Question

Please solve this question by using the strategy Elimination Method or Solve By Substitution. This is the math equation: 1/2x+y=15 and -x-1/3y=-6
2nd Question: 5/6x+1/3y=0 and 1/2x-2/3y=3

Answer 1

Answer with explanation:

First pair of equations :

$\dfrac{1}{2}x+y=15\ ..(i)\\\\-x-\dfrac{1}{3}y=-6\ ..(ii)$

Multiply 2 to equation (i), we get

$x+2y=30\ ..(iii)$

By Elimination Method, Add (i) and (ii) (term with x eliminate), we get

$2y-\dfrac{1}{3}y=30-6\\\\\Rightarrow\ \dfrac{5}{3}y=24\\\\\Rightarrow\ y=\dfrac{24\times3}{5}=14.4$

put y= 14.4 in (iii), we get

$x+2(14.4)=30\Rightarrow\ x=30-28.8=1.2$

hence, x=1.2 and y =14.4

Second pair of equations :

$\dfrac{5}{6}x+\dfrac13y=0\ ..(i)\\\\ \dfrac12x-\dfrac{2}{3}y=3\ ..(ii)$

Multiply 2 to equation (i), we get

$\dfrac{5}{3}x+\dfrac{2}{3}y=0\ ..(iii)$

Elimination Method, Add (i) and (ii) (term with y eliminate) , we get

$\dfrac53x+\dfrac12x=3\Rightarrow\ \dfrac{10+3}{6}x=3\\\\\Rightarrow\ \dfrac{13}{6}x=3\\\\\Rightarrow\ x=\dfrac{18}{13}$

put $x=\dfrac{18}{13}$   in (i), we get

$\dfrac{5}{6}(\dfrac{18}{13})+\dfrac{1}{3}y=0\\\\\Rightarrow\ \dfrac{15}{13}+\dfrac{1}{3}y=0\\\\\Rightarrow\ \dfrac{1}{3}y=-\dfrac{15}{13}\\\\\Rightarrow\ y=-\dfrac{45}{13}$

hence, $x=\dfrac{18}{13}$   and $y=\dfrac{-45}{13}$ .