Question

What is the solution to the following system?

Answer 1

Answer:

(4,3,2)

Step-by-step explanation:

We can solve this via matrices, so the equations given can be written in matrix form as:

$\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]$

Now I will shift rows to make my pivot point (top left) a 1 and so:

$\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]$

Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,

-2R1+R2=R2 , -3R1+R3=R3

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]$

$\frac{R_2}{9}=R_2$

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]$

4R2+R1=R1 , -14R2+R3=R3

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]$

$-\frac{9}{20}R_3=R_3$

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]$

$-\frac{4}{9}R_3+R_2=R2$ , $-\frac{7}{9}R_3+R_1=R_1$

$\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]$

Therefore the solution to the system of equations are (x,y,z) = (4,3,2)

Note: If answer choices are given, plug them in and see if you get what is "equal to".  Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.