Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set x3 = t and solve for x1 and x2.) x1 − 3x3 = −7 3x1 + x2 − 2x3 = −5 2x1 + 2x2 + x3 = −3

x1 = -7+3t

x2 = (11-7t)/2

Step-by-step explanation:

Given the system of equations

x1 − 3x3 = −7 ..., 1

3x1 + x2 − 2x3 = −5 2

2x1 + 2x2 + x3 = −3 3

Setting x3 = t.

Substitute x3 = t into equation 1 to get x1 in terms of t.

x1 − 3x3 = −7

x1-3(t) = -7

x1-3t = -7

Add 3t to both sides;

x1-3t+3t = -7+3t

x1 = -7+3t

To get x2, substitute x1 = -7+3t and x3 = t into equation 3

2x1 + 2x2 + x3 = −3

2(-7+3t)+2x2+t = -3

Open the parenthesis

-14+6t+2x2+t = -3

Collect like terms

2x2+6t+t = -3+14

2x2+7t = 11

Subtract 7t from both sides

2x2+7t-7t = 11-7t

2x2= 11-7t

Divide both sides by 2

2x2/2 = (11-7t)/2

x2 = (11-7t)/2

Hence the solution to the system if equation (x1, x2, x3) = (-7+3t, (11-7t)/2, t)

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step-by-step explanation:

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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