1/2=2/5c-3, I do not get fractions
What do you need to do? Solve for c?
yes, solve for c
Tell you what. If you don't like fractions, we'll just get rid of them. How? Here is how.
1/2 = 2/5(c) -3
multiply both sides by 2 (to get rid of the 1/2).
2*(1/2) = 2*(2/5)c - 2*3
1 = 4/5(c) - 6
Now multiply both sides by 5 (to get rid of the 5).
5*1 = 5((4/5)(c) - 5*6
5 = 4c - 30
can you take it from here?
To solve this equation, let's first get rid of the fractions. We can do this by multiplying both sides of the equation by the denominator of each fraction. In this case, the denominators are 2 and 5c.
Multiplying both sides of the equation by 2 and 5c, we get:
2 * (1/2) = (2/5c) * 2 - 3 * 2
The fraction on the left-hand side simplifies to 1, and on the right-hand side, the 2 in the numerator cancels out with the 2 in the denominator. This leaves us with:
1 = 4/5c - 6
Next, let's isolate the variable term by moving the constant term to the other side of the equation. Add 6 to both sides:
1 + 6 = 4/5c - 6 + 6
7 = 4/5c
Now, to solve for c, we need to isolate the variable. Let's multiply both sides of the equation by the reciprocal of 4/5, which is 5/4:
7 * (5/4) = (4/5c) * (5/4)
Simplifying, we get:
35/4 = 1c
Or, written as a fraction:
c = 35/4
So, the solution to the equation is c = 35/4.