In Exercises 1–26, solve and check each linear equation.25 - [2 +...

Question

In Exercises 1–26, solve and check each linear equation.25 - [2 + 5y - 3(y + 2)] = - 3(2y - 5) - [5(y - 1) - 3y + 3]

Accepted Answer

Hi there. So this problem says we want to solve the given linear equation. So I'm first gonna rewrite this equation. So let's go ahead and see 10 minus six plus two, X minus four X plus two equals negative five two x - minus two two x -3 -1. Now we have a couple of distributive properties we can use here. So look here you have a negative four times X plus two. You can go ahead. And this should be those. We also have a negative five time to two X -4 vetoes. And we also have a two times X -3 which you can distribute these as well. Let's go ahead and rewrite with some things that should be taken care of. We have 10 minus bracket six plus two X. Have a negative four times X which is negative four X negative four times two which is negative eight. Then we go to the other side negative five times two X is negative 10 x negative five times negative four is positive 20. We have a minus two times X X two X two times negative three is negative six minus X plus one. Now that we have that we can go ahead and distribute our negative signs. So if you look here we have a negative in front of this set of brackets and a negative in front of this set of brackets. So then distribute those next make it minus six minus two X plus a four X plus an eight. That's equal to negative 10 x plus 20 minus two X plus six plus x minus one. Now. Because and try and combine our like terms on each side. So if you look on the left side we have a 10 we can take minus six and eight plus eight. Does none of those have Xs on them? We can combine those. We have 10 minus six plus 8. 10 minus six is four. 4-plus 8 is 12. So we have a 12 here. When we combine all of our numbers then we have a negative two X plus a four X. Well negative two plus four is positive two X. So we have 12 plus two X. On the left side. The right side we have 8 Plus A 6 -1. Well 20-plus 6 is 26 -1 is 25. Now we also have our excess, we have negative 10 minus two plus one X. So you have negative 10 X minus two X. Is negative 12 negative 12 X plus one X. Is negative 11 X. Now we've certified that we go ahead and try and solve for X. So we want to move all of our excess on one side. All of our numbers on the other side that aren't exes. So I'm going to move my exes on the left side. Also do that. We already have a two X over here. We have an 11 X. On this other side. Go ahead and add 11 x to this set. And the second do that. That was canceled, meaning we get 12 plus two plus 11 which is X equals 25. Now I can go ahead and move our 12 next the same match. 12. Let me do that. R. 12 and R. 12 go away until we get 13 X. In 25 minus 12 which is 13. Finally We have 13 x equals 13. We can divide both sides by 13. We did that. I get 13/13 on this side. We just get X. 13/13 on the right side. We get one. So the solution to the equation is x equals one which is answer A. Okay thank you for watching. Help you solve the problem. Goodbye.

Struggling with College Algebra?

In Exercises 1–26, solve and check each linear equation.25 - [2 + 5y - 3(y + 2)] = - 3(2y - 5) - [5(y - 1) - 3y + 3]

Watch next

Rational Equations practice set