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4 August, 14:13

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

+2
Answers (2)
  1. 4 August, 15:53
    0
    It is c because it's has no solution
  2. 4 August, 17:07
    0
    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
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