Minimize $$2{x}_{1}+3{x}_{2}$$

Subject to

$$\frac{1}{2}{x}_{1}+\frac{1}{4}{x}_{2}\le 6$$

$${x}_{3}+{x}_{2}\ge 2$$

$${x}_{1}+{x}_{2}=10$$

$${x}_{1}\ge 0,{x}_{2}\ge 0$$

My solution:

$0.5{x}_{1}+0.25{x}_{2}+{x}_{3}=6$

$-{x}_{1}-3{x}_{2}+{x}_{4}=-2$

${x}_{1}+{x}_{2}=10$

Is this correct?