Answer:
The solutions for the given system of equations are:

Step-by-step explanation:
Given the equation system:

We obtain the following matrix:
![\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%261%264%26-3%5C%5C-1%261%264%2617%5Cend%7Barray%7D%5Cright%5D)
<u>Step 1:</u> Multiply the fisrt row by 1/3.
![\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26%5Cfrac%7B1%7D%7B3%7D%20%26%5Cfrac%7B4%7D%7B3%7D%26-1%5C%5C-1%261%264%2617%5Cend%7Barray%7D%5Cright%5D)
<u>Step 2:</u> Sum the first row and the second row.
![\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26%5Cfrac%7B1%7D%7B3%7D%20%26%5Cfrac%7B4%7D%7B3%7D%26-1%5C%5C0%26%5Cfrac%7B4%7D%7B3%7D%20%26%5Cfrac%7B16%7D%7B3%7D%2616%5Cend%7Barray%7D%5Cright%5D)
<u>Step 3:</u> Multiply the second row by 3/4.
![\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26%5Cfrac%7B1%7D%7B3%7D%20%26%5Cfrac%7B4%7D%7B3%7D%26-1%5C%5C0%261%20%264%2612%5Cend%7Barray%7D%5Cright%5D)
<u>Step 4:</u> Multiply the second row by -1/3 and sum the the first row.
![\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%20%260%26-5%5C%5C0%261%20%264%2612%5Cend%7Barray%7D%5Cright%5D)
The result of the reduced matrix is:

This is equal to:

These are the solutions for the system of equations in terms of z, where z can be any number.
It is substitution. Because you are substituting 3=y into x+5=3
Answer:
943,281
Step-by-step explanation:
Just move the 3 up a value
If it's in the ten's place, then move it to the hundred's place value.
If it's in the hundred's place, then move it to thousand's place value.
So on and so on...
You have 20 candies together in abag 10 red 10 green tou mix them yp togethr in a bag how many candies do you have in the bag
G=4ca-3ba <=> G=a(4c-3b)
<=> a=G/(4c-3b)