Question

Solve the system using substitution method
3x+y=6
2x-4y=10

Answer 1

`\begin{cases}&3x + y = 6 \\&2x - 4y = 10\end{cases}`

Multiply 4 to the first equation:

`\begin{cases}&12x + 4y = 24 \\&2x - 4y = 10\end{cases}`

ADD Equation 1 to Equation 2 and find x:

`\begin{aligned}&14x =34 \\&x= 34/14 \\& x = 17/7\end{aligned}`

Substitute x = 17/7 and find y:

`\begin{aligned}&3x + y = 6 \\&3 (17/7) + y = 6 \\& 51/7 + y = 6 \\&y = -9/7\end{aligned}`

Answer: x = 17/7, y = -9/7

Answer 2

`\left\{\begin{array}{ccc}3x+y=6&|-3x\\2x-4y=10\end{array}\right\\\\\left\{\begin{array}{ccc}y=6-3x\\2x-4y=10\end{array}\right\\\\substitute\ y=6-3x\ to\ the\ second\ equation\\\\2x-4(6-3x)=10\\\\2x-4\cdot6-4\cdot(-3x)=10\\\\2x-24+12x=10\\\\14x-24=10\ \ \ |+24\\\\14x=34\ \ \ \ |:14\\\\x=(34)/(14)\to x=(17)/(7)\\\\substitute\ the\ value\ of\ x\ to\ first\ equation\\\\y=6-3\cdot(17)/(7)=6-(51)/(7)=(42)/(7)-(51)/(7)=-(9)/(7)`


`\boxed{\left\{\begin{array}{ccc}x=(17)/(7)\\\\y=-(9)/(7)\end{array}\right}`