Question

What is the solution to the rational equation..

Answer 1

\[\frac{ x }{ x^2-9 }+\frac{ 1 }{ x-3 }=\frac{ 1 }{ 4x-12 }\]

Answer 2

Hint:
first step, factorize the first denominator:
\(\frac{ x }{ (x+3)(x-3) }+\frac{ 1 }{ x-3 }=\frac{ 1 }{ 4x-12 }\)
and add the left-hand side to make one fraction to get:
\(\dfrac{ x + (x+3) }{ (x+3)(x-3) }=\dfrac{ 1 }{ 4x-12 }\)
\(\dfrac{ 2x+3 }{ (x+3)(x-3) }=\dfrac{ 1 }{ 4x-12 }\)
Now cross multiply and solve the resulting quadratic.
However, watch the denominators. All solutions giving x=3 or x=-3 must be rejected

Answer 3

In order to avoid a quadratic, multiply every term by 4x - 12 or 4(x-3)
you should get the equation 4 + (4x)/(x + 3) = 1

Answer 4

Combine the left side using the common denominator (x+3). Solve the linear equation by clearing the fraction by multiplying both sides by (x + 3).

Answer 5

After that is done here is what you have 4(2x+3)=x+3. Simplify and solve for x

Answer 6

Very true, we can take advantage of the fact that 4x-12 = 4(x-3) which is already a factor, so no need to make it into a quadratic.
\(\rm\frac{ x }{ x^2-9 }+\frac{ 1 }{ x-3 }=\frac{ 1 }{ 4x-12 }\)
\(\rm\frac{ 4x }{ 4(x^2-9) }+\frac{ 4(x+3) }{ 4(x^2-9) }=\frac{ x+3 }{ 4(x^2-9) }\)
Extract the numerators and solvefor x:
\(\rm 4x+(4x+12) = x+3\)

Answer 7

dot.