System of Equations

T

TiaharaJBennett

New member
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Jul 10, 2012
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I also would like some help with system of equations.

My first one is:

5x+4y=6
-2x_3y=-1

Completely forgot how to do these.
 
Denis said:
Better if you start a new thread with a new problem...
Assuming equations are (typo again?!):
5x + 4y = 6 [1]
-2x - 3y = -1 [2]

Many ways to solve; here's one:
from [1]:
4y = 6 - 5x
y = (6 - 5x) / 4

Substitute that in [2], then solve for x ; capish?
Click to expand...

Sorry, I still don't understand...
 
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I merged your duplicate threads into a single thread (this one).

Please do not post the same exercise more than once. It confuses readers (like you, for example), and it leads to extra work for the moderators.

Thank you! :cool:



The symbols [1] and [2] are simply names used to reference their respective equations.

In other words, the meaning of [1] is: Equation #1.

Likewise, [2] should be read as: Equation #2.

You may think of them as labels. We get tired of continually writing "equation #1" and "equation #2", so we write [1] and [2] instead.
 
TiaharaJBennett said:
Sorry, I still don't understand...
Click to expand...

This statement is too vague to be of much use.

Please begin explaining either (1) what you are thinking about or (2) specifically why you are stuck.

If you don't tell us, then we have to guess, and that usually wastes time.

Please check the FORUM GUIDELINES for more information about how to ask for help.

Cheers ~ Mark :cool:
 
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mmm4444bot said:
This statement is too vague to be of much use.

Please begin explaining either (1) what you are thinking about or (2) specifically why you are stuck.

If you don't tell us, then we have to guess, and that usually wastes time.

Please check the FORUM GUIDELINES for more information about how to ask for help.

Cheers ~ Mark :cool:
Click to expand...

How do I substitue [1] into [2]? That's the part I'm stuck on.
 
Denis said:
from [1]:

4y = 6 - 5x
y = (6 - 5x) / 4

Substitute that in [2]
Click to expand...

Here is the meaning.

Denis solved equation [1] for y above.

This tells us that we may express the number y using this algebraic fraction:

(6 - 5x)/4

Next, Denis' instruction "Substitute that in [2]" has the following meaning.

Rewrite equation [2], but instead of writing y write (6-5x)/4 in its place.

When you do this, you will end up with a new equation that contains only the symbol x. You may then solve that new equation to discover the value of x.

Once you know the value of x, you can calculate the value of y using (6-5x)/4.
 
Thank you for posting the lesson link, Subhotosh. I was just about to do that, along with the following note to the OP and readers.



NOTE: The FreeMathHelp boards are primarily a place for students who have some idea of what they're trying to do. We provide guidance based on the progress posted by the student.

Our boards do not comprise an on-line classroom because it's too difficult to teach broad topics in the back-and-forth setting of a bulletin board.

We are always happy to answer specific questions, and, when time permits, some "teaching". However, when posters find themselves totally lost on a particular exercise, then our boards are probably not the first place to start.

Please understand, if we refer you to lessons. Feel free to return after some studying and post specific questions about the lessons or show some work.

Thank you! :cool:
 
mmm4444bot said:
Thank you for posting the lesson link, Subhotosh. I was just about to do that, along with the following note to the OP and readers.



NOTE: The FreeMathHelp boards are primarily a place for students who have some idea of what they're trying to do. We provide guidance based on the progress posted by the student.

Our boards do not comprise an on-line classroom because it's too difficult to teach broad topics in the back-and-forth setting of a bulletin board.

We are always happy to answer specific questions, and, when time permits, some "teaching". However, when posters find themselves totally lost on a particular exercise, then our boards are probably not the first place to start.

Please understand, if we refer you to lessons. Feel free to return after some studying and post specific questions about the lessons or show some work.

Thank you! :cool:
Click to expand...

Okay. I'm sorry to bother you with all these questions.

Here's the work I did:

-2x-3(6-5x)/4=1

Added 3 to both sides

-2x/-2=4/-2

x=-2

Is this right?

PS - I'm really sorry, again. I should have read the guidelines first.
 
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TiaharaJBennett said:
I'm sorry to bother you with all these questions.
Click to expand...

Speaking for myself, questions do not bother me per se; it's the appearance of things like: not paying attention; an unwillingness to try harder for oneself; or succumbing for no good reason to some attitude of helplessness that bother me.

:!: I am not suggesting that you have exhibited all of these behaviors.

Having said that, did you study the lesson at the link provided to you by Subhotosh?

I'm really sorry, again.
Click to expand...

Please stop apologizing. Spend that energy on studying, instead. :cool:
 
mmm4444bot said:
Speaking for myself, questions do not bother me per se; it's the appearance of things like: not paying attention; an unwillingness to try harder for oneself; or succumbing for no good reason to some attitude of helplessness that bother me.

:!: I am not suggesting that you have exhibited all of these behaviors.

Having said that, did you study the lesson at the link provided to you by Subhotosh?
Click to expand...

I will. But I don't think it's the same as the problems I have. I don't have to graph it, I just have to solve it.
 
TiaharaJBennett said:
I don't think it's the same as the problems I have. I don't have to graph [my system of equations], I just have to solve [the system].
Click to expand...

It seems to me that you just "judged a book by its cover". It will serve you well in the future, if you begin training yourself now to not jump to conclusions before all of the facts are in.

Graphs appear in that lesson because graphing is a valid solution-method. Once you study the entire lesson, you will see the algebraic methods, too.

Additionally, you will gain a better perspective from studying the graphing section, too, because understanding your exercise from a graphical point-of-view provides a visual aid in your mind to help with the "big picture" when solving a system of two linear equations algebraically.

