# Thread: Solving systems of equations

1. ## Solving systems of equations

Hi,

I'm looking for help with solving these two systems of equations (any method can be used):

1.

-4x - 5y - 8z = 268
-10x + 7y + 6z = -786
-12x + 9y - 2z = -548

2.

-5x - 4y + 12z = -140
9x - 8y + z = -649
3x - 2y + 6z = -290

2. What method have you chosen? Where are you stuck?

3. Originally Posted by stapel
What method have you chosen? Where are you stuck?

I'm pretty sure what I used was substitution when trying to solve them. Here's my work so far for number 2:

-5x - 4y + 12z = -140
9x - 8y + z = -649
3x - 2y + 6z = -290

I multiplied row 1 by -1

5x + 4y - 12z = 140

Then I multiplied row 3 by 2

6x - 4y + 12z = -580

11x = -440, which means x = -40

What I did next was substitute -40 in for x in one of the equations,

9(-40) – 8y + z = -649
-360 – 8y + z = -649
-8y + z = - 289
z = 8y – 289

is what I got. From there I'm a little unsure as to how I should continue.

4. Once you get past two equations in two variables, it's often better to add equations together. For instance, you have:

$\displaystyle R_1:\, -5x\, -\, 4y\, +\, 12z\, =\, 140$

$\displaystyle R_2:\, 9x\, -\, 8y\, +\, 1z\, =\, -649$

$\displaystyle R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290$

I see right away that I can multiply $\displaystyle R_3$ by $\displaystyle -3$ and add this to $\displaystyle R_2$ to get:

$\displaystyle R_1:\, -5x\, -\, 4y\, +\, 12z\, =\, 140$

$\displaystyle R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221$

$\displaystyle R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290$

Because I really would rather leave fractions until the end, and since twice three is close to five, I'll multiply $\displaystyle R_3$ by $\displaystyle 2$ and add the result to $\displaystyle R_1$. This gives me:

$\displaystyle R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440$

$\displaystyle R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221$

$\displaystyle R_3:\, 3x\, -\, 2y\, +\, 6z\, =\, -290$

Now I'll multiply $\displaystyle R_1$ by $\displaystyle -3$ and add the result to $\displaystyle R_3$. This gives me:

$\displaystyle R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440$

$\displaystyle R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221$

$\displaystyle R_3:\, 0x\, +\, 22y\, -\, 66z\, =\, 1030$

For simplicity's sake, I'll divide $\displaystyle R_3$ by $\displaystyle 2$ to get:

$\displaystyle R_1:\, 1x\, -\, 8y\, +\, 24z\, =\, -440$

$\displaystyle R_2:\, 0x\, -\, 3y\, -\, 17z\, =\, 221$

$\displaystyle R_3:\, 0x\, +\, 11y\, -\, 33z\, =\, 515$

The last two rows are now a system of two equations in two unknowns. Solve that, plug the values for $\displaystyle y$ and $\displaystyle z$ into the first equation, and simplify to find the value of $\displaystyle x$.