One common method for solving systems is called substitution,
because information from
one equation is substituted into other equations.
The technique of substitution can be used for a wide variety of systems:
any number of equations or inequalities, any number of unknowns, linear or nonlinear.
However, in this exercise, the technique is only applied to systems of two linear equations
in two unknowns.
The technique of substitution works best when
one of the equations can easily be solved
for one of the variables
.
For example, maybe the first equation can easily be solved for $\,x\,$.
Or, perhaps the second equation can easily be solved for $\,y\,$.
Remember:
to solve for a variable means to get it all by itself, on one side of the equation.
The variable you're solving for is not allowed to appear on the other side.
Here are some systems which are ideally suited to the substitution technique:
Here are the general steps for solving a system of two equations in two unknowns using substitution:
$5x+7( \overset{y}{\overbrace{3x1}})=4$ 
The first equation ($\,y=3x1\,$) is already solved for $\,y\,$. We'll call this the original equation. Substitute this expression for $\,y\,$ into the second equation ($\,5x+7y=4\,$). You now have an equation with only one variable ($\,x\,$). 
$ \begin{gather} \cssId{s47}{5x+21x7=4}\cr \cssId{s48}{26x=11}\cr \cssId{s49}{x=\frac{11}{26}} \end{gather} $  Solve this resulting equation for $\,x\,$. 
$ \begin{gather} \cssId{s51}{y=3\bigl( \overset{x}{\overbrace{\frac{11}{26}}}\bigr)  1}\cr\cr \cssId{s52}{y =\frac{33}{26} \frac{26}{26}}\cr\cr \cssId{s53}{y =\frac{7}{26}} \end{gather} $ 
Substitute this value of $\,x\,$ back into the original equation. Simplify. 
The unique solution is the ordered pair: $$ \cssId{s57}{\bigl(\frac{11}{26},\frac{7}{26}\bigr)} $$  Clearly report the solution. 
Go to wolframalpha.com and type this in
(you can just cutandpaste):
y=3x1, 5x+7y=4
You'll see the two lines graphed, and the solution clearly labeled and reported.
How easy is that!?
$ \begin{gather} \cssId{s71}{q+7u=4}\cr \cssId{s72}{q = 4  7u} \end{gather} $ 
The first equation is simpler. Since $\,q\,$ has a coefficient of $\,1\,$, it's easy to solve for $\,q\,$. Thus, solve the first equation for $\,q\,$. 
$ 4( \overset{q}{\overbrace{47u}})+20=28u $ 
Substitute this expression for $\,q\,$ into the second equation. You now have an equation with only one variable ($\,u\,$). 
$ \begin{gather} \cssId{s79}{16+28u+20=28u}\cr \cssId{s80}{36=0} \end{gather} $ 
Simplify. The resulting equation is always false. There are no solutions. These two lines are parallel. 
Go to wolframalpha.com and type this in
(you can just cutandpaste):
q+7u=4, 4q+20=28u
You'll see the two parallel lines, and you're told that no solutions exist.
Voila!
An equivalent form for this system makes it easier to see that there are no solutions.
Start by renaming the second equation:
$4q+20=28u$  second equation 
$4q28u=20$ 
subtract $\,28u\,$ from both sides; subtract $\,20\,$ from both sides 
$q+7u=5$  divide both sides by $\,4\,$ 
Now, put this equivalent second equation together with the first equation,
to get an equivalent form for the system:
$
\begin{gather}
\cssId{s101}{q + 7u = 4}\cr
\cssId{s102}{q + 7u = 5}
\end{gather}
$
Different names are good for different purposes!
This ‘new name’ for the system makes it easy to see why there are no solutions.
The expression $\,q+7u\,$ cannot simultaneously equal $\,4\,$ and $\,5\,$.
Behold the power of renaming!
$ \begin{gather} \cssId{s114}{t9=4s3}\cr \cssId{s115}{t=4s+6}\end{gather} $ 
The second equation is simpler. Solve it for $\,t\,$. 
$ 2( \overset{t}{\overbrace{4s+6}})=8s+12 $ 
Substitute this expression for $\,t\,$ into the first equation. You now have an equation with only one variable ($\,s\,$). 
$ \begin{gather} \cssId{s121}{8s+12 = 8s+12}\cr \cssId{s122}{0 = 0} \end{gather} $ 
Simplify. The resulting equation is always true. There are infinitely many solutions. 
All points of the form $\,(s,4s+6)\,$ are solutions. 
Report the entire line of solutions. 
Go to wolframalpha.com and type this in
(you can just cutandpaste):
2t=8s+12, t9=4s3
You'll see only one line, but the legend shows that two lines are being graphed.
This is your clue that there is an entire line of solutions in this case!
Again, an equivalent form for this system makes it easy to see that there are infinitely many solutions.
Let's write both of the equations in the form $\,at + bs + c = 0\,$
(with no common factors for $\,a\,$, $\,b\,$, and $\,c\,$):
$2t = 8s + 12$  first equation 
$2t  8s  12 = 0$  get zero on the righthand side 
$t  4s  6 = 0$  divide both sides by $\,2\,$ 
$t  9 = 4s  3$  second equation 
$t  4s  6 = 0$  get zero on the righthand side 
Now, put these equivalent versions of the two equations together
to get an equivalent form for the system:
$
\begin{gather}
\cssId{s147}{t  4s  6 = 0}\cr
\cssId{s148}{t  4s  6 = 0}
\end{gather}
$
Now it's clear what's going on!
Any choices for $\,s\,$ and $\,t\,$ that make the first equation true are also going to make the second equation true,
since they're precisely the same equation.
Thus, the solution set is the entire line $\,t  4s  6 = 0\,$.
You'll learn another method for solving
systems in the next web exercise:
Solving Systems using Elimination
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
