Question

2p+5q-r=25
2p-2q-r=-24
3p-q+5r=4

Answer 1

are those separate questions?
and are you solving for p and q? @Amandasteele

Answer 2

It's all one question and yes solving for both p and q

Answer 3

take first 2 equation

Answer 4

we need p and q

Answer 5

so subtract first 2 from each other and tell me what u got

Answer 6

tell me what u got ??

Answer 7

u will get value of q from here

Answer 8

subtract and post what u got ??

Answer 9

To solve this kind of simultaneous equation, you make one of the variables of the first equation as the subject of the formula. Once you find it, replace such variable in the remaining 2 equations.Thirdly, you make one of the 2 remaining variables as the subject of the formula to replace it in the last equation and surely you will find the value of one of the variables.
Fourthly, you will move backwards, by going to the previous equation to replace the found variable to find the other one and repeat this with the next previous equation and you shall get the value as well. Mathematically I am saying:

Answer 10

\[25p+5p-25=r\]

Answer 11

Replacing it in the 2 other equations...\[2p-2q-2p-5q+25+24=0\]

Answer 12

and \[3p-q+5(2p+5q-25)-4=0\]

Answer 13

can we use a matrix?

Answer 14

If you work out the equation \[2p-2q-2p-5q+25+24=0\], q=7, which you will replace it in the last equation, of \[3p-7+5(2p+5*7-25)-4=0\], which simplifies to \[-3p-7+10p+50-4=0\] and \[p=-\frac{ 39 }{ 7 }\].

Answer 15

Now that we have P and Q, let us replace it in the very first equation where we made r the subject.
\[2(-\frac{ 39 }{ 7 })+5(7)-25=r\] and r=-8/7

Answer 16

There. All the variables were found.

Answer 17

@UnkleRhaukus , I am not familiar with the matrix method that finds 3 variables of polynomials.

Answer 18

\[2p+5q-r=\ \ \ 25\\
2p-2q-r=-24\\
3p-q+5r=\ \ \ \ \ 4\]
\[\begin{bmatrix}2&5&-1\\2&-2&-1\\3&-1&5\end{bmatrix}
\begin{bmatrix}p\\q\\r\end{bmatrix}
=\begin{bmatrix}\ \ \ 25\\-24\\\ \ \ \ \ 4\end{bmatrix}\]
\[\left[\begin{array}{rrr|r}2&5&-1&25\\2&-2&-1&-24\\3&-1&5&4\end{array}\right]\]

Answer 19

\[R_1\to R_3-R_1\]\[\left[\begin{array}{rrr|r}1&-6&6&-21\\2&-2&-1&-24\\3&-1&5&4\end{array}\right]\]

\[R_2\to R_2-2R_1\]
\[\left[\begin{array}{rrr|r}1&-6&6&-21\\0&10&-13&18\\3&-1&5&4\end{array}\right]\]

Answer 20

I think I recall a bit. It looks like Gaussian method.

Answer 21

\[R_2\to R_2/10\]\[R_3\to R_3-3R_1\]\[\left[\begin{array}{rrr|r}1&-6&6&-21\\0&1&-1.3&1.8\\0&17&-13&67\end{array}\right]\]

Answer 22

\[R_1\to R_1+6R_2\]\[R_3\to R_3-17R_2\]\[\left[\begin{array}{rrr|r}1&0&-1.8&10.2\\0&1&-1.3&1.8\\0&0&9.1&36.4\end{array}\right]\]

Answer 23

\[R_3\to R_3/9.1\]\[\left[\begin{array}{rrr|r}1&0&-1.8&10.2\\0&1&-1.3&1.8\\0&0&1&4\end{array}\right]\]
\[R_1\to R_1+1.8R_3\\
R_2\to R_2+1.3 R_3\]\[\left[\begin{array}{rrr|r}1&0&0&17.4\\0&1&0&7\\0&0&1&4\end{array}\right]\]
\[\begin{bmatrix}p\\q\\r\end{bmatrix}
=\begin{bmatrix}17.4\\ 7 \\4\end{bmatrix}\]
hmm `17.4`, i probably made an error, lets check the result anyway...

Answer 24

What we had in common was q.

Answer 25

\[2p+5q-r=\ \ \ 25\\
2p-2q-r=-24\\
3p-q+5r=\ \ \ \ \ 4\\~\\\qquad\qquad\downarrow\\~\\~
2(17.4)+5(7)-(4)=65.8\\
2(17.4)-2(7)-(4)=16.8\\
3(17.4)-(7)+5(4)=65.2\] errors \(\ddot \frown\)

Answer 26

i bet p does not equal 17.4

\[3p-(7)-5(4)=4\]

\[3p=4+7-20=-9\\
p=-3\]

Answer 27

Oh my ...
We have to retry this.
I am restarting now...

Answer 28

do these values solve the system?

\[2p+5q-r=\ \ \ 25\\
2p-2q-r=-24\\
3p-q+5r=\ \ \ \ \ 4\\~\\\qquad\qquad\downarrow\\~\\~
2(-3)+5(7)-(4)=25\\
2(-3)-2(7)-(4)=-24\\
3(-3)-(7)+5(4)=4\]

Answer 29

B
I
N
G
O
!

Answer 30

Right, good!

Answer 31

did you get your method to work @Hoslos ?

Answer 32

Haha. You know? I re-did it and found new values. when plotted in the first 2 equations it went correct, but in the third one did not. Strange. it is my first time missing this method... Maybe because it is a 3-term variable expression, whereas I used this method a lot with 4-term expressions.

Answer 33

i found my error
...
\[R_1\to R_1+6R_2\\\left[\begin{array}{rrr|r}1&0&-1.8&\color{red} -10.2\\0&1&-1.3&1.8\\0&0&9.1&36.4\end{array}\right]\]
...

\[R_1\to R_1+1.8R_3\\
\left[\begin{array}{rrr|r}1&0&0&-3\\0&1&0&7\\0&0&1&4\end{array}\right]\]