Could you help me with a few algebra 1 problems please :)?

1. You substitute whatever you isolated.

For example:

2x + 6y = 18

x + 3y = 9

x = -3y + 9

Now substitute this value for every x.

3x - 2y = 6

3(-3y + 9) - 2y = 6

-9y + 9 - 2y = 6

-11y = -3

11y = 3

y = 3/11

Now substitute this value into any equation:

2x + 6y = 18

2x + 6(3/11) = 18

2x + 18/11 = 18

22x + 18 = 11(18)

22x = 10(18)

22x = 180

11x = 90

x = 90/11

(90/11, 3/11)

Just out of boredom, I'll answer question 3 as well:

x^2 - 3x + 1 = 0

Substitute into the quadratic formula:

x = (-b +or- sqrt(b^2 - 4ac)) / 2a

where a = 1, b = -3 and c = 1

x = (3 +or- sqrt(9 - 4)) / 2

x = 3/2 +or- sqrt(5) / 2

since sqrt(5) < 3, it is true that both solutions are positive and both are irrational.

1. 2x + 6y = 18

2x = 18 - 6y

x = (18 - 6y) / 2

x = 9 - 3y

9 - 3y is what you would substitute for x in the second equation

2. (10a^3 - 15a^2 + 25a) / -5a

-2a^2 + 3a - 5

3. x^2 - 3x + 1 = 0 has solutions that are positive, that, added together equal 3, and that are irrational numbers. I have this sinking feeling that none of these is the answer that will marked as correct.

I'm going to stop here before I hurt myself.

1. To answer this, we need to actually isolate x in the first equation.:) So, let's subtract 6y from both sides:

2x = 18 - 6y

Now divide both sides by 2:

x = 9 - 3y

So, (9-3y) is what you would substitute, instead of x, into the second equation.

2. Ok. To do this, you need to understand what powers are. Read pages 25-32 (only 7 pages) of this book: http://books.google.com/books?id=Z9z7iliyFD0C&prin...

3. Does x2 mean "x squared"? Then this is a quadratic equation with a positive discriminant. Therefore, there are two real, distinct roots.

4. Same reading as for problem 2. Seeing someone else do the problem won't help learn this kind of material.