Beginning Algebra 1: How would you solve this: (5n+1)/8 = (3n-5)/4

How would you solve this: (5n+1)/8 = (3n-5)/4

cmtoland said:

How would you solve this: (5n+1)/8 = (3n-5)/4

Click to expand...

How would you solve this?

. . . . .a. 5n + 1 = 3n - 5

How would you solve this?

. . . . .b. 2(5n + 1) = 3n - 5

What do you get if you multiply your posted equation, on both sides, by the Lowest Common Denominator of the two fractions? ;-)

First we get rid of the denominators by multiplying both sides by 8. Now we get (5n+1)=2(3n-5).

We then expand everything and get 5n+1=6n-10,

this turns into... what?