Variables in Matrices

[lolily]

Hello, in my homework I came across this problem:



The text didn't really go over how to solve it and I'm finding it difficult to understand. Is there anyone who can explain how to get the answer? Thanks!!

 

[lolily]

To give some background on what I don't understand, I do know how to perform basic matrix operations, like addition/subtraction, multiplication, etc. I guessed at this by multiplying the two matrices by the numbers in front, 3 and -4, and then taking them out of the matrix to solve for the variables, but that got me the wrong answer.

 

[Dr.Peterson]

Please show your actual work, not just a general description of it.

What you describe sounds reasonable; so the error will be in the details you left out! Very likely it is just a sign error or something of that sort.

 

[pka]

Hello, in my homework I came across this problem:

View attachment 12028

The text didn't really go over how to solve it and I'm finding it difficult to understand. Is there anyone who can explain how to get the answer?

Here is a start:\(\displaystyle \left\{ \begin{array}{l}3(5a)-4(3a)=a+4\\3d-4(5d)=68\end{array} \right.\)
No you have two more equations to setup and then to solve the four.