Question

Use the row operations tool to solve the following system of equations, obtaining the solutions in fraction form.

12x + 2y + z = 4
3x + 3y - 4z = 5
2x - 2y + 4z = 1
Give the values for x, y, and z with the fractions reduced to lowest terms (for example 4/7 rather than 8/14).
x = ____
y = ____
z = ____

Answer 1

`x=(45)/(4),  y=-(201)/(4), z=-(61)/(2)`

Explanation:

We start by putting our equation in a matricial form:

`\left[\begin{array}{cccc}12&2&1&4\\3&3&-4&5\\2&-2&4&1\end{array}\right]`

Then, we multiply the second row by 4 and substract the first row:

`\left[\begin{array}{cccc}12&2&1&4\\0&10&-17&16\\2&-2&4&1\end{array}\right]`

Now, multiply the third row by 6 and substract the first row:

`\left[\begin{array}{cccc}12&2&1&4\\0&10&-17&16\\0&-14&23&2\end{array}\right]`

Next, we will add
`(7)/(5)` times the second row to the third row:

`\left[\begin{array}{cccc}12&2&1&4\\0&10&-17&16\\0&0&(-4)/(5)&(122)/(5)\end{array}\right]`

Now we can solve
`(-4)/(5) z=(122)/(5)` to obtain

`z=-(61)/(2)`

Then
`10y-17(-61)/(2)=16` wich implies that

`y=(16-(17*61)/(2))/(10) =((32-17*61)/(2))/(10)=(-1005)/(20)=(-201)/(4)`

Finally

`x=(4-2*(-201)/(4)+(61)/(2))/(12) =((8+201+61)/(2))/(12)=(270)/(24)=(135)/(12)=(45)/(4)`.

`z=-(61)/(2)\\ y=-(201)/(4) \\x=(45)/(4)`