Solve each of the following systems of equation.
(b.) 3x + y = 1
6x^2 - y^2 - 2y - 3 = 0

(a) The solution to the given system of equations is x = 1, y = 2, z = -3
(b) The solutions to the system of equations are (3.4142, -9.2426) and (0.5858, -0.7574)
From the question, we are to solve the given system of equations
The given system of equation is
x + y + z = 0 ----------- (1)
2x + z = -1 ----------- (2)
x - y - z = 2 ----------- (3)
Add equations (1) and (3)
x + y + z = 0 ----------- (1)
x - y - z = 2 ----------- (3)
__________
2x = 2
x = 2/2
x = 1
Substitute the value of x into equation (2) to find z
2x + z = -1
2(1) + z = -1
2 + z = -1
z = -1 -2
z = -3
Substitute the values of x and z into equation (1) to determine the value of y
x + y + z = 0
1 + y + -3 = 0
1 + y - 3 = 0
y = 3 -1
y = 2
Hence, the solution to the given system of equations is x = 1, y = 2, z = -3
b.
The given system of equations is
3x + y = 1 --------- (1)
6x² - y² - 2y -3 = 0 --------- (2)
From equation (1)
3x + y = 1
y = 1 - 3x -------- (3)
Substitute into equation (2)
6x² - y² - 2y -3 = 0
6x² - (1 -3x)² -2(1 -3x) -3 = 0
6x² - (1 -3x)(1 -3x) -2 + 6x -3 = 0
6x² - (1 -3x -3x +9x²) -2 +6x -3 = 0
6x² - (1 -6x + 9x²) +6x -5 = 0
6x² -1 +6x -9x² +6x -5 = 0
-3x² +12x -6 = 0
3x² -12x +6 = 0
x² -4x + 2 = 0
Solve the quadratic equation by the formula method,
x = [-b±√(b²-4ac)]/2a
a = 1, b = -4 and c = 2
Thus,
x = [-(-4)±√((-4)²-4(1)(2))]/2(1)
x = [4±√(16-8)]/2
x = (4±√8)/2
x = (4+√8)/2 OR (4 - √8)/2
x = 3.4142 OR x = 0.5858
Substitute the values of x into equation (3)
y = 1 - 3x
When x = 3.4142
y = 1 - 3(3.4142)
y = -9.2426
When x = 0.5858
y = 1 - 3(0.5858)
y = -0.7574
Hence, the solutions to the system of equations are (3.4142, -9.2426) and (0.5858, -0.7574)
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