5 over 2x-5 minus 4 over 5= 1 over 2x-5
I have a quick question... what would i multiply by? I know how to do the rest of the problem. Thank you
Lets' start by writing the equation without words.
5/(2x-5) - 4/5 = 1/(2x-5)
Combine the two terms with 2x-5 in the denominator.
4/(2x-5) = 4/5
20 = 8x -20
8x = 40
x = 5
To solve this equation, we need to eliminate the denominators by finding a common denominator for all the fractions involved. In this case, the common denominator is (2x - 5)(5).
To get the expression 5 over 2x - 5 with a denominator of (2x - 5)(5), we need to multiply both the numerator and the denominator by 5. So the first fraction becomes:
(5 * 5) over (5 * (2x - 5)) = 25 over (10x - 25)
To get the expression 4 over 5 with a denominator of (2x - 5)(5), we need to multiply both the numerator and the denominator by (2x - 5). So the second fraction becomes:
(4 * (2x - 5)) over (5 * (2x - 5)) = (8x - 20) over (10x - 25)
Now, our equation becomes:
25 over (10x - 25) - (8x - 20) over (10x - 25) = 1 over (2x - 5)
Since the denominators are now the same, we can combine the numerators on the left-hand side:
(25 - (8x - 20)) over (10x - 25) = 1 over (2x - 5)
Simplifying further:
(25 - 8x + 20) over (10x - 25) = 1 over (2x - 5)
(45 - 8x) over (10x - 25) = 1 over (2x - 5)
Now that we have eliminated the denominators, we have a single fraction equation. We can cross-multiply:
(45 - 8x) * (2x - 5) = 1 * (10x - 25)
(90x - 225 - 16x^2 + 40x) = 10x - 25
Rearranging the equation:
-16x^2 + 90x + 40x - 10x + 225 - 25 = 0
-16x^2 + 120x + 200 = 0
From here, you can solve the quadratic equation using factoring, the quadratic formula, or completing the square to find the values of x that satisfy the equation.