Which system of equations is consistent and dependent? {3x+2y=3
6x+4y=6
{4x+ 6y= 2
4x+6y=1
{2x+3y=-6. 2x+3y=-12
{5x+4y=-30
3x-9y=-18

Step-by-step explanation:

4x − y = −11

2x + 3y = 5

lets multiply the second equation by -2 and add it to the first:

4x − y = −11

-4x - 6y = -10

0  - 7y = -21

y = -21/-7

y = 3

now we substitute this result in the first equation to find x:

4x − y = −11

4x - 3 = -11

4x = -8

x = -8/4

x = 2

so the solution is y = 3 and x =2

4x − 9y = −21

−10y = −30

we solve for y

−10y = −30

y = -30/-10

y = 3

and substitute in the first equation:

4x − 9y = −21

4x − 9(3) = −21

4x - 27 = -21

4x = 6

x = 6/4 = 3/2

so the solution is x = 3/2 and y = 3

4x + 3y = 5

2y = −6

we solve for y:

2y = −6

y = -6/2

y = -3

we do substitute in the first equation:

4x + 3y = 5

4x + 3(-3) = 5

4x - 9 = 5

4x = 14

x = 14/4

x = 7/2

so the solution is x = 7/2 and y = -3

7x − 3y = −11

9x = −6

we solve for x:

9x = −6

x = -6/9

x = -2/3

then we substitute in the first equation the result found:

7x − 3y = −11

7(-2/3) − 3y = −11

-14/3 - 3y = -11

we multiply by 3 to eliminate fractions:

-14 - 9y = -33

9y = 19

y = 19/9

so the solution is x = -2/3 and y = 19/9

12x − 3y = −33

14x = −28

we solve for x:

14x = −28

x = -28/14

x = -2

then we substitute in the first equation:

12x − 3y = −33

12(-2) − 3y = −33

-24 - 3y = -33

3y = 9

y = 3

then the solution is x = -2 and y = 3

A.

Step-by-step explanation:

Took it on Edge2020 :)

may i see the picture? it would a lot, xox

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.