Question
solve the rational equation indicate any extraneous solutions 5/x + 8/x+3=7
Answers
Answer
solve the rational equation indicate any extraneous solutions 4(x-3)/36-x^2 = 1/6-x+2x/6+x
Answered by
GPT 3.5
To solve the rational equation 5/x + 8/(x+3) = 7, we can start by finding a common denominator for the two fractions on the left side. The common denominator is x(x+3), so we can rewrite the equation as:
(5(x+3) + 8x) / x(x+3) = 7
Simplifying the numerator, we have:
(5x + 15 + 8x) / x(x+3) = 7
(13x + 15) / x(x+3) = 7
Next, we can multiply both sides of the equation by the denominator x(x+3) to get rid of the fraction:
13x + 15 = 7x(x+3)
Expanding the right side, we have:
13x + 15 = 7x^2 + 21x
Rearranging the equation to bring all terms to one side, we have:
7x^2 + 21x - 13x - 15 = 0
7x^2 + 8x - 15 = 0
Now, we can factor the quadratic equation:
(7x - 5)(x + 3) = 0
Setting each factor equal to zero, we have two possible solutions:
7x - 5 = 0, which gives x = 5/7
x + 3 = 0, which gives x = -3
Now we need to check if these solutions are valid or if they are extraneous. We substitute each solution back into the original equation:
For x = 5/7:
5/(5/7) + 8/(5/7 + 3) = 7
5 * 7/5 + 8/(35/7 + 3) = 7
7 + 8/(35/7 + 3) = 7
7 + 8/(8 + 3) = 7
7 + 8/11 = 7
7 + 8/11 = 7
This equation is true, so x = 5/7 is a valid solution.
For x = -3:
5/(-3) + 8/(-3 + 3) = 7
-5/3 + 8/0 = 7
The expression 8/0 is undefined, so x = -3 is an extraneous solution.
Therefore, the only valid solution to the original rational equation 5/x + 8/(x+3) = 7 is x = 5/7.
(5(x+3) + 8x) / x(x+3) = 7
Simplifying the numerator, we have:
(5x + 15 + 8x) / x(x+3) = 7
(13x + 15) / x(x+3) = 7
Next, we can multiply both sides of the equation by the denominator x(x+3) to get rid of the fraction:
13x + 15 = 7x(x+3)
Expanding the right side, we have:
13x + 15 = 7x^2 + 21x
Rearranging the equation to bring all terms to one side, we have:
7x^2 + 21x - 13x - 15 = 0
7x^2 + 8x - 15 = 0
Now, we can factor the quadratic equation:
(7x - 5)(x + 3) = 0
Setting each factor equal to zero, we have two possible solutions:
7x - 5 = 0, which gives x = 5/7
x + 3 = 0, which gives x = -3
Now we need to check if these solutions are valid or if they are extraneous. We substitute each solution back into the original equation:
For x = 5/7:
5/(5/7) + 8/(5/7 + 3) = 7
5 * 7/5 + 8/(35/7 + 3) = 7
7 + 8/(35/7 + 3) = 7
7 + 8/(8 + 3) = 7
7 + 8/11 = 7
7 + 8/11 = 7
This equation is true, so x = 5/7 is a valid solution.
For x = -3:
5/(-3) + 8/(-3 + 3) = 7
-5/3 + 8/0 = 7
The expression 8/0 is undefined, so x = -3 is an extraneous solution.
Therefore, the only valid solution to the original rational equation 5/x + 8/(x+3) = 7 is x = 5/7.
Answered by
GPT 3.5
To solve the rational equation (4(x-3))/(36-x^2) = (1)/(6-x) + (2x)/(6+x), we can follow these steps:
Step 1: Simplify both sides of the equation.
On the left side, we can simplify the expression by factoring the denominator. The denominator 36 - x^2 can be factored as (6 - x)(6 + x). Therefore, the equation becomes:
(4(x-3))/((6 - x)(6 + x)) = (1)/(6-x) + (2x)/(6+x).
Step 2: Find a common denominator.
The common denominator for the fractions on the right side is (6 - x)(6 + x). We can rewrite the equation with the common denominator:
(4(x-3))/((6 - x)(6 + x)) = ((1)(6+x))/((6-x)(6+x)) + ((2x)(6-x))/((6+x)(6-x)).
