Linear Systems in 3 Variables

Question

Solve the system of equations for all three variables

4x+3y-z=10

8x+7y-z=6

6x+y+z=4

Answer 1

eliminates Z

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equation 2 MINUS equation 1:

4x + 4y = -4

x + y = -1

then equation 2 PLUS equation3:

14x + 8y = 10

7x + 4y = 5

the 2x2 system is:

x + y = -1 <-- equation A

7x + 4y = 5 <--- equation B

by elimination:-7 * eqA + eqB:

-3y = 12

y = -4

per equation A, with y=-4, forces x=3

original first equation says:

4x+3y-z = 10

4(3) + 3(-4)-z = 10

12 + -12 - z = 10

0 - z = 10

-z = 10

z = -10

the solution set is {x=3, y=-4, z = -10}

CHECK:

equation 1: 4x+3y-z = 4(3)+3(-4)-(-10) = 12+-12+10 = 10

equation 2: 8x+7y - z = 8(3) + 7(-4) - (-10)

= 24 + -28 + 10

= -4 + 10

= 6

equation 3: 6x + y + z = 6(3)+ -4 + -10

= 18 -4-10

= 4

yes, the solution checks

the solution set is {x=3, y=-4, z = -10}