solving systems of linear equations via substitutio method?

I'll work the first problem for you.

Step 1:

Solve one equation for either x or y. In this case the second equation (8x = -12y + 24) is the easiest; just divide both sides by 8 and we get x=-3/2y + 3.

Step 2:

Substitute the value of x that we just found in to the first equation (2x + 3y = 6); doing so we get 2(-3/2y + 3) + 3y = 6. Now simplifying we get:

2(-3/2y + 3) + 3y = 6

(-6/2y + 6) +3y = 6

(-3y + 6) + 3y = 6

-3y + 3y + 6 = 6

6 = 6

In this case we don't need to solve for y; most systems will require another step. There are two instances where this will happen, 1) We get a true statement, which is what we have here (6 does equal 6, right?) or 2) We get a false statement (perhaps, 6 = 3, which doesn't make sense). The first situation, and the answer to this system, tells us that we have infinitely many solutions. The second situation is an example of an answer to a system with no solutions.

Edit: I was going to post another problem, but all of your problems are similar. Let me know if you need more help.

Hi,

Having gone through college algebra recently, I found that youtube.com had lots of videos from math teachers. My favorite was the guy on yay math. If you type that in, all his videos will come up. He doesn't charge for his knowledge, and is the best at explaining, even to a dinosaur like me.

Most colleges have a tutoring center: make use of it. You paid for it.

I was working on a hybrid course which means I did all the work through a software program. IF that is the case, you will have resources there. And the last piece of advice I can give: find a high school student who would be willing to show you what they know. I did that, and enjoyed being taught the basics.

Good luck!

Having gone through college algebra recently, I found that youtube.com had lots of videos from math teachers.

:up:Youtube is a good resource. There's also Khan Academy for "how to" videos.

purple math is confusing to me too.