If you are serious about learning mathematics, you need to invest the time. People do not learn much, when their first choice is consistently relying on others to lead them by the hand.

Food for thought. Cheers :cool:
 
mmm4444bot said:
It seems to me that you just "judged a book by its cover". It will serve you well in the future, if you begin training yourself now to not jump to conclusions before all of the facts are in.

Graphs appear in that lesson because graphing is a valid solution-method. Once you study the entire lesson, you will see the algebraic methods, too.

Additionally, you will gain a better perspective from studying the graphing section, too, because understanding your exercise from a graphical point-of-view provides a visual aid in your mind to help with the "big picture" when solving a system of two linear equations algebraically.

If you are serious about learning mathematics, you need to invest the time. People do not learn much, when their first choice is consistently relying on others to lead them by the hand.

Food for thought. Cheers :cool:
Click to expand...

Okay, I'll read it through. Mathematics is really the only thing I have problems with in school. Always has been. And I do know that all this can be really easy if you just take the time to learn it. But I've always believed in my mind that it's just too hard and I can't do it. I really need to get over that. :rolleyes: But thank you so much for your help and patience. It is much appreciated.
 
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Denis said:
Capish?
Click to expand...

Sometimes French - sometimes Italian ..... where are you from .... must not be a cricket playing country (of course French/Italians don't play cricket).

By the way, I can cuss-out in seven different languages (including Turkish) - thanks to graduate school.
 
Denis said:
Hope your wife can do same...:roll:
Click to expand...

Nah.... she is mono-linguistic... I can understand all the cuss-words she can pronounce.

.
 
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Getting back to the original question, if that is permitted:

Rather than solve one of
5x+4y=6
-2x-3y=-1
for x or y and then substitute we can also
1) multiply the first equation by 2 to get 10x+ 8y= 12
2) multiply the seond equation by 5 to get -10x- 15y= -5
3) add those two equations to eliminate x and get -7y= 7 and divide both sides of that by 7 so that y= -1.
4) put that value of y back into either of the original equations to get either 5x+ 4(-1)= 5x- 4= 6 so, adding 4 to both sides 5x= 10 and then, dividing both sides by 5, x= 2, or -2x- 3(-1)= -2x+ 3= -1 so, subtracting 3 from both sides, -2x= -4, x= 2.

Or:
1) multiply the first equation by 3 to get 15x+ 12y= 18
2) multiply the second equation by 4 to get -8x- 12y= -4
3) add those two equations to eliminate y and get 7x= 14 and divide both sides of that by 7 to get x= 2.
4) put that value of x back into either of the original equations to get either 5(2)+ 4y= 10+ 4y= 6 so that, subtracting 10 from both sides 4y= -4 so that y= -1, or -2(2)- 3y= -4- 3y= -1 so that, adding 4 to both sides, -3y= 3 and, dividing both sides by -3, y= -1.

There are, in fact, many other ways to solve such equations, but however you do it, you will get x= 2, y= -1 and you can check those in the original
equations: with x= 2, y= -1,
5x+4y= 5(2)+ 4(-1)= 10- 4= 6
-2x-3y= -2(2)- 3(-1)= -4+ 3= -1
 
Denis said:
10 + 7 = 15 + 2
2*5 + 7 = 3*5 + 2
2x + 7 = 3x + 2

x has replaced the "5": see that?
That's the MOST basic thing to realise: a letter replaces a number.

Now you're given it to solve:
3x + 2 = 2x + 7: "solve that for x" says the teacher
Subtract 2x from BOTH sides:
3x - 2x + 2 = 2x - 2x + 7
x + 2 = 7
Subtract 2 from each side:
x + 2 - 2 = 7 - 2
x = 5
Capish?
Click to expand...

Yes! THANK YOU. :grin:
 
HallsofIvy said:
Getting back to the original question, if that is permitted:

Rather than solve one of
5x+4y=6
-2x-3y=-1
for x or y and then substitute we can also
1) multiply the first equation by 2 to get 10x+ 8y= 12
2) multiply the seond equation by 5 to get -10x- 15y= -5
3) add those two equations to eliminate x and get -7y= 7 and divide both sides of that by 7 so that y= -1.
4) put that value of y back into either of the original equations to get either 5x+ 4(-1)= 5x- 4= 6 so, adding 4 to both sides 5x= 10 and then, dividing both sides by 5, x= 2, or -2x- 3(-1)= -2x+ 3= -1 so, subtracting 3 from both sides, -2x= -4, x= 2.

Or:
1) multiply the first equation by 3 to get 15x+ 12y= 18
2) multiply the second equation by 4 to get -8x- 12y= -4
3) add those two equations to eliminate y and get 7x= 14 and divide both sides of that by 7 to get x= 2.
4) put that value of x back into either of the original equations to get either 5(2)+ 4y= 10+ 4y= 6 so that, subtracting 10 from both sides 4y= -4 so that y= -1, or -2(2)- 3y= -4- 3y= -1 so that, adding 4 to both sides, -3y= 3 and, dividing both sides by -3, y= -1.

There are, in fact, many other ways to solve such equations, but however you do it, you will get x= 2, y= -1 and you can check those in the original
equations: with x= 2, y= -1,
5x+4y= 5(2)+ 4(-1)= 10- 4= 6
-2x-3y= -2(2)- 3(-1)= -4+ 3= -1
Click to expand...

This helped me so much. Thank you lots!!!
 
TiaharaJBennett said:
HallsofIvy: Can I use this ^ for any system of equations?
Click to expand...

I'm not sure what you mean by "this". What I did was use the basic properties of algebra- which are really basic properties of numbers. And they are true for any equation or system of equations.
 
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