Answer:
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.
