i tried and tried and i cant find the answer. is it not a solution.
(3/x)-1=7/(x-20)
(3/x)-1 = 7/(x-20)
Looks nasty, I agree, but it's not as bad as it looks!
Multiply across by x(x-20) and you are left with
(3-x)(x-20) = 7x
which expands out into a neat quadratic with two friendly numbers as your solutions.
ahhh, I thought so the last time you posted that question.
so this time we multiply each term by x(x-2) to get
3(x-20 - x(x-20) = 7x
3x - 60 - x^2 + 20x = 7x
x^2 -16x + 60 = 0
(x-10)(x-6) = 0
x = 10 or x = 6
To find the solution to the equation (3/x) - 1 = 7/(x - 20), we can follow these steps:
Step 1: Clear the fractions by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD is x(x - 20).
(x(x - 20) * (3/x)) - (x(x - 20) * 1) = (x(x - 20) * 7/(x - 20))
Simplifying this equation, we get:
3(x - 20) - x(x - 20) = 7x
Step 2: Expand the equation by multiplying the terms inside the parentheses.
3x - 60 - x^2 + 20x = 7x
Combining like terms, we get:
2x^2 + 10x - 7x - 3x + 60 = 0
2x^2 + 0x + 60 = 0
Step 3: Rearrange the equation into the standard quadratic form, ax^2 + bx + c = 0, where a, b, and c are constants.
2x^2 + 0x + 60 = 0
Since there is no x term, b = 0.
The equation becomes:
2x^2 + 60 = 0
Step 4: Solve the quadratic equation. We can use the quadratic formula to find the values of x.
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 0, and c = 60.
Substituting these values into the quadratic formula:
x = (0 ± √(0 - 4*2*60)) / (2*2)
Simplifying further:
x = (±√(-480)) / 4
Since we cannot take the square root of a negative number in the real number system, it means that this equation has no real solutions.
Therefore, the equation (3/x) - 1 = 7/(x - 20) does not have a solution in the real number system.