asked 142k views
2 votes
Solve the system of equation
4x+y−z=13
3x+5y+2z=21
2x+y+6z=14

asked
User Maestro
by
7.1k points

1 Answer

5 votes

Answer:

x = 3, y = 2 and z = 1.

Explanation:

4x+y−z=13

3x+5y+2z=21

2x+y+6z=14

Subtract the third equation from the first:

2x - 7z = -1 ........... (A)

Multiply the first equation by - 5:

-20x - 5y + 5z = -65

Now add the above to equation 2:

-17x + 7z = -44 ...... (B)

Now add (A) and (B)

-15x = -45

So:

x = 3.

Substitute x = 3 in equation A:

2(3) - 7z = -1

-7z = -7

z = 1.

Finally substitute these values of x and z in the first equation:

4x+y−z=13

4(3) +y - 1 = 13

y = 13 + 1 - 12

y = 2.

Checking these results in equation 3:

2x+y+6z=14:-

2(3) + 2 + 6(1) = 6 + 2 + 6 = 14

- checks out.

answered
User Valepu
by
8.7k points
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