Question

Solve the following system of equations using the

Answer 1

Answer:

x = 3, y = - 1, z = 1

Step-by-step explanation:

listing the coefficients

1   - 2   1   6 ← R1

3    1    - 1  7 ← R2

4   - 1     2   15 ← R3

We require the first entry in R2 to be 0 while retaining R1 and the first 2 entries of R3 to be 0, thus

R2 - 3R1 and R3 - 4R1

1   - 2   1    6 ← R1

0    7   - 4  - 11 ← R2  

0    7    - 2  - 9 ← R3

Now R3 - R2

1    - 2    1     6 ← R1

0     7   - 4   - 11 ← R2

0    0      2    2 ← R3

From R3

2z = 2 ⇒ z = 1

Substitute z = 1 into R2

7y - 4(1) = - 11

7y - 4 = - 11 ( add 4 to both sides )

7y = - 7 ⇒ y = - 1

Substitute y = - 1, x = 1 into R1

x - 2(- 1) + 1 = 6

x + 2 + 1 = 6

x + 3 = 6 ( subtract 3 from both sides )

x = 3

Solution is (3, - 1, 1 )