This might not be correct but....
I’m going to use rounding,
42 x 2 = 84
84 divides by 3 = 28,
Which is near 30......So the number of people in the group is rounding to about 2 people.
If they need 22 boxes for every 6 kids the they would need 44 boxes for every 12 kid
solution is (x, y ) → (- 1, 4 )
y + 2x = 2 → (1)
y + 4x = 0 → (2)
subtract equation (1) from (2) to eliminate term in y
(y - y ) + (4x - 2x ) = (0 - 2 )
2x = - 2 ( divide both sides by 2 )
x = - 1
substitute x = - 1 in either (1) or (2) and solve for y
(1) → y - 2 = 2 ( add 2 to both sides )
y = 4
solution is (x, y ) → (- 1, 4 )
Y=-4x+3-x because if it's -y=4-3+x y is not alone so you need to make y positive and switch the rest of the operations so it can be y=-4+3-x
Answer:
see explanation
Step-by-step explanation:
The inverse of a matrix A = ![\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5C%5C%5Cend%7Barray%7D%5Cright%5D)
is
= 
![\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D)
If ad - bc = 0 then the matrix has no inverse
A = ![\left[\begin{array}{ccc}11&-5\\3&-1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-5%5C%5C3%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
ad - bc = (11 × - 1) - (- 5 × 3) = - 11 - (- 15) = - 11 + 15 = 4 , then
= 
![\left[\begin{array}{ccc}-1&3\\-5&11\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%263%5C%5C-5%2611%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-0.25&0.75\\-1.25&2.75\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-0.25%260.75%5C%5C-1.25%262.75%5C%5C%5Cend%7Barray%7D%5Cright%5D)
