Answer:
x = 1.5
y = 3
Step-by-step explanation:
to solve this system of equation using the substitution methods we say let
2x+3y=12................................... equation 1
y=x+1.5..................................... equation 2
substitute equation 2 into equation 1
2x+3y=12................................... equation 1
2x + 3( x + 1.5) = 12
2x + 3x + 4.5 = 12
5x + 4.5 = 12
collect the like terms
5x = 12 - 4.5
5x = 7.5
divide both sides by the coefficient of x which is 5
5x/5 = 7.5/5
x = 1.5
put x = 1.5 into equation 2
y=x+1.5..................................... equation 2
y = 1.5 + 1.5
y = 3
therefore the value of x = 1.5 and y = 3 respectively.
Answer:
use demos
Step-by-step explanation:
Data from random samples will not always include the same values. Values are chosen randomly and they may or may not be the same. So means will vary among samples.
I think 1 is the only common factor of both 0.20 and 215 but I'm not 100% certain
Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is ![\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D3%266%266%26-9%5C%5C-2%26-3%26-3%263%5Cend%7Barray%7D%5Cright%5D)
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is
,
,
then