Hmm
what you do is try to eliminate 1 variable in 2 equaions first
we can elminate x's in first and 2nd equation
add first and 2nd equation
x+3y-2z=10
<u>-x-2y+z=-7 +</u>
0x+y-z=3
y-z=3
multiply 2nd equation by 3 and add to last one
-3x-6y+3z=-21
<u>3x+9y-5z=28 +</u>
0x+3y-2z=7
3y-2z=7
we now have
y-z=3
3y-2z=7
multiply first equation by -2 and add to 2nd
-2y+2z=-6
<u>3y-2z=7 +</u>
y+0z=1
y=1
now we can sub back
y-z=3
1-z=3
minus 1
-z=2
times -1
z=-2
sub baack into any equation
x+3(1)-2(-2)=10
x+3+4=10
x+7=10
minus 7
x=3
x=3
y=1
z=-2
(3,1,-2)
Y + 7 = 3x.....y = 3x - 7
6x - 2y = 12
6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
14 = 12 (incorrect)
when ur variables cancel out and u r left with an incorrect statement, this means there is no solution because ur lines are parallel
so ur answer is : no solution
Answer:
If 4 more kids came to the zoo with Liam and his two parents, and so did 4 more adults, the price they would pay would be:
5.3(5) + 10.7(6)
90.7
So, it would cost a total of $90.70 dollars for Liam, his two parents, four other friends, and four other parents, to come to the zoo.
Let me know if this helps!
Answer:
2
Step-by-step explanation:
The side has to be between the angles.
So, Angle1 - Side - Angle2
So, we can only swap the angles. So no other permutation is possible with the given conditions.
[If there's any problem with the solution, please inform me]
Part a: subtract 48-30=18
Part b: i found it by subtracting 48 -30 because it says left over