please help!! 1. 3(y-5) + 25 = 3y + 10 2. -5s - 14 = 2(2s+3) - 9s

Hi there!

1.
[tex]3(y - 5) + 25 = 3y + 10[/tex]
First work out the parenthesis.

[tex]3y - 15 + 25 = 3y + 10[/tex]
Now collect terms.

[tex]3y + 10 = 3y + 10[/tex]
Subtract 10 from both sides.

[tex]3y = 3y[/tex]
As you can see, both sides of the equation are the same. Therefore, it does not matter whatever number you fill in for the variable y. Since both sides are expressed by 3y, filling in a number will always make the equation true. All values of y are a solution.

2.
[tex] - 5s - 14 = 2(2s + 3) - 9s[/tex]
First work out the parenthesis.

[tex] - 5s - 14 = 4s + 6 - 9s[/tex]
Now collect terms.

[tex] - 5s - 14 = - 5s + 6[/tex]
Add 14 to both sides of the equation.

[tex] - 5s = - 5s + 20[/tex]
Now we can conclude that this equation does not have any solutions, because the left side of the equation will always be 20 bigger. There are no values of s that make the equation true.

~ Hope this helps you!