1. ## Linear Equations

It turns out that I need to have these questions done by 6:30 (EST) so Im moving it to the Urgent Thread, ( I have to go to my school for a class event and my teacher wants me to come in for the test due to her being absent on monday). Sorry for the thread switch, Thanks in advance!
Here are the questions:
1)
2x - 3y = 5
4y + 2z = -6
5x + 7z = -15
Solve for X Y and Z using substitution or elimination.
EDIT :I Recently tried to use a homework help online service who gave me the answer ( I didnt understand their work though) For this they got x=4 y=1 and z=-5

2)
2x + 3y - z = 9
x - 3y + z = -6
3x + y - 4z = 31
Solve for X Y and Z using substitution or elimination. (Same as before)

3)
6/x - 4/y = 1
-3/x + 3/y = 2
Solve for X and Y (6/x, 4/y , -3/x and 3/y and fractions) using substitution
EDIT: Same homework help (Once again Didn't Understand what the heck their work meant but..) They Got for X=6/11 and Y=2/5

2. 1) We have the following equations:

$\displaystyle 2x - 3y = 5$

$\displaystyle 4y + 2z = -6$

$\displaystyle 5x + 7z = -15$

Notice that the final two equations allow for x and y to be given in terms of z, all we do is solve the 2nd equation for y and the 3rd one for x:

$\displaystyle 4y + 2z = -6$

$\displaystyle 4y = -6 - 2z$

$\displaystyle y = -\frac{3}{2} - \frac{1}{2}z$

$\displaystyle 5x + 7z = -15$

$\displaystyle 5x = -15 - 7z$

$\displaystyle x = -3 - \frac{7}{5}z$

Now, we take the 1st equation and plug in what we know for x and y:

$\displaystyle 2x - 3y = 5$

$\displaystyle 2(-3 - \frac{7}{5}z) - 3(-\frac{3}{2} - \frac{1}{2}z) = 5$

$\displaystyle -6 - \frac{14}{5}z + \frac{9}{2} + \frac{3}{2}z = 5$

$\displaystyle \frac{9}{2} - \frac{12}{2} - \frac{14}{5}z + \frac{3}{2}z = 5$

$\displaystyle -\frac{3}{2} - \frac{28}{10}z + \frac{15}{10}z = 5$

$\displaystyle -\frac{3}{2} - \frac{13}{10}z = 5$

$\displaystyle -\frac{13}{10}z = \frac{10}{2} + \frac{3}{2}$

$\displaystyle -\frac{13}{10}z = \frac{13}{2}$

$\displaystyle z = \frac{\frac{13}{2}}{-\frac{13}{10}}$

$\displaystyle z = \frac{13}{2} *-\frac{10}{13}$

$\displaystyle z = -\frac{10}{2} = -5$

Now we sub in the z-value:

$\displaystyle 4y + 2z = -6$

$\displaystyle 4y - 10 = -6$

$\displaystyle 4y = 4$

$\displaystyle y = 1$

$\displaystyle 5x + 7z = -15$

$\displaystyle 5x - 35 = -15$

$\displaystyle 5x = 20$

$\displaystyle x = 4$

And there you go.

3. Originally Posted by ChrisBarrett
It turns out that I need to have these questions done by 6:30 (EST) so Im moving it to the Urgent Thread, ( I have to go to my school for a class event and my teacher wants me to come in for the test due to her being absent on monday). Sorry for the thread switch, Thanks in advance!
Here are the questions:

(1) 2x - 3y = 5
(2) 4y + 2z = -6
(3) 5x + 7z = -15
Solve for X Y and Z using substitution or elimination.
EDIT :I Recently tried to use a homework help online service who gave me the answer ( I didnt understand their work though) For this they got x=4 y=1 and z=-5
Well, it's almost 6:30 somewhere, but it may be too late for you.
Multiply (1) by 5 and multiply (3) by -2. Add the equations to eliminate x.

$\displaystyle 10x-15y=25$ [1]

$\displaystyle -10x-14z=30$ [3]
--------------------------------------
$\displaystyle 15y-14z=55$ [4]

$\displaystyle 28y+14z=-42$ [5] Multiply (2) by 7 and add to [4].
-------------------------------------
$\displaystyle 13y=13$
$\displaystyle \boxed{y=1}$

Use (1): $\displaystyle 2x-3(1)=5$
$\displaystyle 2x=8$
$\displaystyle \boxed{x=5}$

Use (2): $\displaystyle 4(1)+2z=-6$
$\displaystyle 2z=-10$
$\displaystyle \boxed{x=-5}$

I'll hold up solving the others to see if this was any help to you or if you still have questions. Good luck.