Chase2001

Note, the written (f) above is off though the equation used later is correct. Likewise the 'z' written at the end should be labeled as y.

 

Plugging your numbers back into the equations results in:

 

(14 - 1 + 3)/16 = 16/16 = 1(Correct)

 

(28 + 1 + 3)/16 = 32/16 = 2(Correct)

 

(7 - 1 - 2(3))/16 = 0/16 = 0(Correct)

 

Your answer is correct, the labeling issue I suspect occurred during transposition. All in all a nice bit of work.

Thanks

Solve the following system of linear equations.

a) [latex]2x-y+z=1[/latex]

b) [latex]4x+y+z=2[/latex]

c) [latex]x-y-2z=0[/latex]

The y's cancel in equations a and b to give e) [latex]6x+2z=3[/latex]

The y's cancel in equations b and c to give f) [latex]5x-z=3[/latex]

I now multiply f by 2 and add it to equation e to eliminate z giving me [latex]16x=7 \Rightarrow x=\frac{7}{16}[/latex]

Now I plug [latex]\frac{7}{16}[/latex] back into equation f giving me [latex]5\left(\frac{7}{16}\right)-z=2 \Rightarrow \frac{35}{16}-2=z \Rightarrow z=\frac{3}{16}[/latex]

Then finally I plug x and z back into equation b giving me [latex]4\left(\frac{7}{16}\right)+\frac{3}{16}+z=2 \Rightarrow \frac{35}{16}+\frac{3}{16}+z=2 \Rightarrow \frac{31}{16}-2=-z \Rightarrow z=\frac{1}{16}[/latex]

I find it difficult to keep track of these long winded questions so I took the time to learn latex so it should be easier for you guys to read and in turn easier to help me if need be.

Just out of curiosity what are linear equations used for in the real world? I would quite like to be a physicist or a structural engineer when I grow up

A little judicious examination of the coefficients in the equations can reduce the work and increase the accuracy of this method.

 

equ1.jpg

Thanks this makes it a bit clearer. Much faster than my method. Direct substituation works for all linear equations right?

There is at least one mistake in the calculations you made (I stopped reading at some point). But the approach is correct.

Oh yeh I see my error but the approach is ok which is what I wanted to know. It just gets a bit confusing because I end up with so many equations on my page lol

Hello everyone. My name is Chase and I'm here to get some help with my math. I have the following problem. Solve the following system of linear equations:

1) 2x-3y+z=0

2) x+y+z=1

3) x-2y-4z=2

From what I have learnt in my book. I can take two of the equations and get rid of a variable, then I need to take two more equations and get rid of the same variable. This is what I have so far.

I take equation 1 and 3 and multiply the first by 4

1) 8x-12y+4z=0

3) x-2y-4z=8

This gives me 9x+14y=9

Now I take equations 2 and 3 and multiply equation 2 by 4.

2) 4x+4y+4z=4

3) x-2y-4z=2

This gives me 5x+2y=6. Now if I understand I now need to take these 2 new equations and solve for x or y? I guess i'll multiply the second equation by 7 to get rid of the y. Then i'll change the signs to make the subtraction

9x+14y=9

-35x-14y=-42

This gives me

-26x=-33 which is x=33/26 so now I have found the value of x which I can plug back into the equations to find the other variables but I'm not sure if this is even correct? It seems very long just to get the answers.