Math help again, please

[JacksonsMom]

Originally Posted by Laurelin

2x-y = 2
-2x -2x (subtract 2x from both sides, the space looks funny for some reason when I post it)

-y = 2 -2x (now multiply by -1 on both sides)

y = 2x -2

Slope is 2 which is the same as 2/1. So when you count, your intercept is (0,-2) and then go up 2 and right one when you graph.

If that makes sense.

Gotcha. Ok so I screwed up on my negatives again lol. I was sorta close...

Ugh hate not having a math brain.

[JacksonsMom]

So my solution is (3,4) because that's where the lines intersect. I think.

[JessLough]

Originally Posted by JacksonsMom

So my solution is (0,4) because that's where the lines intersect. I think.

Test it by plugging those values into the equations

[Laurelin]

(0,4) doesn't work with either equation. One way to quick check is to take each equation and then plug in your answer for x (0) and y (4).

Ex:
2x-y =2

2(0)-4(4)=2

-16 =2, which is false.

Is there any way you can show us what you've graphed?

[JacksonsMom]

I mean 3 and 4, lol. just changed it.

[Laurelin]

Oh haha that makes a difference. Yeah that should be right.

[JacksonsMom]

I effed it up again... grr... got a 62.5%. There's only 8 questions on this quiz, so it sucks because even when you get two wrong, you're screwed. I have one more attempt :\ and if I fail, I'm stuck yet again and have like four more sections to do before I can do test #8. (I did test 7 yesterday).

Sucks, can't wait for this to be over and get a small break lol. and it's even noncredit *sigh* Don't know what I'll do in regular college algebra. Can't believe I put this off for so long. Just beyond frustrating because I work so hard at it, and feel like all I ever do nowadays is math homework, and I'll finally get the hang of one section and then we move on to another.

Class ends May 8th, and I'm there Tues & Thursdays, but I can go in there other days too to catch up but it's down to crunch time. there is 12 tests total.

I keep getting these same ones wrong/half wrong.




And the original one I posted.

[Laurelin]

On the first one I must be blind because I can't tell what you missed. (I'm really tired)

2nd one, do the slope intercept again. So make each equation y=mx+b

If the slope (m) is different then the lines intersect at one point. The system then only has one solution.

If the slope (m) is the same on both lines then they are either parallel or the same. If the intercept is the same (b) then the lines are the same. If the lines are the same then there is infinite solutions. If the intercept is different on the two equations then the lines are parallel and have zero solutions.

[JacksonsMom]

I think it may have changed the graph to the correct one, and not mine.
But thanks for all your help It's def helping me!

[Laurelin]

When making the equation into slope intercept form, just go SLOW and one step at a time. Whatever you do to the left side of the equation, you have to do to the right side.

Ex:

10x -y =10
(subtract 10x from both sides)
-y = 10-10x
(multiply both sides by -1)
y = -10+ 10x
(put into y form y=mx+b)
y = 10x+10

So the slope of this line is 10 and the y intercept is (0,10)

2nd equation:
(1/5)y=-2+2x
(divide by 1/5, which is the same thing as multiplying by 5)
y=-10+10x
(put into slope intercept form)
y= 10x-10

So the slope of this equation is 10 and the y intercept is (0,-10)

Since the slope is the same but the y intercepts are not, the two lines are parallel. They will never intersect so there is no solution to this set of equations.