Answer:
y=1x-4
Wait sorry I did it wrong the first time lol.
In standard Form, it should be 4x-4y=16
Step-by-step explanation:
yes yes hand me brain
Answer:
![\binom{4}{0}](https://tex.z-dn.net/?f=%20%5Cbinom%7B4%7D%7B0%7D%20)
Step-by-step explanation:
A column vector is shown in the form
where x is how many spaces to move in the positive x direction y is how many spaces move in the positive y direction.
The vector b moves 4 spaces to the right (which is in the positive direction of x) and doesn't move up or down in the y direction which is 0, so the column vector is ![\binom{4}{0}](https://tex.z-dn.net/?f=%20%5Cbinom%7B4%7D%7B0%7D%20)
Answer:
your answer is going to be (-2,2)
Step-by-step explanation:
Answer:
Choice A
Step-by-step explanation:
Hopefully this helps!
Answer:
Step-by-step explanation:
Given is a system of equations as
![x1-4x2 - x4 = -7\\ \\x2 - 2x4 = 3 \\\\x4 + 2x5 = 3](https://tex.z-dn.net/?f=x1-4x2%20-%20x4%20%3D%20-7%5C%5C%20%5C%5Cx2%20-%202x4%20%3D%203%20%5C%5C%5C%5Cx4%20%2B%202x5%20%3D%203)
We have 5 variables and 3 equations
a) coefficient matrix of this system is
1 -4 0 -1 0\\
0 1 0 -2 0\\
0 0 0 1 2\\
We find that x3 has no coefficient in any of the equations so we can omit x3 and write as equations for 4 variables as
1 -4 -1 0\\
0 1 -2 0\\
0 0 1 2\\
b) Augmented matrix is
1 -4 -1 0\\ 7
0 1 -2 0\\
3
0 0 1 2\\3
c) For row operations to ehelon form
we can do R1+4R2 = R1
We get
1 0 -9 0 \\ 19
0 1 -2 0 \\ 3
0 0 1 2 \\ 3
Now let us do R1 = R1+9R3 and R2 = R2+2R3
1 0 0 0 \\ 46
0 1 0 0 \\ 9
0 0 1 2 \\ 3
d) We find that there are infinite solutions to the system in parametric form, since x4 and x5 are linked with only one equation
e) x1 = 46, x2 = 9, x4+2x5 =3
Or x1 =46, x2 =9, x4 = 3-2x5, x5 = x5 is the parametric solution