Answer: C) [tex](-\dfrac{1}{2},\dfrac{3}{2})[/tex]
Step-by-step explanation:
The given system of equations :
[tex]2(y-x) = 5+2x\ \ ...(i)\\\\ 2(y+x)=5-2y\ \ ..(ii)[/tex]
Simplify left side, we get
[tex]2y-2x=5+2x\Rightarrow\ 2y-4x=5\ \ ...(iii)\\\\ 2y+2x=5-2y\Rightarrow\ 4y+2x=5\ \ ...(iv)[/tex]
Multiplying 2 on equation (iii), we get
[tex]4y-8x=10\ \ ...(v)[/tex]
Subtracting (v) from (iv) , we get
[tex]2x-(-8x)=5-10\\\\\Rightarrow\ 2x+8x=-5\\\\\Rightarrow\ 10x=-5\\\\\Rightarrow\ x=-\dfrac{5}{10}=-\dfrac{1}{2}[/tex]
Put value of [tex]x=-\dfrac{1}{2}[/tex] in (v), we get
[tex]4y-8(-\dfrac{1}{2})=10\\\\\Rightarrow \ 4y+4=10\\\\\Rightarrow\ 4y=10-4=6\\\\\Rightarrow\ y=\dfrac{6}{4}=\dfrac{3}{2}[/tex]
hence, the solution to the system is [tex](x,y)=(-\dfrac{1}{2},\dfrac{3}{2})[/tex]
