solving rational equation
trying to figure out where i did wrong
x/x-4 -4/x-5 = 4/x^2-9x +20
x/x-4 -4/x-5= 4/(x-4)(x-5)
domain x not equal +- 4,5
x(x-4)(x-5)/x-4 - 4(x-4)(x-5)/x-5= 4(x-4)(x-5)/(x-4)(x-5)
x(x-5)-4(x-4)=4
x^2-5x-4x+16=4
x^2-9x+16=4
x^2-9x+16-4=4
x^2-9x+16-4=0
x^2-9x+12=0
(x-2)(x-6)
x^2-9x+12=0***right so far
(x-2)(x-6) ***recheck this. What two numbers multiply to 12 and add to 9?
1 x 12
2 x 6
3 x 4
it would be none of these
The world is not made of whole numbers. Use the quadratic equation.
And, recheck the math. I checked it twice, and that was the only error I could find.
Y=7x-16, -7x+y=3 are the lines perpendicular, parallel or niether
To solve the rational equation:
x/(x-4) - 4/(x-5) = 4/(x^2-9x+20)
First, find the least common denominator (LCD) of the fractions on the left side, which is (x-4)(x-5) since both denominators are different.
Rewrite the equation with the common denominator:
x(x-5)/(x-4)(x-5) - 4(x-4)/(x-4)(x-5) = 4/(x^2-9x+20)
Now, simplify the equation:
[x(x-5) - 4(x-4)]/(x-4)(x-5) = 4/(x^2-9x+20)
Expand and combine like terms:
[x^2 - 5x - 4x + 16] / (x-4)(x-5) = 4/(x^2-9x+20)
Simplify further:
(x^2 - 9x + 16) / (x-4)(x-5) = 4/(x^2-9x+20)
To solve for x, cross multiply:
(x^2 - 9x + 16) * (x^2-9x+20) = 4 * (x-4)(x-5)
Expand the equation:
(x^2 - 9x + 16)(x^2 - 9x + 20) = 4(x^2 - 9x + 36)
Now, simplify this quadratic equation:
x^4 - 18x^3 + 82x^2 - 570x + 320 = 4x^2 - 36x + 144
Combine like terms:
x^4 - 18x^3 + 78x^2 - 534x + 176 = 0
Now, we need to solve this equation. You can use factoring, the quadratic formula, or other methods to solve for x. In this case, it looks like you've already factored it correctly:
(x-2)(x-6) = 0
This gives two possible solutions for x: x = 2 and x = 6.
Note: It's always a good practice to check for extraneous solutions by plugging the found values back into the original equation to ensure they satisfy the domain restrictions and don't create any undefined solutions.