Question

Solve the system.

2/3 x - 1 /2 y = 1
1/ 4 x + 3/8 y = -1
A) (3, 2)
B) (9, 10)
C) ( 9/ 2 , 5)
D) (- 1 /3 , - 22 /9 )

Answer 1

`\large\boxed{D)\ \left(-(1)/(3),\ -(22)/(9)\right)}`

Explanation:

`\left\{\begin{array}{ccc}(2)/(3)x-(1)/(2)y=1&\text{multiply both sides by 6}\\(1)/(4)x+(3)/(8)y=-1&\text{multiply both sides by 8}\end{array}\right\\\left\{\begin{array}{ccc}6\!\!\!\!\diagup^2\cdot(2)/(3\!\!\!\!\diagup_1)x-6\!\!\!\!\diagup^3\cdot(1)/(2\!\!\!\!\diagup_1)y=6\cdot1\\8\!\!\!\!\diagup^2\cdot(1)/(4\!\!\!\!\diagup_1)x+8\!\!\!\!\diagup^1\cdot(3)/(8\!\!\!\!\diagup_1)y=8\cdot(-1)\end{array}\right`

`\underline{+\left\{\begin{array}{ccc}4x-3y=6\\2x+3y=-8\end{array}\right}\qquad\text{add both saides of the equations}\\.\qquad6x=-2\qquad\text{divide both sides by 6}\\.\qquad x=-(2)/(6)\\\\.\qquad\boxed{x=-(1)/(3)}\\\\\text{Put the value of x to the second equation:}\\\\2\left(-(1)/(3)\right)+3y=-8\\\\-(2)/(3)+3y=-8\qquad\text{add}\ (2)/(3)\ \text{to both sides}\\\\3y=-(24)/(3)+(2)/(3)`

`3y=-(22)/(3)\qquad\text{divide both sides by 3}\ /multiply\ both\ sides\ by\ (1)/(3)/\\\\\boxed{y=-(22)/(9)}`