I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
Answer:
5 hamburger packages and 6 bun packages
Step-by-step explanation:
It is asking for the highest common multiple of 12 and 10. the first time a multiple of 12 is divisible by 10 is at 12x5, which is 60. now divide 60 by 12 and 10, 5 and 6.
"Will is twice as old as Jill."
Jill's age . . . . . J
Will's age . . . . 2J .
"Three years ago . . .
Jill's age then . . . . . J - 3
Will's age then . . . . 2J - 3
". . . Jill's age then was 2/5 of Will's age then."
J - 3 = (2/5) (2J - 3)
Multiply
each side by 5 : 5J - 15 = 2 (2J - 3)
Divide
each side by 2 : 2.5 J - 7.5 = 2J - 3
Subtract 2J
from each side: 0.5 J - 7.5 = -3
Add 7.5
to each side: 0.5 J = 4.5
Multiply
each side by 2 : J = 9
Jill is 9 y.o. now.
Will is 18 y.o. now.