Step 3: Simplify and combine like terms.
The equation becomes:
(4(x-3))/((6 - x)(6 + x)) = (6+x + 12x - 2x^2)/((6+x)(6-x)).
Step 4: Multiply both sides of the equation by the common denominator.
By multiplying both sides by (6 - x)(6 + x), we can eliminate the denominators:
(4(x-3)) = (6+x + 12x - 2x^2).
Step 5: Simplify and bring all terms to one side.
Expand and simplify the right side:
4x - 12 = 18x - 2x^2.
Rearrange the equation:
2x^2 - 14x + 12 = 0.
Step 6: Factor the quadratic equation.
To factor the quadratic equation 2x^2 - 14x + 12 = 0, we can divide each term by 2 to simplify the equation:
x^2 - 7x + 6 = 0.
Factor the simplified equation:
(x - 1)(x - 6) = 0.
Step 7: Set each factor equal to zero and solve for x.
Set x - 1 = 0, which gives x = 1.
Set x - 6 = 0, which gives x = 6.
Step 8: Check for extraneous solutions.
We need to check if these solutions are valid or if they are extraneous by substituting them back into the original equation.
For x = 1, substituting into the original equation gives:
(4(1-3))/(36-1^2) = (1)/(6-1) + (2(1))/(6+1).
(-4)/35 = 1/5 + 2/7.
The equation is not true, so x = 1 is an extraneous solution.
For x = 6, substituting into the original equation gives:
(4(6-3))/(36-6^2) = (1)/(6-6) + (2(6))/(6+6).
(4)(3)/(36-36) = 1/12 + 12/12.
12/0 is undefined, so x = 6 is also an extraneous solution.
Therefore, there are no valid solutions to the equation.
Step 1: Simplify both sides of the equation.
On the left side, we can simplify the expression by factoring the denominator. The denominator 36 - x^2 can be factored as (6 - x)(6 + x). Therefore, the equation becomes:
(4(x-3))/((6 - x)(6 + x)) = (1)/(6-x) + (2x)/(6+x).
Step 2: Find a common denominator.
The common denominator for the fractions on the right side is (6 - x)(6 + x). We can rewrite the equation with the common denominator:
(4(x-3))/((6 - x)(6 + x)) = ((1)(6+x))/((6-x)(6+x)) + ((2x)(6-x))/((6+x)(6-x)).
Step 3: Simplify and combine like terms.
The equation becomes:
(4(x-3))/((6 - x)(6 + x)) = (6+x + 12x - 2x^2)/((6+x)(6-x)).
Step 4: Multiply both sides of the equation by the common denominator.
By multiplying both sides by (6 - x)(6 + x), we can eliminate the denominators:
(4(x-3)) = (6+x + 12x - 2x^2).
Step 5: Simplify and bring all terms to one side.
Expand and simplify the right side:
4x - 12 = 18x - 2x^2.
Rearrange the equation:
2x^2 - 14x + 12 = 0.
Step 6: Factor the quadratic equation.
To factor the quadratic equation 2x^2 - 14x + 12 = 0, we can divide each term by 2 to simplify the equation:
x^2 - 7x + 6 = 0.
Factor the simplified equation:
(x - 1)(x - 6) = 0.
Step 7: Set each factor equal to zero and solve for x.
Set x - 1 = 0, which gives x = 1.
Set x - 6 = 0, which gives x = 6.
Step 8: Check for extraneous solutions.
We need to check if these solutions are valid or if they are extraneous by substituting them back into the original equation.
For x = 1, substituting into the original equation gives:
(4(1-3))/(36-1^2) = (1)/(6-1) + (2(1))/(6+1).
(-4)/35 = 1/5 + 2/7.
The equation is not true, so x = 1 is an extraneous solution.
For x = 6, substituting into the original equation gives:
(4(6-3))/(36-6^2) = (1)/(6-6) + (2(6))/(6+6).
(4)(3)/(36-36) = 1/12 + 12/12.
12/0 is undefined, so x = 6 is also an extraneous solution.
Therefore, there are no valid solutions to the equation.
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