Which system will have no solution?

A.

4y = 5x - 12

y = 5/4x - 3

B.

5x - 4y = 12

4x - 5y = 12

C.

5x - 4y = 12

y = 5/4x - 3

D.

5x - 4y = 3

y = 5/4x - 3

A. plug the second equation into the first to solve for x

4 (5/4x-3) = 5x-12

distribute the 4: 5x-12=5x-12

0=0 all real numbers

b. cancel out a variable of your choice by multiplying both equations to get a common factor. i multiplied the first one by 4 and the second by - 5 to cancel out the x

(5x-4y=12) 4

(4x-5y=12) - 5

20x-16y=48

-20x+25y=-60

add the two new equations, which cancels out the x's and leaves you with 9y=-12

divide both sides by 9, y=-1.3

plug y back into one of the equations to solve for x, i did the first

5x-4y=12

5x-4 (-1.3) = 12

5x+5.3=12

5x=6.7

x=1.34 - > (-1.34,-1.3) = solution

**c. similar to a, plug the second equation into the first

5x-4 (5/4x-3) = 12

5x-5x+12=12

5x-5x=0

12=12

subtract 12 from both sides

0=0 all real numbers

d. same as a&c, plug the second equation into the first

5x-4 (5/4x-3) = 3

5x-5x-12=3

-12 doesnt = 3

no solution

It is c because it's has no solution