elimination method,please,answer,anyone with working brains,PLEASE?

Question

use the elimination method to solve the system:

{2x + 3y + 5z = 12
{4x + 2y + 4z = -2
{ 5x + 4y + 7z = 7


thanks! godbless!

Answer 1

I hope i'm not to late. (cus this response took me a while)lol.. I'll give you the answer, but because the answer has little long term value if you don't know how solve it on your own, (u know the whole give a man a fish he eats for a day, teach him to fish and he eats for a lifetime) so i'll do my best to make this look easy...lol.


First, the answer is:
x = -4
y = 5
z = 1

Now let me try to explain this...it's my first time helping someone with math on a computer and the first time helping someone with math period! But I think I can do a good job.


To make it easier for me to explain, i'll number each equation.


2x + 3y + 5z = 12.........(1)
4x + 2y + 4z = -2..........(2)
5x + 4y + 7z = 7...........(3)

First, pair up equations (1) and (2). Like this:
2x + 3y + 5z = 12........(1)
4x + 2y + 4z = -2.........(2)

Next, find a variable to eliminate. I'll choose the X variable since it looks like the easiest one to eliminate to me. So we must multiply equation (1) by -4 and then combine the two equations to get:
-4x - 6y - 10z = -24
4x +2y + 4z = -2
________________
- 4y - 6z = -26 ...........lets # this new equation (4)

Then go back to the original 3 equations and pair up equations (2) and (3) and eliminate the x variable in those two equations.
4x + 2y + 4z = -2..........(2)
5x + 4y + 7z = 7...........(3)

In order to eliminate the x variable in these two equations, it is easiest (for me) to multiply equation (2) by -5 and equation (3) by 4. Then we go ahead and combine the two equations to get:

-20x -10y - 20z = 10
20x+16y +28z = 28
__________________
6y + 8z = 38 .........lets # this new equation (5)


So far we created two new equations, (4) and (5), that don't have X variables by combining the original equations to cancel out the Xs. Lets pair up new equations (4) and (5)

- 4y - 6z = -26........(4)
6y + 8z = 38........(5)

Now we choose another variable to eliminate from (4) and (5). I choose to eliminate the y variable because it looks easier to me. To do that, we multiply equation (4) by 6 and equation (5) by 4 and then combine them to get:

-24y - 36z = -156
24y - 32z = 152
______________
-4z = -4

z = 4!

Okay so we found one variable so far. The easiest way for me to find the next variable is to combine equations (4) and (5) again.

- 4y - 6z = -26........(4)
6y + 8z = 38........(5)

Now instead of eliminating the y, we need to eliminate the z variables. To do that we multiply equation (4) by 8 and equation (5) by 6 and then combine to get:

-32y - 48z = -208
36y +48z = 228
______________
4y = 20

y = 5!

Okay so we found our second variable. Now since we have two variables found, all we do is plug them into one of the original equations. I choose to plug them into equation (1) because guess why.....because it just looks easier to me.

2x + 3*5 + 5*1 = 12
which becomes:
2x + 15 + 5 = 12
which becomes:
2x + 20 = 12
then subtract 20 from both sides to get:
2x = -8
x = -4!!!!!!


So all the variables are
x = -4
y = 5
z = 1

Answer 2

{2x + 3y + 5z = 12 | *(-2) then add to the second
{4x + 2y + 4z = -2 | *(-5) thenn add to the third *(4)
{ 5x + 4y + 7z = 7 | *4

-4y-6z=-26 |*3
6y+4z=34 |*2 then add

-10z=-10
z=1
y=(34-4)/6 = 5
x=(12-15-5)/2 = -4

x=-4, y=5, z=1

Answer 3

clean up via removing 2y + x = 3 ===> Eq. a million 4y - 3x = a million ====> Eq. 2 Multiply Eq. a million via 3 3( 2y + x = 3) = 6y + 3x = 9 ===> Eq. 3 upload Eq. 2 and Eq. 3 4y - 3x = a million ===> Eq. 2 + (6y + 3x = 9) ===> Eq. 3 ------------------------------------- 10 y = 10 y = 10/10 y = a million replace the fee of y into Eq. a million ( 2 or 3) to discover x 2(a million) + x = 3 2 + x = 3 x = 3 - 2 x = a million =========== to income x = a million, y = a million 2y + x = 3 ==> Eq. a million 2(a million) + a million = 3 2 + a million = 3 3 = 3 4y - 3x = a million ==> Eq. 2 4(a million) - 3(a million) = a million 4 - 3 = a million a million = a million

Answer 4

in elimination try this..
first get rid of 1st unknown "x" from 2nd and 3rd equations by subtracting the 1st equation from each one(2Xeq1-eq2)and(5Xeq1-2Xeq3)
after you have done this u'll have two equations with 2 unknowns (y and z)
do the same procedure to get rid of 2nd unknown "y" from eq3 by subtracting it from eq2(7Xeq2-4Xeq3 i think)..and you will get the answer for "z" which you'll sustitute in prvious equations to get other unknowns

Answer 5

4y+6z=26
x=11+z
4y-6x=92

solving them will giv:
x=-4
y=5
z=1