x=(c-b)/a where ax+b=c and the quadratic formula.

Quadratic equations are characterized by this equation: ax ² + bx + c = 0 where a, b, and c are all real numbers not equaling zero. These are called coefficients, quadratic coefficients, to be precise. In order to have a quadratic, you must have something that looks like that equation, or something that can be arranged to look like that equation. If you just have one variable with no powers, then you have a standard algebraic equation. E.g. 2x ² + 4x + 3 = 0 <--quadratic; 2x + 4 = 3x - 1 <-- standard algebraic. Hope that answers your first question--if not, please restate your question again so I can better understand and answer it.

Now, all the equations you listed were standard algebraic equations. I'm not sure how you learned how to solve them, but I always learned to get the numbers with the variables attached to them(x, t, y, etc.) on one side of the equation, and the normal numbers on another. Let's take that first question that you got wrong as an example:

3x+4= 4x+3

First, let's subtract 3x from both sides to get the "x's" on one side. 3x's on the left cancel and we're left with: 4 = x + 3 (Subtract 3x from 4x)

From there, we subtract three from both sides, leaving us with 1 = x...and there you have it! 1 = x. If you plug 1 back in for x, you'll see that it lines up.

We'll do one more, and then I'll let you do the rest by yourself, using this method.

-2(x-3)=-12

First, we want to eliminate any parentheses from the equation. That's pretty much a general rule for algebraic equations. So, multiplying -2 throughout the parentheses, we get: -2x + 6 = -12

Now, let's subtract 6 from both sides to get the number with x attached to it by itself. We end up with: -2x = -18

From here, we divide both sides by - 2 to get x = ___ Working it out, we find that x = 9 (-18/-2 = 9 [two negatives cancel, leaving us with a positive])

Thus, x = 9. Once again, you will find that the answer works with 9 as x.

So, if you identify your equation as a standard algebraic equation, then go ahead and try to get all the numbers with variables attached on one side and the numbers without variables on the other. From there, it's pretty easy to solve for your lone variable. Hope that helps! I suggest you try this out a couple of times to get the hang of it. If it doesn't work, just repost and I"ll try to help you further!

Best of luck in you algebraic pursuits!

Warm Regards,
Grammar Fascist

Grammar Fascist has given you some really good input.
I'll just take a look and see if I can help too

Stu:

Regarding the maths
x = (c-b)/a where ax+b = c (what is the name for this formula?) ..
You should not remember this as an equation. You have simply rearranged ax+b=c and you need to know how to do it not just memorize the answer. Grammar Fascist has given you some really good pointers for this.
and
Quadratic formula where ax²+-bx+-c=0 => x = (-b+-Sqrt( b2 -4ac))/2a Those brackets need to be added.
This might help you remember the quadratic formula. Remember, the equation must be of the form ax²+bx+c = 0

http://www.youtube.com/watch?v=O8ezDEk3qCg

Give me as much help as you can. Specially identifying what is able to be made into an equation for a quadratic formula and what is to be left as a std x= c-b/a.
So for the following questions.. the difficulty I have is with the sign values of a/b/c and transposition of terms. I get how to transpose. but when working backwards/ slotting the answer I get in as x, it is often not correct. So I'd like a little help with steps.

1)
3x-8=1
x= 1- (-8) / 3
x= 1+8 /3
x= 9/3
x= 3
3*3-8=1 (correct.)
1)
3x-8=1
You want the lots of x by themselves so you need to add the 8, but if you do it to one side you must do it to the other side as well. The equation must always 'balance'
3x-8+8=1+8
3x=9
Now you need to get rid of the 3 so divide both sides by 3
3x/3=9/3
x=3
This is how it should be done.
2)
7t-5= 4t+7
7t-5-4t= +7
7t-4t-5= +7
3t-5= +7
Forget your next 3 steps do it the same as the last one
x= (7- (-5))/3
x=12/3
x= 4
7*4-5= 4*4+7
28-5= 16+7
23= 23 (correct)

3x+4= 4x+3
3x-4x+4=3
-1x+4=3
Good so far now do it the same as the one i showed you
x= (3- (+4)/-1
x= (3-4)/-1
x= -1/-1 = -1?
(3*-1)+4= (4*-1)+3
-3+4= -4+3
1= -1 (incorrect)

-3x+4?=4x+3
-3x-4x+4= +3
(-7x)+4 ≠ to +3 --> to move the +4 from the front of equation to the back of the equation do I need to alter its symbol?
Now i am lost
Start from here
3x+7= 7x+2
3x-7x+7= +2
-4x+7= +2
Forget the next 3 lines do it the other way
x= 2-(+7)/-4
x= 2-7/-4
x= -5/-4
(3/1*5/4)+7/1= (7/1*5/4)+2/1
(3*5/4*1)+7/1=(7*5/4*1)+2/1
15+28/4= 35+8/4
43/4= 43/4 (correct)
Please show simpler way to the above~ thanks.

(x+7)= 7*(x+2)
(x+7)= 7x+14 --> can't work out steps to make equation. >>
x+7 = 7x + 14 The brackets had no added meaning
Now work it the same as you did for the others

Try cleaning up everything you have done and represent a tidied up version. Thanks Stu.
(x+7)= (7x+14)
(x+7) (-7x+14) =0
(x+7) -7(x-2) =0 --> yep just can't get it.
-2(x-3)=-12
-2x+6= -12
x= -12-6/2
x=-18/2
x=9
-2*9-3= -12 (not correct.)


or<-2x+6+12=0 >
2x+18=0 ( I can't see how it works, even if 2*-9 or 2*+8 we can not +18 to get 0 unless x is = to 0 its self but which in my calculation I couldn't solve to be 0.

3(2t+1)= 4(t+2)


5m-3= 5(m-3) +2m


17(x-2)+3(x-1)=x+24 (please do entire maths expanded through all steps.)

Yes, you raised two key points. One was that two halves of a fraction at negative make a positive overall fraction. This was a question I was wondering about but didn't state.
Second was that this is easy, but I was trying to fit to the equation x=c-b/a, and trying to get it to look like ax+-b=+-c. when doing so, I guess a and b, could be either number value or x term values, and it wouldn't matter. That is 2t+5t=15 would fit in the equation the same as 7+8=7t, a,b,c respectively to both terms.

Thanks again for a good explanation.

At melody. The post has been answered now and very well. I think there is not a real need to repost my comment. All that has been answered is enough that anyone who wants to conclude a way to work these problems can solve them, identify them and move on, irrelevant of my post. Which a repost imo would prove to messy up the thread a little.

Stu:

Yes, you raised two key points. One was that two halves of a fraction at negative make a positive overall fraction. This was a question I was wondering about but didn't state.
Second was that this is easy, but I was trying to fit to the equation x=c-b/a, and trying to get it to look like ax+-b=+-c. when doing so, I guess a and b, could be either number value or x term values, and it wouldn't matter. That is 2t+5t=15 would fit in the equation the same as 7+8=7t, a,b,c respectively to both terms.

Thanks again for a good explanation.

You are welcome Stu.