Answer:
<h2>x = 4, y = 3 → (4, 3)</h2>
Step-by-step explanation:
![\left\{\begin{array}{ccc}2x+3y=17&\text{multiply both sides by (-2)}\\3x+6y=30\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-4x-6y=-34\\3x+6y=30\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad-x=-4\qquad\text{change the signs}\\.\qquad x=4\\\\\text{put the value of x to the first equation:}\\\\2(4)+3y=17\\8+3y=17\qquad\text{subtract 8 from both sides}\\3y=9\qquad\text{divide both sides by 3}\\y=3](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D2x%2B3y%3D17%26%5Ctext%7Bmultiply%20both%20sides%20by%20%28-2%29%7D%5C%5C3x%2B6y%3D30%5Cend%7Barray%7D%5Cright%5C%5C%5Cunderline%7B%2B%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D-4x-6y%3D-34%5C%5C3x%2B6y%3D30%5Cend%7Barray%7D%5Cright%7D%5Cqquad%5Ctext%7Badd%20both%20sides%20of%20the%20equations%7D%5C%5C.%5Cqquad-x%3D-4%5Cqquad%5Ctext%7Bchange%20the%20signs%7D%5C%5C.%5Cqquad%20x%3D4%5C%5C%5C%5C%5Ctext%7Bput%20the%20value%20of%20x%20to%20the%20first%20equation%3A%7D%5C%5C%5C%5C2%284%29%2B3y%3D17%5C%5C8%2B3y%3D17%5Cqquad%5Ctext%7Bsubtract%208%20from%20both%20sides%7D%5C%5C3y%3D9%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%203%7D%5C%5Cy%3D3)
Answer: You will spend 25.94$
Step-by-step explanation:
6.5 yards by 3.99 per yard = 6.5 times 3.99 = 25.935 witch rounds too 25.94$
If we divide 12.6/4.5 by 3 we receive 4.2/1.5. So yes they are proportional. <span />
8 - 8x is already in its simplest form.
Without knowing the value of x, you cannot find what is equivalent to 8 - 8x, but you can it if is an equation, such as 8 - 8x = 88 (x would be -10).
However, you can still rewrite it as -8x + 8, which is equivalent to 8 - 8x.
9514 1404 393
Answer:
(a, b) = (-2, -1)
Step-by-step explanation:
The transpose of the given matrix is ...
![A^T=\left[\begin{array}{ccc}1&2&a\\2&1&2\\2&-2&b\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26a%5C%5C2%261%262%5C%5C2%26-2%26b%5Cend%7Barray%7D%5Cright%5D)
Then the [3,1] term of the product is ...
![(A\cdot A^T)_{31}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}1&2&2\end{array}\right]=a+2b+4](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B31%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%262%5Cend%7Barray%7D%5Cright%5D%3Da%2B2b%2B4)
and the [3,2] term is ...
![(A\cdot A^T)_{32}=\left[\begin{array}{ccc}a&2&b\end{array}\right]\cdot\left[\begin{array}{ccc}2&1&-2\end{array}\right]=2a-2b+2](https://tex.z-dn.net/?f=%28A%5Ccdot%20A%5ET%29_%7B32%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%262%26b%5Cend%7Barray%7D%5Cright%5D%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%26-2%5Cend%7Barray%7D%5Cright%5D%3D2a-2b%2B2)
Both of these terms in the product matrix are 0. We can solve the system of equations by adding these two terms:
(a +2b +4) +(2a -2b +2) = (0) +(0)
3a +6 = 0
a = -2
Substituting for 'a' in term [3,1] gives ...
-2 +2b +4 = 0
b = -1
The ordered pair (a, b) is (-2, -1).