Solve the following system of equations using the
1 answer:
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 )
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The is number 3 or c. I hope this helps.
Answer:
X = -7
Step-by-step explanation:
-6x + 5 = 47
-6x +5 -5 = 47 - 5
-6x = 42
-6x/-6 = 42/-6
X = -7
The function is
y =
----1.
Plug x = 2x in 1.
∴y` =
----- 2.
Divide 1 and 2. You will get,
y = 4 y`
Therefore, y is quartered
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