I can't remember how to continue?

[4smiles]

a+5=-5a+5
-4a=5=5
-4 -4

The 2nd 4 ^^^ belongs under the 5 closest to the end, it just won't stay
I'm am trying to find A
Well actually i know what A is because my teacher gave us the answers to check them, but i forgot how to solve equations over summer please help?
A=0

 

[BigGlenntheHeavy]

\(\displaystyle a+5 \ = \ 5a+5, \ solve \ for \ a.\)

\(\displaystyle a-a+5 \ = \ 5a-a+5\)

\(\displaystyle 5 \ = \ 4a+5\)

\(\displaystyle 5-5 \ = \ 4a+5-5\)

\(\displaystyle 0 \ = \ 4a\)

\(\displaystyle \frac{0}{4} \ = \ \frac{4a}{4}\)

\(\displaystyle 0 \ = \ a\)

 

[Mrspi]

a+5=-5a+5
-4a=5=5
-4 -4

The 2nd 4 ^^^ belongs under the 5 closest to the end, it just won't stay
I'm am trying to find A
Well actually i know what A is because my teacher gave us the answers to check them, but i forgot how to solve equations over summer please help?
A=0

a + 5 = -5a + 5

Let's eliminate the variable from one side of the equation. You can ADD 5a to both sides; that will eliminate the term containing "a" from the right-hand side.

a + 5 + 5a = -5a + 5 + 5a

Simplify by combining like terms:

6a + 5 = 5

Now, we want the term containing the variable to be BY ITSELF, so we need to eliminate the " + 5" from the left-hand side of the equation. Add -5 to both sides to accomplish this:

6a + 5 + (-5) = 5 + (-5)

Again, simplify by combining like terms:

6a = 0

Finally, we want 1a, or just "a" on the left side. But we have 6 times a....

To "undo" that multiplication, divide both sides by 6:

(6a) / 6 = 0 / 6

a = 0