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This module shows how it can be helpful, on occasion, to use letters as numbers in order to quickly find solutions to variations on a simultaneous equation problem.

Toward the end of this chapter, there are some problems in substitution and elimination where letters are used in place of numbers . For instance, consider the following problem:

2y ax = 7 size 12{2y - ital "ax"=7} {}
4y + 3 ax = 9 size 12{4y+3 ital "ax"=9} {}

What do we do with those "a"s? Like any other variable, they simply represent an unknown number. As we solve for x size 12{x} {} , we will simply leave a size 12{a} {} as a variable.

This problem lends itself more naturally to elimination than to substitution, so I will double the top equation and then subtract the two equations and solve.

4y 2 ax = 14 4y + 3 ax = 9 ̲ 0y 5 ax = 5 size 12{ matrix { 4y - 2 ital "ax"="14" {} ##{underline { - left (4y+3 ital "ax"=9 right )}} {} ## 0y - 5 ital "ax"=5} } {}
x = 5 5a = 1 a size 12{x= { {5} over { - 5a} } = { { - 1} over {a} } } {}

As always, we can solve for the second variable by plugging into either of our original equations.

2y a 1 a = 7 size 12{2y - a left ( { { - 1} over {a} } right )=7} {}
2y + 1 = 7 size 12{2y+1=7} {}
y = 3 size 12{y=3} {}

There is no new math here, just elimination. The real trick is not to be spooked by the a , and do the math just like you did before.

And what does that mean? It means we have found a solution that works for those two equations, regardless of a. We can now solve the following three problems (and an infinite number of others) without going through the hard work.

If a = 5 , If a = 10 , If a = - 3 ,
The original equations become: The original equations become: The original equations become:
2y 5x = 7 4y + 15 x = 9 size 12{ matrix { 2y - 5x=7 {} ##4y+"15"x=9 } } {} 2y 10 x = 7 4y + 30 x = 9 size 12{ matrix { 2y - "10"x=7 {} ##4y+"30"x=9 } } {} 2y + 3x = 7 4y - 9x = 9 size 12{ matrix { 2y - 3x=7 {} ##4y+9x=9 } } {}
And the solution is: And the solution is: And the solution is:
x = 1 5 , y = 3 size 12{x= { { - 1} over {5} } ,y=3} {} x = 1 10 , y = 3 size 12{x= { { - 1} over {"10"} } ,y=3} {} x = 1 3 , y = 3 size 12{x= { {1} over {3} } ,y=3} {}

The whole point is that I did not have to solve those three problems—by elimination, substitution, or anything else. All I had to do was plug a size 12{a} {} into the general answer I had already found previously. If I had to solve a hundred such problems, I would have saved myself a great deal of time by going through the hard work once to find a general solution!

Mathematicians use this trick all the time. If they are faced with many similar problems, they will attempt to find a general problem that encompasses all the specific problems, by using variables to replace the numbers that change. You will do this in an even more general way in the text, when you solve the “general” simultaneous equations where all the numbers are variables. Then you will have a formula that you can plug any pair of simultaneous equations into to find the answer at once. This formula would also make it very easy, for instance, to program a computer to solve simultaneous equations (computers are terrible at figuring things out, but they’re great at formulas).

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Source:  OpenStax, Advanced algebra ii: conceptual explanations. OpenStax CNX. May 04, 2010 Download for free at http://cnx.org/content/col10624/1.15
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