Answer
- The solution is: $(x , \frac {2}{3}x)$; x arbitrary, or $(\frac {3} 2 y, y)$; y arbitrary.
Work Step by Step
1. Notice that, if we add the two equations, neither $x$ or $y$ will be eliminated. But, if we multiply the first equation by -12, the $x$ unknown will be eliminated:
$-12(\frac 1 2x - \frac 3 4y) = -12(0) $
$6x - 9y = 0$
$-6x + 9y = 0$
$6x - 9y = 0$
Adding the equations:
$-6x + 6x + 9y - 9y = 0$
$0 = 0$
2. Apparently, the system is redundant. So, solve for $y$ and for $x$ in the second equation:
$6x - 9y = 0$
$6x = 9y$
$\frac 6 9 x = \frac 9y 9$
$\frac 2 3 x = y$
- Solving for $x$:
$6x = 9 y$
$\frac {6x} 6 = \frac {9y} 6$
$x = \frac {3y} 2$
- Therefore, the solution is: $(x , \frac {2}{3}x)$; x arbitrary, or $(\frac {3} 2 y, y)$; y arbitrary.
