Math Help - A "system of equations" question and....

1. A "system of equations" question and....

First my math question....

I am asked to solve this

$2x+5y=-2$
$15y=-6-6x$

I followed these examples Systems of Equations - Solve by addition or subtraction and thus this occurred...

$6(2x+5y=-2)$
$2(6x+15y=-6)$

$12x+10y=12$
$(-1)(12x+30y=-12)$

$12x+10y=-12$
$-12x-30y=12$

$20y=0$

$\frac{20y}{20}=\frac{0}{20}$
cancel the $20$s and I'm left with
$y=0$

Now to solve for $x$

I plug in $y=o$ into the second equation $6x+15y=-6$

$6x+15(0)=-6$
$6x=-6$

$\frac{6x}{6}=\frac{-6}{6}$
cancel...

and $x=-1$

Ok now that I have the $x$s and the $y$s I plug those into the first equation....right so far?!

$2(-1)+5(0)=-2$

$-2=-2$

Did I do this right?! if not, then where did I go wrong, I have a hard time with these types of equations because I can never figure out elimination/addition nor graphing!!

NOW FOR MY SECOND QUESTION what has happened to the Message: options that used to be above the text box!?

Many thanks!!!

2. Its right, using the subtitution method I get the same answer.

3. Originally Posted by Hoopy
Its right, using the subtitution method I get the same answer.

hmmmm.....ok then so then could it possibly be said that it has multiple answers!? or just one!?

Thanks

4. Originally Posted by Morzilla
hmmmm.....ok then so then could it possibly be said that it has multiple answers!? or just one!?

Thanks
you have linear equations. there are three possibilities: it can have no solution (that is, the lines don't intersect), it can have one solution (that is, the lines cross paths once), or it can have an infinitely many solutions (that is, both equations represent the same line, so they are equal at all points on either line)

5. Originally Posted by Jhevon
you have linear equations. there are three possibilities: it can have no solution (that is, the lines don't intersect), it can have one solution (that is, the lines cross paths once), or it can have an infinitely many solutions (that is, both equations represent the same line, so they are equal at all points on either line)

Yes I remember about the 3 solutions, according to the answer key it says that it has Infinite number of solutions. Now by this does the problem i presented have an infinite number of solutions seeing how the answer is -2=-2 ?!

thanks once again!!!

6. Originally Posted by Morzilla
Yes I remember about the 3 solutions, according to the answer key it says that it has Infinite number of solutions. Now by this does the problem i presented have an infinite number of solutions seeing how the answer is -2=-2 ?!

thanks once again!!!
no. you have infinitely many solutions when one line is a constant times the other. that is the case here. note that the second line is 3 times the first. thus without doing any calculations, we know that we will have an infinite number of solutions ....(by the way, you made your life way too complicated by multiplying the first equation by 6 and the second by 2, rethink this strategy)

7. Originally Posted by Jhevon
no. you have infinitely many solutions when one line is a constant times the other. that is the case here. note that the second line is 3 times the first. thus without doing any calculations, we know that we will have an infinite number of solutions ....(by the way, you made your life way too complicated by multiplying the first equation by 6 and the second by 2, rethink this strategy)
Let me see if I understand this right. When you say that the second line is 3 time the first you mean that the first equation multiplied by 3 will give me as a product the second equation, i.e (5y)(3)=15y and (2x)(3)=6x ?!

So then I would use substitution to solve this problem instead of elimination/addition?

Thanks

8. Originally Posted by Morzilla
Let me see if I understand this right. When you say that the second line is 3 time the first you mean that the first equation multiplied by 3 will give me as a product the second equation, i.e (5y)(3)=15y and (2x)(3)=6x ?!

So then I would use substitution to solve this problem instead of elimination/addition?

Thanks
if you multiply each term in the first equation by 3, you get exactly the second equation. thus the are lines that lie on top of each other, there are infinitely many solutions. for the math you're doing, i don't think you have to do anything once you realize that. just state that it has infinitely many solutions and move on. finding formulas that gives you the solutions is more a topic for linear algebra