Math help again, please

[JacksonsMom]

Grrrr. I've seemingly breezed through the last few chapters and now for some reason I'm stuck on something that SHOULD be simple.

Why can't I get this? can anyone explain this to me a simpler way than my MathLab program is?

Here is the question...

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I'm on attempt 2/3 on this quiz-thing, and if I don't get an 80%, it's not going to let me move on which is a pain because then I have to wait for the teacher (who is super cool) to give me another attempt, but I need to be ready to take a test tomorrow and I've been working on this ALL day.

My brain is freaking fried.

[Laurelin]

What I would do is put each equation into slope intercept form (y=mx+b where m =slope and b= y intercept) and then graph each. Start with the y intercept then just count the slope.

The solution will be where the two lines intersect. If the two lines are the same (ie: the two equations are the same) they have infinite solutions. If the lines are parallel and never cross then they have zero solutions.

[JacksonsMom]

Ok, because this program has us doing these charts, and each problem has it's own x/y chart with 3 different (seemingly random to me) numbers, that they're plugging in for x and y to graph it. Unless that was for a different kind of problem *sigh*

I don't know...

So I'm gonna try it this way.

[CaliTerp07]

A system of equations is just like it sounds--multiple equations. Your goal is to find that magic point (or points, in some cases), where the two equations intersect.

In order to graph them, you have a couple options (I'm not sure how you learned it). The most common way is to put the equations into slope-intercept form (y=mx+b). Basically, solve for y. Then m tells you the slope, and b is the y-intercept.

Eq 1:
2x-3y=-6
-3y=-2x-6
y=2/3x+2

So for the first equation, the line crosses the y-axis at 2, and the slope is 2/3 (so you'd move up 2, right 3 to find the next point).

Do the same for line two, and look to see where they intersect.

The second way to graph them is to find the intercepts. If y is 0 (plug in 0 for y), what value of x would make the equation true? That's your x-intercept, put a point there. Do the same if x is zero to find the y-intercept. Connect your two intercepts, and you'll see the line.

Do you have a graphing calculator? If you have a TI (82/83/84), you can enter both equations into the y= menu once you have them in slope intercept form, hit "second" then "calc" and go to "intersect". It will calculate the exact coordinate pair for you.

Finally...if you have learned other methods of solving systems, I would use elimination/combination to solve this one. Your x terms both have the same coefficient, so it's easy to subtract one equation from the other and solve for y. That's only if the computer doesn't care if you don't graph it though.

[Laurelin]

Ex: First equation

2x - 3y = -6
-2x -2x
_______________
-3y = -6 -2x

divide by -3

y = (2/3)x +2

y intercept is (0,2) and the slope is 2/3. So when you're graphing start at (0,2) then count up two and right three. Do the same thing with the second equation.

[CaliTerp07]

Originally Posted by JacksonsMom

Ok, because this program has us doing these charts, and each problem has it's own x/y chart with 3 different (seemingly random to me) numbers, that they're plugging in for x and y to graph it. Unless that was for a different kind of problem *sigh*

I don't know...

So I'm gonna try it this way.

So the computer is just having you plot 3 points that are on the line, and then connect them. That works too--it's just slightly more "guess and check" strategy.

[JessLough]

Solve both equations for y. Then graph using y=mx+b, m is the slope and b is the y intercept.

If the lines intercept, that is the solution. If they don't, there is no solution. If they are the same, there is infinite number of solutions

Edit: I got distracted and you got answers

[JacksonsMom]

Originally Posted by CaliTerp07

So the computer is just having you plot 3 points that are on the line, and then connect them. That works too--it's just slightly more "guess and check" strategy.

Yes, and it takes me forever, LOL

I like the way you guys posted. THANK YOU!

Lifesaver.

[JacksonsMom]

Ok, so for the second part of the equation...

2x-y=2
turns into
y= -2x+2 ???

so my slope is -2? or do I do -2/1? and my y-intercept is (0,2).

[Laurelin]

Originally Posted by JacksonsMom

Ok, so for the second part of the equation...

2x-y=2
turns into
y= -2x+2 ???

so my slope is -2? or do I do -2/1? and my y-intercept is (0,2).

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.