Answer:
<h3>Each of the given matrix equations does not represent this system of equations.</h3>
Step-by-step explanation:
![\left\{\begin{array}{ccc}2x-3=2y\\y-5x=14\end{array}\right\\\\\text{Let's convert the system of equations to form}\\\\\left\{\begin{array}{ccc}a_1x+b_1y=c_1\\a_2x+b_2y=c_2\end{array}\right\\\\\left\{\begin{array}{ccc}2x-3=2y&(1)\\y-5x=14&(2)\end{array}\right\\\\(1)\ 2x-3=2y\qquad\text{add 3 to both sides}\\2x=2y+3\qquad\text{subtract}\ 2y\ \text{From both sides}\\\boxed{2x-2y=3}\\(2)\ y-5x=14\\\boxed{-5x+y=14}\\\\\text{We get the system of equations in the form we need.}](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D2x-3%3D2y%5C%5Cy-5x%3D14%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7BLet%27s%20convert%20the%20system%20of%20equations%20to%20form%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Da_1x%2Bb_1y%3Dc_1%5C%5Ca_2x%2Bb_2y%3Dc_2%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D2x-3%3D2y%26%281%29%5C%5Cy-5x%3D14%26%282%29%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%281%29%5C%202x-3%3D2y%5Cqquad%5Ctext%7Badd%203%20to%20both%20sides%7D%5C%5C2x%3D2y%2B3%5Cqquad%5Ctext%7Bsubtract%7D%5C%202y%5C%20%5Ctext%7BFrom%20both%20sides%7D%5C%5C%5Cboxed%7B2x-2y%3D3%7D%5C%5C%282%29%5C%20y-5x%3D14%5C%5C%5Cboxed%7B-5x%2By%3D14%7D%5C%5C%5C%5C%5Ctext%7BWe%20get%20the%20system%20of%20equations%20in%20the%20form%20we%20need.%7D)
![\left\{\begin{array}{ccc}2x-2y=3\\-5x+y=14\end{array}\right\\\\\text{The first matrix is the matrix of coefficients at x and y.}\\\\\left[\begin{array}{ccc}a_1&b_1\\a_2&b_2\end{array}\right] \Rightarrow\left[\begin{array}{ccc}2&-2\\-5&1\end{array}\right]\\\\\text{The second matrix is the matrix:}\\\\\left[\begin{array}{ccc}x\\y\end{array}\right]\\\\\text{The third matrix is the matrix of numbers from the right side of the equation.}\\\\\left[\begin{array}{ccc}c_1\\c_2\end{array}\right]\Rightarrow\left[\begin{array}{ccc}3\\14\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D2x-2y%3D3%5C%5C-5x%2By%3D14%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7BThe%20first%20matrix%20is%20the%20matrix%20of%20coefficients%20at%20x%20and%20y.%7D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26b_1%5C%5Ca_2%26b_2%5Cend%7Barray%7D%5Cright%5D%20%5CRightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-2%5C%5C-5%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Ctext%7BThe%20second%20matrix%20is%20the%20matrix%3A%7D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Ctext%7BThe%20third%20matrix%20is%20the%20matrix%20of%20numbers%20from%20the%20right%20side%20of%20the%20equation.%7D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc_1%5C%5Cc_2%5Cend%7Barray%7D%5Cright%5D%5CRightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C14%5Cend%7Barray%7D%5Cright%5D)
![\text{Therefore we have:}\\\\\left[\begin{array}{ccc}2&-2\\-5&1\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}3\\14\end{array}\right]](https://tex.z-dn.net/?f=%5Ctext%7BTherefore%20we%20have%3A%7D%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26-2%5C%5C-5%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C14%5Cend%7Barray%7D%5Cright%5D)
⓵
When multiplying several factors together, an even number of negative signs will result in a positive answer.
↠ Because when you multiply two negative signs together, the negative signs turn into a positive sign, giving you a positive answer.
↠ If you have an even number of negative signs to multiply, then all your negative signs will be turned into positive signs, giving you a positive answer!
⓶
When multiplying several factors together, an odd number of negative signs will result in a negative answer.
↠ Because when you multiply a negative sign by a positive sign, the negative sign always wins, giving you a negative answer.
↠ If you har an odd number of negative signs to multiply, this means that once you are done multiplying all of the negative signs together except for the last one, you will be multiplying a positive sign with a negative sign.
↠ Here’s another way to see it : if you have 5 negative signs to multiply together, you will start by multiplying the first 2, which will give you a positive result. Then you will multiply the third and the fourth negative signs together, which will also give you a positive result. By that time, you will have two positive signs and one negative sign left, which means you will have no choice but to multiply a positive sign by the last negative sign, resulting in a negative answer!
I hope this helps! If you need a visual example, just tell me, it will be my pleasure! ☻
Step-by-step explanation:
total ratio is 5
first find the total amount
bob's ratio is 2/5
so 2/5 ×total amount=8 pounds
0.4 ×total amount=8 pounds
total amount=8/0.4=20 pounds
so chris receives: 20-8=12 pounds
Answer:
For x = 1, y = 2× 1² y = ![2^1](https://tex.z-dn.net/?f=2%5E1)
= 2 × 1 = 2
= 2
For x = 2, y = 2× 2² y = 2²
= 2 × 4 = 4
= 8
Answer: 8x+2
Step-by-step explanation:
Distribute 2 to 4x and 1 by multiplying them.