need help with 2 line gaussian elimination probs

caseyn2008

here are the problems i am disabled and on bed rest i do my school over the net and i have no books nor teachers and i have no idea how to do the 2 line ones.

-5x +y=11
-4x +5y=13

-x -y = 7
-3x +5y= -11

5x +y= -8
5x +2y= -6

i would just like some one to explain to me how to solve these and if i get stuck help me on one of them i have a lil less than 2 months till graduation.
3x +5y= 13
-5x-4y =-13


-x -2y= 9
3x -5y = - 16

5x+5y =0
3x +y =-10

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

-x +y = -2
-4x -4y =16


-x -5y = -1
2x +3y= -5

3x - 2y = 15
3x+ 3y= 0

LeixlipRed

To solve systems like that there is no need to even use Gaussian elimination. You simply add one multiple of one equation to another with the aim of cancelling a variable when you add the equations. For example in your first example if you multiply the first equation by minus 5 then when you add both equations the "y"s cancel. Then you can solve for "x". Can you make a stab at it from there?

caseyn2008

its been over a year since i done algebra, so if i had 4x +y = 0 i would get rid of y using x-1y right and x with x-4x?

LeixlipRed

Not exactly. Think of it this way. x is equal to some number. And when we talk about "cancelling a variable" we mean adding some number of multiples so we get zero. So what would you add to 5 to make 0? Minus five. I dont want to give it away too easy for you

caseyn2008

what do u mean by ad both equations? that will leave me with -4x +10y =24 answer ur ? -5
i added 5x to the first line and heres what i came up with its wrongs so tell me what i did wrong ill put up all the work i did

-5x +y=11
-4x +5y=13

x = -10.5 y = 11
52.5 + 11 = 11
i used a calcultor to do 13-55 which gave me -10.5

caseyn2008

what would be the easiest way to eliminate a varible from one of the lines? bc i did The simplest kind of linear system involves two equations and two variables:

\begin{alignat}{5} 2x &&\; + \;&& 3y &&\; = \;&& 6 & \\ 4x &&\; + \;&& 9y &&\; = \;&& 15 \end{alignat}

One method for solving such a system is as follows. First, solve the top equation for x in terms of y:

x = 3 - \frac{3}{2}y

Now substitute this expression for x into the bottom equation:

4\left( 3 - \frac{3}{2}y \right) + 9y = 15

that gave me a -31/25 which didnt end up working

LeixlipRed

caseyn2008 wrote: »

what do u mean by ad both equations? that will leave me with -4x +10y =24 answer ur ? -5
i added 5x to the first line and heres what i came up with its wrongs so tell me what i did wrong ill put up all the work i did

-5x +y=11
-4x +5y=13

x = -10.5 y = 11
52.5 + 11 = 11
i used a calcultor to do 13-55 which gave me -10.5

Ok, I think you need to step back a little. What you have here in each of these examples is two equations in two variables. So you have two lines on a flat page say and you wish to find where is their point of intersection (if they are parallel they wont have such a point). I can see you're having difficulties with the method so I'll do one example:

-5x +y=11
-4x +5y=13

Ok so the method I will use involved first multipying one equation on both sides by a number (if you don't understand why you can do this then just accept it on faith for the moment). You choose the number so that you can then add the equations together (again faith ) and one of your variables will disappear (ie 5 plus minus 5 equals zero as an example). Sometimes you will have to multiply both equation by a number to achieve this.

Ok I will multiply the first equation by minus 5.

So now i have

25x - 5y=-55
-4x + 5y=13

Now adding both equations I get

(25x -4x) + (-5y +5y) =-55 + 13

Which gives you:

21x = -42

Dividing both sides by 21 you get

x=-2

Now finally you return to one of your original equations and plug in x=-2 to find the corresponding value of y.

So taking the first eqn ( -5x +y=11):

-5(-2) + y = 11

which gives you:

y = 1 when you rearrange

Any help?

sd123

LeixlipRed wrote: »

Ok, I think you need to step back a little. What you have here in each of these examples is two equations in two variables. So you have two lines on a flat page say and you wish to find where is their point of intersection (if they are parallel they wont have such a point). I can see you're having difficulties with the method so I'll do one example:

-5x +y=11
-4x +5y=13

Ok so the method I will use involved first multipying one equation on both sides by a number (if you don't understand why you can do this then just accept it on faith for the moment). You choose the number so that you can then add the equations together (again faith ) and one of your variables will disappear (ie 5 plus minus 5 equals zero as an example). Sometimes you will have to multiply both equation by a number to achieve this.

Ok I will multiply the first equation by minus 5.

So now i have

25x - 5y=-55
-4x + 5y=13

Now adding both equations I get

(25x -4x) + (-5x +5x) =-55 + 13

Which gives you:

21x = -42

Dividing both sides by 21 you get

x=-2

Now finally you return to one of your original equations and plug in x=-2 to find the corresponding value of y.

So taking the first eqn ( -5x +y=11):

-5(-2) + y = 11

which gives you:

y = 1 when you rearrange

Any help?

You should turn those x's into y's, or you're going to really confuse him.

