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

The last one bc that's a calendar