solutions for x3. We assume that the denominators are not equal to zero, as division by zero is undefined

STEP 2

First, we simplify the right side of the equation. The expression $x^{2}-25$ is a difference of squares and can be factored into $(x-5)(x+5)$ .
$\frac{x^{2}-25}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{x^{2}-9 x+18}$

STEP 3

The denominator $x^{2}-9 x+18$ can also be factored into $(x-3)(x-6)$ .
$\frac{(x-5)(x+5)}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{(x-3)(x-6)}$

STEP 4

Now we rewrite the original equation using the factored forms.
$\frac{4}{x-3}+\frac{}{x-6}=\frac{(x-)(x+)}{(x-3)(x-6)}$

STEP 5

We can now multiply each term by $(x-3)(x-)$ to eliminate the denominators.
$(x-3)(x-)\left(\frac{4}{x-3}\right) + (x-3)(x-)\left(\frac{5}{x-}\right) = (x-3)(x-)\left(\frac{(x-5)(x+5)}{(x-3)(x-)}\right)$

STEP 6

implify each term.
$4(x-6) +5(x-3) = (x-5)(x+5)$

STEP 7

Expand each term.
$4x -24 +5x -15 = x^{2} -25$

STEP 8

Combine like terms.
$x -39 = x^{2} -25$

STEP 9

Rearrange the equation to set it equal to zero.
$x^{2} -9x -14 =$

STEP 10

This is a quadratic equation in the form $ax^{2} + bx + c =0$ . We can solve it by factoring.
$(x -7)(x -2) =0$

STEP 11

Set each factor equal to zero and solve for x.
$x -7 =0 \Rightarrow x =7$ $x - =0 \Rightarrow x =$ So, the solution set is $\{,7\}$ .

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Solve the rational equation $\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{2}-25}{x^{2}-9x+18}$ and simplify the solution set.

STEP 1

Assumptions1. The equation to solve is $\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{}-25}{x^{}-9 x+18}$ . We are looking for the set of real number solutions for x3. We assume that the denominators are not equal to zero, as division by zero is undefined

STEP 2

First, we simplify the right side of the equation. The expression $x^{2}-25$ is a difference of squares and can be factored into $(x-5)(x+5)$ .
$\frac{x^{2}-25}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{x^{2}-9 x+18}$

STEP 3

The denominator $x^{2}-9 x+18$ can also be factored into $(x-3)(x-6)$ .
$\frac{(x-5)(x+5)}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{(x-3)(x-6)}$

STEP 4

Now we rewrite the original equation using the factored forms.
$\frac{4}{x-3}+\frac{}{x-6}=\frac{(x-)(x+)}{(x-3)(x-6)}$

STEP 5

We can now multiply each term by $(x-3)(x-)$ to eliminate the denominators.
$(x-3)(x-)\left(\frac{4}{x-3}\right) + (x-3)(x-)\left(\frac{5}{x-}\right) = (x-3)(x-)\left(\frac{(x-5)(x+5)}{(x-3)(x-)}\right)$

STEP 6

implify each term.
$4(x-6) +5(x-3) = (x-5)(x+5)$

STEP 7

Expand each term.
$4x -24 +5x -15 = x^{2} -25$

STEP 8

Combine like terms.
$x -39 = x^{2} -25$

STEP 9

Rearrange the equation to set it equal to zero.
$x^{2} -9x -14 =$

STEP 10

This is a quadratic equation in the form $ax^{2} + bx + c =0$ . We can solve it by factoring.
$(x -7)(x -2) =0$

STEP 11

Set each factor equal to zero and solve for x.
$x -7 =0 \Rightarrow x =7$ $x - =0 \Rightarrow x =$ So, the solution set is $\{,7\}$ .

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Solve the rational equation 4x−3+5x−6=x2−25x2−9x+18\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{2}-25}{x^{2}-9x+18}x−34​+x−65​=x2−9x+18x2−25​ and simplify the solution set.

STEP 1

STEP 2

STEP 3

The denominator x2−9x+18x^{2}-9 x+18x2−9x+18 can also be factored into (x−3)(x−6)(x-3)(x-6)(x−3)(x−6).(x−5)(x+5)x2−9x+18=(x−5)(x+5)(x−3)(x−6)\frac{(x-5)(x+5)}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{(x-3)(x-6)}x2−9x+18(x−5)(x+5)​=(x−3)(x−6)(x−5)(x+5)​

STEP 4

Now we rewrite the original equation using the factored forms.4x−3+x−6=(x−)(x+)(x−3)(x−6)\frac{4}{x-3}+\frac{}{x-6}=\frac{(x-)(x+)}{(x-3)(x-6)}x−34​+x−6​=(x−3)(x−6)(x−)(x+)​

STEP 5

STEP 6

implify each term.4(x−6)+5(x−3)=(x−5)(x+5)4(x-6) +5(x-3) = (x-5)(x+5)4(x−6)+5(x−3)=(x−5)(x+5)

STEP 7

Expand each term.4x−24+5x−15=x2−254x -24 +5x -15 = x^{2} -254x−24+5x−15=x2−25

STEP 8

Combine like terms.x−39=x2−25x -39 = x^{2} -25x−39=x2−25

STEP 9

Rearrange the equation to set it equal to zero.x2−9x−14=x^{2} -9x -14 =x2−9x−14=

STEP 10

This is a quadratic equation in the form ax2+bx+c=0ax^{2} + bx + c =0ax2+bx+c=0. We can solve it by factoring.(x−7)(x−2)=0(x -7)(x -2) =0(x−7)(x−2)=0

STEP 11

Set each factor equal to zero and solve for x.x−7=0⇒x=7x -7 =0 \Rightarrow x =7x−7=0⇒x=7x−=0⇒x=x - =0 \Rightarrow x =x−=0⇒x=So, the solution set is {,7}\{,7\}{,7}.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Solve the rational equation $\frac{4}{x-3}+\frac{5}{x-6}=\frac{x^{2}-25}{x^{2}-9x+18}$ and simplify the solution set.

The denominator $x^{2}-9 x+18$ can also be factored into $(x-3)(x-6)$ .
$\frac{(x-5)(x+5)}{x^{2}-9 x+18} = \frac{(x-5)(x+5)}{(x-3)(x-6)}$

Now we rewrite the original equation using the factored forms.
$\frac{4}{x-3}+\frac{}{x-6}=\frac{(x-)(x+)}{(x-3)(x-6)}$

implify each term.
$4(x-6) +5(x-3) = (x-5)(x+5)$

Expand each term.
$4x -24 +5x -15 = x^{2} -25$

Combine like terms.
$x -39 = x^{2} -25$

Rearrange the equation to set it equal to zero.
$x^{2} -9x -14 =$

This is a quadratic equation in the form $ax^{2} + bx + c =0$ . We can solve it by factoring.
$(x -7)(x -2) =0$

Set each factor equal to zero and solve for x.
$x -7 =0 \Rightarrow x =7$ $x - =0 \Rightarrow x =$ So, the solution set is $\{,7\}$ .