LeixlipRed

Haha, I knew in a post that long I'd mess up. Cheers

caseyn2008

i got x = -2 and y = 1 for the first one is that right?
i got that b4 i read your post but the 2nd one is giving me x= 2.5 and y = -4.5
i got it down now 2nd ones answer is x = -28 y = 35
im having trouble with the ones with co effecients and no common factors, also i cannot get #3 and 4 idk y my brother showed me how to do it but it just isnt working out for me now

caseyn2008

LeixlipRed wrote: »

Ok, I think you need to step back a little. What you have here in each of these examples is two equations in two variables. So you have two lines on a flat page say and you wish to find where is their point of intersection (if they are parallel they wont have such a point). I can see you're having difficulties with the method so I'll do one example:

-5x +y=11
-4x +5y=13

Ok so the method I will use involved first multipying one equation on both sides by a number (if you don't understand why you can do this then just accept it on faith for the moment). You choose the number so that you can then add the equations together (again faith ) and one of your variables will disappear (ie 5 plus minus 5 equals zero as an example). Sometimes you will have to multiply both equation by a number to achieve this.

Ok I will multiply the first equation by minus 5.

So now i have

25x - 5y=-55
-4x + 5y=13

Now adding both equations I get

(25x -4x) + (-5y +5y) =-55 + 13

Which gives you:

21x = -42

Dividing both sides by 21 you get

x=-2

Now finally you return to one of your original equations and plug in x=-2 to find the corresponding value of y.

So taking the first eqn ( -5x +y=11):

-5(-2) + y = 11

which gives you:

y = 1 when you rearrange

Any help?

ill try your method hopefully my meds arent making me make stupid mistakes

i used your method on 3rd equation and got this
5x +y= -8
5x +2y= -6

-5x -y = 8
5x + 2y = -6

y = 2
x = - 2

-10 + 4 = -6
checks out

caseyn2008

i can only solve a few of these are decimals and fractions a possibility? i hate them well the never ending ones anyways

i got this it all checks out except one the top line one is negative where it should be positive
-x -2y= 9
3x -5y = - 16

-3x -6y = 27
3x -5y = - 16

-11y = 11
y = -1
x = 7

-21 + 6 = -16
-7 -2

caseyn2008

(3x +5y= 13)5
(-5x-4y =-13)3
15x +25y = 65
-15x - 12y = -39
13y=26
y=2
x=1



i got 7 out of 10 answers right i need at least 8 right, so could someone point out my mistakes it could be in the 2nd one 3rd one or any in this post


(-x -2y= 9)3
3x -5y = - 16

-3x -6y = 27
3x -5y = - 16
-11y=11
y=-1
x=--7



5x+5y =0
(3x +y =-10)-5
5x+5y =0
-15x + -5y = -50

10x=-50
x=-5
y=5









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

-y=3
y=-3
x=-2






(-x +y = -2)4
-4x -4y =16
-4x +4y = -8
-4x -4y = 16
-8x=8
x=-1
y=-3











(-x -5y = -1)2
2x +3y= -5

-2x -10y = -2
2x + 3y =-5

-7y=-7
y=1
x=-4



(3x - 2y = 15)-1
3x+ 3y= 0

-3x -2y =-15
3x + 3y = 0
y=-15
x=15

hm i think i got down except for one

Crumbs

caseyn2008 wrote: »

i got this it all checks out except one the top line one is negative where it should be positive
-x -2y= 9
3x -5y = - 16

-3x -6y = 27
3x -5y = - 16

-11y = 11
y = -1
x = 7

-21 + 6 = -16
-7 -2

Should be x = -7 then it will check out.

caseyn2008 wrote: »

(3x - 2y = 15)-1
3x+ 3y= 0

-3x -2y =-15
3x + 3y = 0
y=-15
x=15

hm i think i got down except for one

The -2y should become +2y after multiplying across by -1.

caseyn2008

thank you o so much arnt you glad that i used a couple techniques to figure these out? the 2nd one i think i multiplied both lines or maybe that was the third one and the 1st one i did by making it y= yadda yadda yadda then i did the rest with his technique

LeixlipRed

If you need anymore help just ask. And fractions and decimals are a possibility. As I said, think of the answer as the point at which two lines meet. The point could take on any value be it a whole number/fraction/decimal. Though I have no idea if any of the examples above have nasty answers. And at 3:15 in the morning I'll pass on checking But reply here if you cant get a particular one out

caseyn2008

caseyn2008 wrote: »

here are the problems i am disabled and on bed rest i do my school over the net and i have no books nor teachers and i have no idea how to do the 2 line ones.

-5x +y=11
-4x +5y=13

-x -y = 7
-3x +5y= -11

5x +y= -8
5x +2y= -6

i would just like some one to explain to me how to solve these and if i get stuck help me on one of them i have a lil less than 2 months till graduation.
3x +5y= 13
-5x-4y =-13


-x -2y= 9
3x -5y = - 16

5x+5y =0
3x +y =-10

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

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




-x +y = -2
-4x -4y =16


-x -5y = -1
2x +3y= -5

(3x - 2y = 15)-1
3x+ 3y= 0

5y=-15y=-3
x=3