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3 years ago

14

Systems of 3 equations by using a combination of both substitution and elimination

1 answer:

LiRa [457]3 years ago

6 0

X-2y+z=3
2x+y-2z=31
-x+2y+3z=-23
multiply the first equation by -2 and add the results to the second equation:
x-2y+z=3
+5y-4z=25
-x+2y+3z=-23
multiply the first equation by 1 and the results to the third equation:
x-2y+z=3
5y-4z=25
4z=-20
solve for z:
4z=-20
z=-5
solve for y:
5y-4z=25,z=-5
5y-(4*-5)=25
5y+20=25
5y+20-20=25-20
5y=5
y=1
solve for x by substituted y and z into the first equation:
x-2y+z=3, y=1, z=-5
x-(2*1)-5=3
x-2-5=3
x-7=3
x-7+7=3+7
x=10
the answer are x=10, y=1, z=-5

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Part b) The radius of the film canister is $2\ cm$

Step-by-step explanation:

<u><em>The complete question is</em></u>

A film canister in the shape of a cylinder has a height of 8 centimeters and a volume of 32π cubic centimeters.

a. Write an equation for the volume of the film canister.

b. What is the radius of the film canister?

Part a) Write an equation for the volume of the film canister

we know that

The volume of a cylinder is equal to

$V=\pi r^{2} h$

where

r is the radius of the circular base

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Part b) What is the radius of the film canister?

we have

$32\pi=\pi r^{2}(8)$

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square root both sides

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