Maria's problem from Facebook (system of equations)

[Jameson]

Hello :) I'm learning about systems of equations and substituting and stuck on this problem. It seems pretty simple but I keep getting trampled by small things and I can't find out what I'm doing wrong.

6y=x+18
2y-x=6

 

[Jameson]

Hi Maria,

There are a few ways to solve this problem, so I'll just pick the one that looks the quickest.

We have (1) $6y=x+18$ and (2) $2y-x=6$. If add (1) and (2) we get $8y=24$ which means that $y=3$. Now let's use this fact to find $x$, but substituting $3$ for $y$ in equation (1). $6(3)=x+18$ means $x=0$.

$x=0,y=3$ is the solution.

Jameson

 

[soroban]

Hello, Maria!

I'm learning about systems of equations and substituting.

. . $\begin{array}{cc} 6y\:=\:x+18 & [1] \\ 2y-x\:=\:6 & [2] \end{array}$

Solve [1] for $x\!:\;\;x \:=\:6y - 18\;\;[3]$

Substitute into [2]: .$2y - (6y-18) \:=\:6 \quad\Rightarrow\quad 2y - 6y + 18 \:=\:6$

. . . . . . . . . . . . . . $-4y \:=\:-12 \quad\Rightarrow\quad \boxed{y \:=\:3}$

Substitute into [3]: .$x \:=\:6(3) - 18 \quad\Rightarrow\quad \boxed{x \:=\:0}$