Solving a system of 4 simultaneous equations- why is my answer wrong?

[moogle12]

I'm not sure if I understand the technique correctly because multiple times with different problems I have found my answers to be wrong when solving larger sets of simultaneous equations. I just want to solve the equations written at the top of the page (excuse my poor handwriting). The solutions I obtained I know to be wrong and I know what the right answers are (w=2,x=1,y=-1,z=1). I want to know WHY I'm wrong and for someone to explain what it is I should be doing to solve these- this is semi-urgent as while the specific problem I have posted is arbitrary, I need to be able to solve simultaneous equations like this consistently and I need to learn to do this quickly, as this is something I'm expected to be able to do while at university, and it isn't being taught- this knowledge is assumed. As such I need to urgently figure out where I'm going wrong so I can avoid making such mistakes in future.

Thanks.

 

[skeeter]

Yes, your handwriting is poor ... try typing out the system of equations to help someone help you.

 

[moogle12]

Yes, your handwriting is poor ... try typing out the system of equations to help someone help you.

x+z+2w=6

y-2z=-3

x+2y-z=-2

2x+y+3z-2w=0

I hope that's a little clearer. I tried to solve it with substitution, I'm aware either elimination or substitution should work but for some reason when I solve equations like this it just doesn't work for me. I'm unsure why this is the case- if anyone can decipher my handwriting and point out my flawed technique it would be appreciated, but otherwise an explanation of exactly how to go about solving the equations I just typed out would be appreciated.

Thanks!

 

[skeeter]

(1) 2w + x + 0y + z = 6
(2) 0w + 0x + y - 2z = -3
(3) 0w + x + 2y - z = -2
(4) -2w + 2x + y + 3z = 0

from equation (2), y = 2z - 3
substitute for y in equation (3) ...

0w + x + 2(2z - 3) - z = -2
x + 4z - 6 - z = -2
x + 3z = 4
x = 4 - 3z

add equations (1) and (4) to eliminate w ...

3x + y + 4z = 6

substitute for x and y ...

3(4 - 3z) + (2z - 3) + 4z = 6
12 - 9z + 2z - 3 + 4z = 6
-3z + 9 = 6
-3z = -3
z = 1

you should be able to finish up

 

[Otis]

Hi. When adding 6 to the right-hand side, you'd forgotten the -2 there.

x = 4 - 3z

?

 

[moogle12]

Solved it. Yeah, I just made a series of silly mistakes over and over- I get simultaneous equations just fine and doubted myself. Thanks guys!

 

[lookagain]

. . . (excuse my poor handwriting).

For many people a handwritten "2" and "z" look pretty close. You may want to write
a "z" with a horizontal cross-bar to more readily distinguish it from all other characters.

 

[Otis]

made a series of silly mistakes over and over

Train yourself to mentally check steps, as you go. With some involved exercises, it's even a good idea to occasionally substitute dummy values for variables and evaluate, to ensure that newly-formed equation(s) are still true. When allowed, a graphing calculator is a good time-saving tool for verifying certain written steps/results.

 

[Steven G]

You say that you have the solution.
To find your 1st error, plug in the values for for the variables and see if you get equality.

For example, if the solution is x=2 and y=7 and you have an equation x+2y = 16 you know it is a valid equation since (2) + 2(7) = 16.
However you you have the equation 5x-y=2 you know it is wrong since 5(2) - (7) = 3 not 2.

This technique above will catch ALL your errors if you have the solution