Solve the following systems of equations by elimination. Verify the solutions. The first question I came across was $r + 2s + 1 = 0$ and $r + 5s + 28 = 0$ and I had no trouble solving this and I verified the solution of s = -9 and r = 15.
The next system is $4m - 3n = 27$ and $8m - 6n = 18$, which solving and I found that there is no solution.
The third is $0.6x=1.2+0.3y$ and $2.2x-1.8y-1.6=0$. The first step I think I must do is rearrange the equation so $0.6x=1.2+0.3y$ looks like $0.6x+0.3y=1.2$.
I have no problem with solving systems using substitution, but using elimination I'm having some difficulty with. If someone could explain how to solve the last two systems it would be greatly appreciated. Thank you!
 A: We have $0.6x=1.2+0.3y$ and $2.2x-1.8y-1.6=0$.
Multiply the equation on the left by $6$ to get $3.6x=7.2+1.8y$; rearrange to get $1.8y=3.6x-7.2$
Now from the other equation we had $1.8y=2.2x-1.6$
Thus $3.6x-7.2=2.2x-1.6$
So $1.4x=5.6$
etc.
A: Let me explain an easier way in details so as you would be able to solve the system of linear equations in two variables easily without re-arrangement.  
Notice, the linear equation in two variables say $x$ & $y$ shows the equation of a straight line. 
In general, the solution of the system of linear equations in two variables: $$a_1x+b_1y=c_1$$ $$a_2x+b_2y=c_2$$
is given by Cramer's rule as follows 
$$\frac{x}{\left|\begin{array}\\c_1&b_1\\c_2&b_2 \end{array}\right|}=\frac{y}{\left|\begin{array}\\a_1&c_1\\a_2&c_2 \end{array}\right|}=\frac{1}{\left|\begin{array}\\a_1&b_1\\a_2&b_2 \end{array}\right|}$$


*

*if the determinant $\left|\begin{array}\\a_1&b_1\\a_2&b_2 \end{array}\right|\neq0$ The system will have a unique solution as the lines represented by the equations are intersecting each other at a unique point. 

*if the determinant $\left|\begin{array}\\a_1&b_1\\a_2&b_2 \end{array}\right|=0$
Then the system will have no solution as the lines represents by the equations are parallel to each other i.e. they do not intersect to each other.
We have $$0.6x=1.2+0.3y\iff 0.6x-0.3y=1.2$$ 
$$2.2x-1.8y-1.6=0\iff 2.2x-1.8y=1.6$$ 
Using Cramer's rule & substituting the corresponding values, we get
$$\frac{x}{\left|\begin{array}\\1.2&-0.3\\1.6&-1.8 \end{array}\right|}=\frac{y}{\left|\begin{array}\\0.6&1.2\\2.2&1.6 \end{array}\right|}=\frac{1}{\left|\begin{array}\\0.6&-0.3\\2.2&-1.8 \end{array}\right|}$$
$$\frac{x}{-1.68}=\frac{y}{1.68}=\frac{1}{-0.42}$$ $$x=\frac{-1.68}{-0.42}\ \ \vee \ \ \ y=\frac{1.68}{-0.42}$$
$$\color{red}{x=4}\ \ \vee \ \ \ \color{red}{y=-4}$$
A: Elimination works as follows in the general case, for
$$a'x+b'y=c',\\a''x+b''y=c''.$$
You normalize the equations so that the coefficient of $x$ is $1$:
$$x+\frac{b'}{a'}y=\frac{c'}{a'},\\x+\frac{b''}{a''}y=\frac{c''}{a''}.$$
And you subtract the first from the second to let $x$ disappear:
$$x+\frac{b'}{a'}y=\frac{c'}{a'},\\\left(\frac{b''}{a''}-\frac{b'}{a'}\right)y=\frac{c''}{a''}-\frac{c'}{a'}.$$
The system is now said to be in triangular form (the array of coefficients was a square and you reduced it to a triangle).
You can compute $y$, and then get $x$ from the first equation. This process is called backsubstitution.
This is a systematic approach that generalizes to systems of larger size. It is much more efficient than Cramer's rule in terms of the number of operations to be performed.
[In practice, you only normalize the first equation to spare a few divisions. You can also reduce to a pure diagonal form with extra combinations, which will eliminate the backsubstitution phase, but this is not the preferred method.]
See https://en.wikipedia.org/wiki/Gaussian_elimination.

Application:
$$4m - 3n = 27,\\8m - 6n = 18$$
then
$$m - \frac34n = \frac{27}4,\\m - \frac34n = \frac94$$
then
$$m - \frac34n = \frac{27}4,\\- 0n = -\frac92$$
which is not possible.
