a=2a/(a-5)
How do I work this out?
In the division problem, the a in the numerator and the a in the denominator cancel each other.
a = 2/-5
a = -(2/5)
a=2a/(a-5)
*note: don't cancel the variable a (though it can be canceled since a can be factored from the left side and right side of equation),,
a^2 - 5a = 2a
a^2 - 7a = 0
a = 0 and a = 7
*notice that if you cancel a, equation becomes:
1=2/(a-5)
a - 5 = 2
a = 7
..in which you did not get the root a = 0
so there,, :)
To work out this equation, "a=2a/(a-5)", you must follow these steps:
1. Simplify the equation by multiplying both sides of the equation by the denominator, (a-5), to eliminate the fraction. This leads to: (a-5)*a = 2a.
2. Expand the equation: a^2 - 5a = 2a.
3. Move all the terms to one side of the equation by subtracting 2a from both sides: a^2 - 5a - 2a = 0.
4. Combine like terms: a^2 - 7a = 0.
5. Factor the equation, if possible: a(a - 7) = 0.
6. Set each factor equal to zero: a = 0 or a - 7 = 0.
7. Solve for 'a' in each equation: a = 0 or a = 7.
Therefore, the solutions to the equation are a = 0 or a = 7.