simultaneous equation using the substitute method I have the following 2 equations:
${6x + 9y = 3}$
${6x -3y = -2}$
The textbook asks to use the substitution method so I would appreciate if we stuck to this method, I could use the addition method but the book asks for this method.
So my steps of doing this are to isolate x in the first equation:
${6x + 9y = 3}$
I then divide both sides by 3 to give
${2x + 3y = 1}$
I then get:
${x = {-3y + 1\over 2}}$
I'm not entirely sure you to take this further using the substitute method.
 A: I prefer short, sweet, and direct methods when they make themselves possible. For example your problem could have quickly reduced to one with one variable to solve for that first variable. The second variable is also quickly solved once we know the value for the first.
Your system is:
$$Eq1: {6x + 9y = 3}$$
$${Eq2: 6x - 3y = -2}$$
Notice how we can immediately solve for $6x$ in either equation and insert that into the other equation. This qualifies as a substitution method.
For example lets solve for $6x$ in $Eq2$ and insert that into $Eq1$
$$6x=3y-2$$
$$3y-2 + 9y=3$$
Now that we have the system reduced to one equation in one variable therefore we can solve for $y$:
$$12y=5$$
$$y=\frac{5}{12}$$
Now take either equation and insert $y$ to determine $x$. We shall use $Eq1$:
$${6x + 9y = 3}$$
$${6x = 3 - 9y}$$
$$x=\frac{3-9y}{6}=\frac{1}{2}-\frac{9(\frac{5}{12})}{6}=\frac{12}{24}-\frac{15}{24}=\frac{-3}{24}$$
$$x=\frac{-1}{8}$$
Our solution set is the vector $[\frac{-1}{8} \frac{5}{12}]$.
See that wasn't difficult at all.
A: Notice, we have $$6x+9y=3\tag 1$$
$$6x-3y=-2\tag 2$$
From, eq(1), we get $$y=\frac{3-6x}{9}=\frac{1-2x}{3}$$ substituting this value of $x$ in (2), we get $$6x-3\left(\frac{1-2x}{3}\right)=-2$$
$$6x-(1-2x)=-2$$ $$8x=-2+1=-1$$ $$x=-\frac{1}{8}$$
Now, substituting the value of $x$, we get $$y=\frac{1-2x}{3}=\frac{1-2\left(-\frac{1}{8}\right)}{3}=\frac{5}{12}$$
Hence, the solution is 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\color{blue}{-\frac{1}{8}}, \ y=\color{blue}{\frac{5}{12}}}}$$
A: plugging the equation $$x=\frac{1-3y}{2}$$ in the second one we obtain
$$6\left(\frac{1-3y}{2}\right)-3y=-2$$ multiplying by $2$ we get
$$6(1-3y)-6y=-4$$ expanding
$$6-18y-6y=-4$$
does this help you?
A: $x = \frac{-3y + 1}{2}$ as you say. You put this into the second equation and get that $6(\frac{-3y + 1}{2}) -3y = -2 \iff -12y + 3 = -2 \iff y = \frac{5}{12} \implies x = - \frac{1}{8}$.
You can check these solutions by putting values of $x$ and $y$ into both equations.
