Answer with explanation:
First pair of equations :
[tex]\dfrac{1}{2}x+y=15\ ..(i)\\\\-x-\dfrac{1}{3}y=-6\ ..(ii)[/tex]
Multiply 2 to equation (i), we get
[tex]x+2y=30\ ..(iii)[/tex]
By Elimination Method, Add (i) and (ii) (term with x eliminate), we get
[tex]2y-\dfrac{1}{3}y=30-6\\\\\Rightarrow\ \dfrac{5}{3}y=24\\\\\Rightarrow\ y=\dfrac{24\times3}{5}=14.4[/tex]
put y= 14.4 in (iii), we get
[tex]x+2(14.4)=30\Rightarrow\ x=30-28.8=1.2[/tex]
hence, x=1.2 and y =14.4
Second pair of equations :
[tex]\dfrac{5}{6}x+\dfrac13y=0\ ..(i)\\\\ \dfrac12x-\dfrac{2}{3}y=3\ ..(ii)[/tex]
Multiply 2 to equation (i), we get
[tex]\dfrac{5}{3}x+\dfrac{2}{3}y=0\ ..(iii)[/tex]
Elimination Method, Add (i) and (ii) (term with y eliminate) , we get
[tex]\dfrac53x+\dfrac12x=3\Rightarrow\ \dfrac{10+3}{6}x=3\\\\\Rightarrow\ \dfrac{13}{6}x=3\\\\\Rightarrow\ x=\dfrac{18}{13}[/tex]
put [tex]x=\dfrac{18}{13}[/tex] in (i), we get
[tex]\dfrac{5}{6}(\dfrac{18}{13})+\dfrac{1}{3}y=0\\\\\Rightarrow\ \dfrac{15}{13}+\dfrac{1}{3}y=0\\\\\Rightarrow\ \dfrac{1}{3}y=-\dfrac{15}{13}\\\\\Rightarrow\ y=-\dfrac{45}{13}[/tex]
hence, [tex]x=\dfrac{18}{13}[/tex] and [tex]y=\dfrac{-45}{13}[/tex] .
