Solve by elimination.

2/3x+4y=-1
1/2x-3y=-9/4

\( \begin{array}{lrcrcrl} (1) & \frac23x &+& 4y &=& -1 \\ (2) & \frac12x &-& 3y &=& -\frac94 \\ \\ \hline \\ (1) & \frac23x &+& 4y &=& -1 \qquad &|\qquad \cdot 3\\ (2) & \frac12x &-& 3y &=& -\frac94 \qquad &|\qquad \cdot 4\\ \\ \hline \\ (1) & 2x &+& 12y &=& -3 \\ (2) & 2x &-& 12y &=& -9 \\ \\ \hline \\ (1)+(2) & 2x+ 2x &+& 12y- 12y &=& -3-9 \\ & 4x && &=& -12 \qquad &|\qquad :4\\ & x && &=& -\frac{12}{4} \\ & \mathbf{x} && & \mathbf{=}& \mathbf{-3} \\ \\ \hline \\ (1) & 2x &+& 12y &=& -3 \\ & 2\cdot(-3) &+& 12y &=& -3 \\ & -6 &+& 12y &=& -3 \qquad &|\qquad +6\\ & & & 12y &=& -3 +6\\ & & & 12y &=& 3\qquad &|\qquad :12\\ & & & y &=& \frac{3}{12}\\ & & & y &=& \frac{1}{4}\\ & & & \mathbf{y}&\mathbf{=}& \mathbf{ 0.25 }\\ \end{array}\)

2/3x+4y=-1
1/2x-3y=-9/4

Solve the following system:
{(2 x)/3+4 y = -1 |     (equation 1)
x/2-3 y = -9/4 |     (equation 2)
Subtract 3/4 × (equation 1) from equation 2:
{(2 x)/3+4 y = -1 |     (equation 1)
0 x-6 y = (-3)/2 |     (equation 2)
Multiply equation 1 by 3:
{2 x+12 y = -3 |     (equation 1)
0 x-6 y = -3/2 |     (equation 2)
Multiply equation 2 by -2/3:
{2 x+12 y = -3 |     (equation 1)
0 x+4 y = 1 |     (equation 2)
Divide equation 2 by 4:
{2 x+12 y = -3 |     (equation 1)
0 x+y = 1/4 |     (equation 2)
Subtract 12 × (equation 2) from equation 1:
{2 x+0 y = -6 |     (equation 1)
0 x+y = 1/4 |     (equation 2)
Divide equation 1 by 2:
{x+0 y = -3 |     (equation 1)
0 x+y = 1/4 |     (equation 2)
Collect results:
Answer: | 
| {x = -3
y = 1/4