simultaneous equations

Question

Equation 1: 3x + y - 2z = -7

Equation 2: -az = x + y + 6

To form the third equation use the digits 2,2,1,9 will be the coefficients of x,y,z, and the constant term in that order, given that the equation is in the form ax + by + cz = d

*Write down Equation 3 and solve the set of three equations when a = -3

*Find the values of a , for which the system is consistent. Give reasons for your answer and interpret the solution geometrically.

Second Part:

Jack wants to experiment with the equations still further continuing to use a = -3

1) A new set of three equations is formed using the original EQUATION 1 and EQUATION 2 and the third equation is formed by multiplying all the coefficients and the constant in Equation 1 by 3.

2) A new set of three equations is formed using the original EQUATION 1 and EQUATION 2 and the third equation is formed by using Equation 1 with the constant changed from -7 to 12.

Answer 1

Equation 3: 2x + 2y + z = 9

With a = -3, the set of equations to solve is

Equation 1: 3x + y - 2z = -7

Equation 2: -x - y + 3z = 6

Equation 3: 2x + 2y + z = 9

First, eliminate both x and y from Eqn 2 and 3. To do that, multiply eqn 2 on both sides by 2. Then add the two equations. The result is

z = 3.

Then, substitute the above value of z in equations 1 and 3.

Equation 1: 3x + y = -1

Equation 3: x + y = 3

Subtract Equation 3 from Equation 1. You should get

x = -2

Then substitute for x in Equation 3 and solve for y.

y = 5

Values 'a' for which the equations are consistent: Solve the above three equations similar to the above, except keeping 'a' as an unknown numerical constant. Then you will find,

z = 21/(1 - 2a)

Clearly, this fraction becomes indeterminate when the denominator is zero.

i.e. 1 - 2a = 0

So, for a=1/2, the equations will not have a consistent solution.