Need some aid solving this system of equations...

L

Ladybug

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Jul 11, 2007
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Okay, I have a problem that is perplexing me! :? I understand all of my math homework so far except this.

Here I'm told to solve this system of equations using the substitution method. It solves for the point where two lines on a graph intersect. The one I'm having trouble with is as follows:

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

So, I decide to get rid of all fractions by multiplying each equation by the LCD, which brings me to:

6(1/3x - 1/6x = -1) 2x-y=-1
2(3/2x + 1/2y = 1/2) 3x+y=1

If I just use the normal addition method, all the y's cross out and the right side ends up being zero. Thus I get both y and x = 0. But using the substitution method, I did this:

2x-y=-1 -> y=1-3x
3x+y=1 -> 3x+1-3x=1 1=1

Now what do I do there? And why does it look different then if I used the substitution method? Is this an identity problem, or a contradiction with the graph having parallel lines and the system inconsistent? The solution set would be empty then, right?

This is my first session with graphing in school, that's why I have so many questions. :p

Oh, and I forgot to add: I verified all the answers I could think of on my TI83 and none of them were correct for both equations.
 
Ladybug said:
Okay, I have a problem that is perplexing me! :? I understand all of my math homework so far except this.

Here I'm told to solve this system of equations using the substitution method. It solves for the point where two lines on a graph intersect. The one I'm having trouble with is as follows:

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

So, I decide to get rid of all fractions by multiplying each equation by the LCD, which brings me to:

6(1/3x - 1/6x = -1) 2x-y=-1
2(3/2x + 1/2y = 1/2) 3x+y=1

If I just use the normal addition method, all the y's cross out and the right side ends up being zero. Thus I get both y and x = 0. But using the substitution method, I did this:

2x-y=-1 -> y=1-3x
3x+y=1 -> 3x+1-3x=1 1=1

Now what do I do there? And why does it look different then if I used the substitution method? Is this an identity problem, or a contradiction with the graph having parallel lines and the system inconsistent? The solution set would be empty then, right?

This is my first session with graphing in school, that's why I have so many questions. :p

Oh, and I forgot to add: I verified all the answers I could think of on my TI83 and none of them were correct for both equations.
Click to expand...

Here's where your problem lies.

6(1/3 x - 1/6 y = -1)

This becomes
2x - y = -6
You forgot to multiply BOTH SIDES of the equation by 6.

And you made the same mistake here:
2(3/2 x + 1/2 y = 1)

This should be
3x + y = 2

So.....your correct system, with fractions eliminated, should be

2x - y = -6
3x + y = 2

Can you take it from there?
 
Mrspi said:
Here's where your problem lies.

6(1/3 x - 1/6 y = -1)

This becomes
2x - y = -6
You forgot to multiply BOTH SIDES of the equation by 6.

And you made the same mistake here:
2(3/2 x + 1/2 y = 1)

This should be
3x + y = 2

So.....your correct system, with fractions eliminated, should be

2x - y = -6
3x + y = 2

Can you take it from there?
Click to expand...

Oh dear. Something that small, and I spent so much time working on it. :( Thanks for pointing it out.

I'm figuring it out and still having trouble: with the substitution method, I got 10 and 4. I did it another way and got...whoa...-4/5 and -22/5! I plugged them both in my calc and none are right. This is escaping me! Usually it isn't this hard... :oops: Can you give me any further pointers? Here's what I worked out:

2x - y = -6
3x + y = 2 Easier to solve for y here, so I did y=3x-2. Then:

2x - 3x - 2 = -6 -> -x-2=-6 -x=-4 or x=4.

I plugged it back in: y=3(4) or 12 - 2. That got me the 10 and 4. -4/5 and -22/5 I got when I did y=2x+6 and plugged that in.

What am I screwing up here??
 
2x - y = -6
3x + y = 2

Solve the second equation for y.....

y = -3x + 2

Substitute (-3x + 2) for y in the first equation:

2x - (-3x + 2) = -6

2x + 3x - 2 = -6
5x - 2 = -6
5x = -4

x = -4/5

Now, you know that y = -3x + 2
y = -3(-4/5) + 2
y = 12/5 + 2
y = 12/5 + 10/5
y = 22/5

Looks like you need to be extra-careful with signs.
 
> 2x - y = -6
> 3x + y = 2 : I make that one 3x + y = 1 (typo, MrsPi?)

So 5x = -5 ; x = -1 ; y = 4
 
Denis said:
> 2x - y = -6
> 3x + y = 2 : I make that one 3x + y = 1 (typo, MrsPi?)

So 5x = -5 ; x = -1 ; y = 4
Click to expand...

No typo, Denis.

The original second equation was

(3/2)x + (1/2)y = 1

Multiply both sides of that by 2, and you have

3x + y = 2

Your answers are "nicer"......but I'll stand by mine.

Janet
 
Denis said:
> 2x - y = -6
> 3x + y = 2 : I make that one 3x + y = 1 (typo, MrsPi?)

So 5x = -5 ; x = -1 ; y = 4
Click to expand...

Hey, you got it right! -4/5 and 22/5 did not verify on my TI83. Plus I looked at the exercise helps in my book, and x=-1 while y=4. 2*1/2 is one, that's how you got to 3x+y=1, right? (The second original equation being 3/2x + 1/2y = 1/2)

Thanks for all your help guys!
 
1/3x - 1/6x = -1
3/2x + 1/2y = 1/2

Janet, that appears in 1st post; was there an "edit"? Oh Ladybug??
 
Denis said:
1/3x - 1/6x = -1
3/2x + 1/2y = 1/2

Janet, that appears in 1st post; was there an "edit"? Oh Ladybug??
Click to expand...

No, I didn't edit it. But nevermind, I got the question right for class, that's what matters. :wink:
 
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