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How to solve simultaneous equations.

Normally, an equation has a single solution when it contains only one undefined variable.  For example, take the equation 3x + 7 = 19.
 
3x + 7 = 19     [original equation]
3x = 12     [subtracted 7 from both sides]
x = 4     [divided both sides by 3]
 
This is one case of a larger trend in algebra.  As I've already said, you can solve an equation for one answer when it contains a single variable.  However, this is derived from the larger rule that you can solve a set of equations where there are as many distinct equations as there are variables.  These are called simultaneous equations, and occur any time that two equations are both true over a certain domain.  In the more practical sense, this is what you should do if an exam asks you to solve for a value and gives you two different equations to use.
 
To solve simultaneous equations, we can use three strategies.
  1. Addition
  2. Subtraction
  3. Substitution
 
The first two strategies are easier, faster, and cleaner to use.  Unfortunately, they don't always work.  By contrast, using substitution is a more time consuming method and won't always be as clean-looking, but will always work.
 
Addition and subtraction work when you compare the two equations and see the same value appear in both.  Let's look at an example.
 
4x+3y=23
7x-3y=-1
 
In these, we see that the value 3y appears in both equations.  In this case, it's positive in one equation and negative in the other.  When we see one positive and one negative value, we use Strategy 1: Addition.  To do this, we add the left sides of both equations, add the right side of the two equations, and set the two sums equal to each other.  Let's take a look:
 
Left sides
4x+3y+(7x-3y)
4x+3y+7x-3y     [distributed across the quantity (7x-3y)]
11x+3y-3y     [algebraic combination of 4x+7x]
11x     [algebraic combination of 3y-3y]
 
Right sides
23+(-1)
23-1
22
 
And now we set the two values equal and solve for the remaining variable:
 
11x = 22
x = 2
 
But we're not done yet!  We've solved for x, and now we can use that to solve for y.  What we do here is take one of the two original equations and plug in the value x=2 to solve for y.  It doesn't matter which equation we use, so I'll use the first simply because it looks easier to work with.
 
4x+3y=23     [original equation]
4(2)+3y=23     [plugged in the value x=2]
8+3y=23     [multiplied 4×2]
3y=15     [subtracted 8 from both sides
y=5     [divided both sides by 3]
 
That gives us our final answer: the ordered pair (2,3).  If you were to plot the two equations on the same graph, you would find that the point (2,3) is where the two lines overlap.
 
So that's one strategy.  What about the others?  We'll look at subtraction next.  This strategy works almost the same way, with one exception.  It's used when the identical term in both equations is either positive in both equations or negative in both equations.  Take this pair of equations:
 
2x-4y=6
7x-4y=31
 
We see 4y in both equations, and both times the value is negative.  This tells us to use subtraction.  The process is almost identical to the previously described addition method with the exception that we're subtracting values.  It's important to note that we can subtract the first equation from the second or the second equation from the first, so long as we do the same operation to both sides.  In this case, I'm going to subtract the first from the second simply because the second equation has larger values.
 
Left side
7x-4y-(2x-4y)     [equation 2 - equation 1]
7x-4y-2x+4y     [distributed over the quantity (2x-4y)]
5x-4y+4y     [algebraic combination of 7x-2x]
5x     [algebraic combination of -4y+4y]
 
Right side
31-6     [equation 2 - equation 1]
25
 
Set the two sides equal to each other, and solve for the remaining variable:
 
5x = 25
x = 5     [divided both sides by 5]
 
Again, we now take the value x=5 and plug it in to one of the original equations.  It doesn't matter which equation we use, so I'll arbitrarily choose the first.
 
2x-4y=6     [original equation]
2(5)-4y=6     [plugged in the value x=5]
10-4y=6     [multiplied 2×5]
-4y=-4     [subtracted 10 from both sides]
y=1     [divided both sides by -4]
 
Again, we have an ordered pair: (5,1).  As with the previous problem, if we plotted the two equations on one graph, the point (5,1) would be where they intersect.
 
The last method of solving a pair of simultaneous equations is substitution.  It would also have worked for either of the first two problems, but we didn't use it simply because the first two methods are easier and faster.  Where we need to use substitution is cases that don't have an identical variable in both equations.  Let's take this set of equations:
 
3x+5y=38
8x-2y=40
 
Seeing that we don't have any shared values between the two equations, we need to use substitution.  There are a few steps involved here.  First, we need to choose either of the equations to start with.  I'm going to use the second one for the reason that its numbers are all multiples of 2, which will make it easier to manipulate.  Ultimately, I could have used the first equation successfully as well.  The second step is to isolate the value x or y.  I'm going to isolate y because at first glance, doing so will mean I don't have to deal with fractions, which makes my life easier.
 
8x-2y=40     [original equation]
-2y=40-8x     [subtracted 8x from both sides]
y=-20+4x     [divided both sides by -2]
y=4x-20     [rearranged the right side to make it easier to work with]
 
Now we'll take this equation, and substitute it into the other equation for y.  In this case the other equation means the first.  If I solved for x, I would substitute in for x and things still would have worked out fine.
 
3x+5y=38     [original equation]
3x+5(4x-20)=38     [substituted the quantity (4x-20) for y]
3x+20x-100=38     [distributed 5 across the quantity (4x-20)]
23x-100=38     [algebraic combination of 3x+20x]
23x=138     [added 100 to both sides]
x=6     [divided both sides by 23]
 
Don't forget, we're only halfway done!  Now we take the value x=6 and substitute it into one of the original equations to solve for y.  As before, it doesn't matter which of the two original equations so I'll arbitrarily choose the first.
 
3x+5y=38     [original equation]
3(6)+5y=38     [plugged in the value x=6]
18+5y=38     [multiplied 3×6]
5y=20     [subtracted 18 from both sides]
y=4     [divided both sides by 5]
 
And there we are.  Our final answer is the point (6,4).  Like in the previous examples, this is the point where the two lines intersect if the equations were plotted on a graph.
 
There is one exception to the rule about choosing one of the three methods.  If you can manipulate or both equation so that they have an identical value, you can use either addition or subtraction instead of substitution.  For example, take these simultaneous equations:
 
x-5y=40
7x-2y=9
 
In this case, substitutions would be the best method, particularly because of how easy it would be to isolate the variable x in the first equation.  However, if you wanted, you could also manipulate one or both equations so that they contain an identical variable.  Looking at both equations I immediately see that multiplying the first equation by 7 would give us the value 7x in both equations.  So let's do that.  Remember, you need to multiply both sides of the equation by the same value to preserve its validity.
 
7(x-5y)=7(40)
7x-35y=280
 
Which now gives us a new pair of "original" equations.
 
7x-35y=280
7x-2y=9
 
These equations may now be solved using subtraction.

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