What is the solution of this system of linear equations? 3y=3/2x+6 1/2y-1/4x=3.

No solution

Step-by-step explanation:

Please, separate equations with a comma, or the word "and," or a semicolon.

Next, determine the LCD and multiply both equations by it, so as to elimiinate the fractional coefficients.

3y=3/2x+6 1/2y-1/4x=3

should be written as the system

   3y=(3/2)x+6         Use parentheses around the fraction for clarity

(1/2)y-(1/4)x=3            Same:  use parentheses

The LCDs here are 2 (for the first equation) and 4 (for the second equation).  Multiply the first equation by 2 to eliminate the fractions:

   3y=(3/2)x+6         →   6y = 3x + 12, and

4[ (1/2)y-(1/4)x=3 ]     →   2y  -  x   = 12

Let's use the substitution method to solve this system.  Solve the 2nd equation for x, obtaining x = 2y - 12.  Substitute 2y - 12 for x in the 1st equation 6y = 3x + 12:

6y = 3(2y - 12) + 12, or

6y = 6y - 36 + 12

This reduces to 0 = -24, which is never true.  Thus,

this system has NO solution.

4th option is correct xy=-10

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the correct option is b

step-by-step explanation: