Solve the system of linear equations and check any solutions algebraically. (If there is no solution, enter NO SOLUTION. If the system is dependent, express x, y, and z in terms of the parameter a.)
3x-3y+6z=15
x+2y-z=11
5x-8y+13z=19
Eq1: 3x-3y+6z = 15
Eq2:-3x-6y+3z = -3(times-3).
sum: =9y+9z = -18, y + z = -2.
Eq2: -5x-10y+5z = -55(times-5).
Eq3: +5x-8y+13z = 19
Sum: -18y+18z =-36, y - z = 2.
y + z = -2.
y - z = +2.
Sum: 2y = 0, y = 0. z = -2.
Eq1: 3x-3y+6z = 15.
3x - 3*0 + 6(-2) = 15,
3x = 27, x = 9.
Solution: (x, y, z) = (9,0,-2).
To check your solutions, plug the values of x, y, and z into the given Eqs.
To solve the system of linear equations, we will use the method of elimination or substitution.
Method of Elimination:
Step 1: Multiply the second equation by 3 to make the coefficients of x the same as the first equation.
3(x + 2y - z) = 3(11) => 3x + 6y - 3z = 33
Step 2: Add Equation 1 and 2 to eliminate x.
(3x - 3y + 6z) + (3x + 6y - 3z) = 15 + 33
6x + 3z = 48 ----> Equation 3
Step 3: Multiply the first equation by 2 to make the coefficients of x the same as the third equation.
2(3x - 3y + 6z) = 2(15) => 6x - 6y + 12z = 30
Step 4: Add Equation 3 and 4 to eliminate x.
(6x + 3z) + (6x - 6y + 12z) = 48 + 30
12z - 6y + 12z = 78
-6y + 24z = 78 ----> Equation 4
Step 5: Multiply Equation 2 by 4 to make the coefficients of y the same as Equation 4.
4(x + 2y - z) = 4(11) => 4x + 8y - 4z = 44
Step 6: Add Equation 4 and 5 to eliminate y.
(-6y + 24z) + (4x + 8y - 4z) = 78 + 44
4x + 20z = 122 ----> Equation 5
Step 7: Solve the equations 5 and 3 to find the values of x and z.
Equation 5: 4x + 20z = 122
Divide by 4:
x + 5z = 30 ----> Equation 6
Equation 3: 6x + 3z = 48
Divide by 3:
2x + z = 16 ----> Equation 7
Step 8: Solve Equations 6 and 7 simultaneously to find the values of x and z.
Multiply Equation 6 by 2 and subtract Equation 7:
(2x + 10z = 60) - (2x + z = 16)
9z = 44
Divide by 9:
z = 44/9
Substitute z = 44/9 into Equation 7:
2x + 44/9 = 16
2x = 16 - 44/9
2x = (144 - 44)/9
2x = 100/9
Divide by 2:
x = (100/9)/2
x = 100/18
x = 50/9
Step 9: Substitute the found values of x and z into Equation 6 to solve for y.
x + 5z = 30
50/9 + 5(44/9) = 30
50/9 + 220/9 = 270/9
y = (270 - 220)/9
y = 50/9
Therefore, the solution to the system of linear equations is:
x = 50/9, y = 50/9, z = 44/9
Now let's check if these values satisfy all the original equations.
Substituting the values into the first equation:
3x - 3y + 6z = 15
3(50/9) - 3(50/9) + 6(44/9) = 15
150/9 - 150/9 + 264/9 = 15
(150 - 150 + 264)/9 = 15
264/9 = 15
29.33 = 15 This equation is not satisfied.
Hence, the system of linear equations has NO SOLUTION.