The solution to the variables using equations 1, 2 and 3 is x = 2, y = -1 and z = 3 and the solutions in part b hold for equation 4
There are three variables in the system
To solve the three variables, there must be at least three equations in the system of equations
Hence, the variables cannot be solved because the number of equations is not enough
We have:
3x + z + y = 8
5y - x = -7
3z + 2x - 2y = 15
Make x the subject in (2)
x = 5y + 7
Substitute x = 5y + 7 in (1) and (3)
3(5y + 7) + z + y = 8
15y + 21 + z + y = 8
16y + z = -13
3z + 2(5y + 7) - 2y = 15
3z + 10y + 14 - 2y = 15
3z + 8y = 1
Multiply 16y + z = -13 by 3
48y + 3z = -39
Subtract 3z + 8y = 1 from 48y + 3z = -39
48y - 8y + 3z - 3z = -39 - 1
40y = -40
Divide by 40
y = -1
Substitute y = -1 in x = 5y + 7
x = 5(-1) + 7
x = 2
Make z the subject in 16y + z = -13
z = -13 - 16y
Substitute y = -1
z = -13 - 16(-1)
z = 3
Hence, the solution to the variables using equations 1, 2 and 3 is x = 2, y = -1 and z = 3
We have:
4x + 5y - 2z = -3
Substitute x = 2, y = -1 and z = 3
4(2) + 5(-1) - 2(3) = -3
Evaluate
-3 = -3 --- this is true
Hence, the solutions in part b hold for equation 4
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Missing part of the question
The following equations are given:
Equation #1: 3x + z + y = 8
Equation #2: 5y - x = -7
Equation #3: 3z + 2x - 2y = 15
Equation #4: 4x + 5y - 2z = -3
