Solve the following system of equations. x + 2y − 6 = z 3y − 2z = 7 4 + 3x = 2y − 5z

X+2y-6=z, 3y-2z=74, 4+3x=2y-5z  rearrange all to the form ax+by+cz=n

x+2y-z=6, 3y-2z=74, 3x-2y+5z=-4

eliminate x from the 1st and 3rd equations...

-3(x+2y-z=6)+(3x-2y+5z=-4)

-3x+3x-6y-2y+3z+5z=-18-4

-8y+8z=-22  and the 2nd equation was 3y-2z=7, now we can eliminate z

(-8y+8z=-22)+4(3y-2z=7)

-8y+12y+8z-8z=-22+28

4y=6

y=1.5, making 3y-2z=7 become:

3(1.5)-2z=7

4.5-2z=7

-2z=2.5

z=-1.25, making x+2y-z=6 become

x+2(1.5)-(-1.25)=6

x+3+1.25=6

x+4.25=6

x=1.75

So (x,y,z)=(1.75, 1.5, -1.25)

Draw a diagram to illustrate the problem as shown below.

Calculate the volume of the empty cone.
V₁ = (1/3)π*(6 in)²*(10 in) = 120π in³

Calculate the volume of the sphere.
V₂ = (4/3)π*(1.5 in)³ = 4.5π in³

The volume that can be filled with flavored ice is 
V = V₁ - V₂ = 115.5π in³ = 362.85 in³

Th volume is 115.5π in³ or 362.9 in³ (nearest tenth)