1-- What type of system is 3x – y = 4, and -6x + 2y = 5? Select one: a. Independ

Question

1-- What type of system is 3x – y = 4, and -6x + 2y = 5?

Select one:

a. Independent

b. Inconsistent

c. Dependent

d. Non-linear

e. None of these

2-- What kind of system is 2x + y = 3, -6x - 3y = -9?

Select one:

a. Independent

b. Inconsistent

c. Dependent

d. Non-linear

e. None of these

3-- What is the solution to 3x – y = 4, x + 6y = -5?

Select one:

a. (1, -1)

b. (-1, 1)

c. The system is dependent

d. No solution

e. None of these

4-- Solve the system: y = x2 – 4,   y = 3x - 6

Select one:

a. (2, 0)

b. (0, 2)

c. 2 and -1

d. (2, 0) and (-1, -3)

e. None of these

5-- Solve the system: x + y + z = 6,   2x – y + z = 3, x + y – z = 0

Select one:

a. (0, 0, 6)

b. (3, 2, 1)

c. 1

d. (1, 2, 3)

e. None of these

Explanation / Answer

If 3x-y = 4, then, on multiplying both the sides by -2, we get -6x+2y = -8. In view of the 2nd equation being 6x + 2y = 5, the given system of equations is Inconsistent. Option b) is the correct answer. If 2x+y = 3, then, on multiplying both the sides by -3, we get -6x - 3y = -9. This means that both the equations are same. Thus, the system has infinite number of solutions, so that the equations are dependent. Option c) is the correct answer. If 3x -y = 4, then, on multiplying both the sides by 6, we get 18x-6y =24. Now, on adding this equation to the 2nd equation, we get 18x-6y+x+6y = 24-5 or, 19x = 19 so that x = 19/19 = 12. Then y = 3x-4 = 3-4 = -1. Thus, the given system of equations has a unique solution i.e. (1,-1). Option a) is the correct answer. If y = x2 -4 and y = 3x-6, then x2 -4 = 3x-6 or, x2-3x +2 = 0 or, (x-2)(x-1)= 0 so that either x = 2 or, x = 1. When x = 2, y = 3x-6 = 3*2-6 = 0 and when x = 1, y = 3x-6 = 3*1-6 = -3. Thus, (x,y) = (2,0) and (1,-3). Option d) is the correct answer. We have x + y + z = 6…(1),   2x – y + z = 3…(2), and x + y – z = 0…(3). On adding the 1st and the 3rd equations ( to eliminate z), we get x+y+z+x+y-z = 6 or, 2x+2y =6 or, ( on dividing both the sides by 2), x+y =3…(4). Further, on subtracting the 1st equation from the 2nd equation (to eliminate z), we get 2x-y+z-x-y-z = 3-6 or, x-2y = -3…(5). Now, on subtracting the 5th equation from the 4th equation (to eliminate x), we get x+y –(x-2y) = 3-(-3) or, 3y = 6 so that y = 2. Then, from the 4th equation, we get x +2= 3 so that x = 3-2 =1. On substituting x =1 and y = 2 in the 1st equation, we get 1+2+z= 6 so that z = 3. Thus, (x,y,z) = (1,2,3). Option d) is the correct answer.

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