Answer:
y = -140
Step-by-step explanation:
Y varies directly with x
y = kx
We know y = -20 when x =2, we can solve for k
-20 = k*2
Divide each side by 2
-20/2= 2k/2
-10 =k
y = -10x
Now we want to find y when x = 14
y = -10 *14
y = -140
Since -6 +6 can be rearranged to 6-6 using the communicative property, we get 6-6 =0 . In addition, -120+(-6) is essentially having -120+ 1*(-6), and 1 times a number is simply that number, so we have -120-6=-126
Just put the coefients in to a matrix
1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C-2%262%26-3%7C-14%5Cend%7Barray%7D%5Cright%5D%20)
mulstiply 2nd row by -1 and add to 3rd
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%262%260%7C-6%5Cend%7Barray%7D%5Cright%5D)
divde last row by 2
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 2rd row by 6 and add to top one
![\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 1st row by -1 and add to 2nd
![\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-3%260%260%7C6%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 2nd row by -3
![\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
mulstiply 2nd row by -1 and add to 1st row
![\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-3%7C-12%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 1st row by -3
![\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%7C4%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
rerange
![\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%7C-2%5C%5C0%261%260%7C-3%5C%5C0%260%261%7C%204%5Cend%7Barray%7D%5Cright%5D)
x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)
B is answer
Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0