Answer:
x = 13
y = -10
z = -7
Step-by-step explanation:
x+y+z=−4
2x+3y−2z=10
−x+2y−3z=−12
Add the first and the third equations to eliminate x
x+y+z=−4
−x+2y−3z=−12
--------------------
3y -2z = -16 Equation A
Add the second and twice the third equation to eliminate x
2x+3y−2z=10
−2x+4y−6z=−24
----------------------------
7y -8z = -14 Equation B
Take Equation A and multiply by -4
-4*( 3y -2z) = (-16)*-4
-12y + 8z = 64
Add this to Equation B
-12y + 8z = 64
7y -8z = -14
-----------------------
-5y = 50
Divide by -5
-5y/-5 = 50/-5
y = -10
Now using equation A
3y -2z = -16
3*-10 -2z = -16
-30 -2z = -16
Add 30 to each side
-2z = -16+30
-2z = 14
Divide by -2
-2z/-2 = 14/-2
z = -7
Now find x using the first equation
x+y+z = -4
x -7-10 = -4
x -17 = -4
Add 17 to each side
x-17+17 = -4+17
x = 13
Step-by-step explanation:
I guess you just slove the exponent can I see the actual question
Answer:
Step-by-step explanation:
Given that a basketball coach will select the members of a five-player team from among 9 players, including John and Peter.
Out of nine players five are chosen at random.
The team consists of John and Peter.
Hence we can sort 9 players as I group, John and Peter and II group 7 players.
Now the selection is 2 from I group and remaining 3 from II group.
Hence no of ways of selecting a team that includes both John and Peter=
=35
Total no of ways = 
=126
= 
=
It’s D
...........................
<u>27 = 6x + 4y </u>= 1 7/20 = 3 4/5<u>
</u>20 = 2x + 5y<u>
</u>
