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Simultaneous Equations Homework -- The “Generic” Simultaneous Equations

Module by: Kenny M. Felder. E-mail the author

Summary: Generalizing simultaneous equations.

Here are the generic simultaneous equations.

  • ax + by = e ax + by = e size 12{ ital "ax"+ ital "by"=e} {}
  • cx + dy = f cx + dy = f size 12{ ital "cx"+ ital "dy"=f} {}

I call them “generic” because every possible pair of simultaneous equations looks exactly like that, except with numbers instead of aa size 12{a} {}, bb size 12{b} {}, cc size 12{c} {}, dd size 12{d} {}, ee size 12{e} {}, and ff size 12{f} {}. We are going to solve these equations.

Very important!!!

When I say “solve it” I mean find a formula x=x= size 12{x={}} {}blah-blah where the blah-blah has only aa size 12{a} {}, bb size 12{b} {}, cc size 12{c} {}, dd size 12{d} {}, ee size 12{e} {}, and ff size 12{f} {}: no xx size 12{x} {} or yy size 12{y} {}. And, of course, y=y= size 12{y={}} {} some different formula with only aa size 12{a} {}, bb size 12{b} {}, cc size 12{c} {}, dd size 12{d} {}, ee size 12{e} {}, and ff size 12{f} {}. If we can do that, we will be able to use these formulas to immediately solve any pair of simultaneous equations, just by plugging in the numbers.

We can solve this by elimination or by substitution. I am going to solve for yy size 12{y} {} by elimination. I will use all the exact same steps we have always used in class.

Step 1: Make the coefficients of x line up

To do this, I will multiply the top equation by cc size 12{c} {} and the bottom equation by aa size 12{a} {}.

acx + bcy = ec acx + bcy = ec size 12{ ital "acx"+ ital "bcy"= ital "ec"} {}

acx + ady = af acx + ady = af size 12{ ital "acx"+ ital "ady"= ital "af"} {}

Step 2: Subtract the second equation from the first

This will make the xx size 12{x} {} terms go away.

acx + bcy = ec ( acx + ady = af ) ̲ bcy ady = ec af acx + bcy = ec ( acx + ady = af ) ̲ bcy ady = ec af alignl { stack { size 12{" " ital "acx"+ ital "bcy"= ital "ec"} {} # size 12{ {underline { - \( "acx"+ ital "ady"= ital "af" \) }} } {} # size 12{" " ital "bcy" - ital "ady"= ital "ec" - ital "af"} {} } } {}

Step 3: Solve for y

This is something we’ve done many times in class, right? First, pull out a yy size 12{y} {}; then divide by what is in parentheses.

bcy ady = ec af bcy ady = ec af size 12{ ital "bcy" - ital "ady"= ital "ec" - ital "af"} {}

y = ( bc ad ) = ec af y = ( bc ad ) = ec af size 12{y= \( ital "bc" - ital "ad" \) = ital "ec" - ital "af"} {}

y = ec af bc ad y = ec af bc ad size 12{y= { { ital "ec" - ital "af"} over { ital "bc" - ital "ad"} } } {}

So what did we do?

We have come up with a totally generic formula for finding yy size 12{y} {} in any simultaneous equations. For instance, suppose we have…

3x + 4y = 18 3x + 4y = 18 size 12{3x+4y="18"} {}

5x + 2y = 16 5x + 2y = 16 size 12{5x+2y="16"} {}

We now have a new way of solving this equation: just plug into y=ecafbcady=ecafbcad size 12{y= { { ital "ec" - ital "af"} over { ital "bc" - ital "ad"} } } {}. That will tell us that y=(18)(5)(3)(16)(4)(5)(3)(2)=9048206=4214=3y=(18)(5)(3)(16)(4)(5)(3)(2)=9048206=4214=3 size 12{y= { { \( "18" \) \( 5 \) - \( 3 \) \( "16" \) } over { \( 4 \) \( 5 \) - \( 3 \) \( 2 \) } } = { {"90" - "48"} over {"20" - 6} } = { {"42"} over {"14"} } =3} {}

Didja get it?

Here’s how to find out.

  1. Do the whole thing again, starting with the generic simultaneous equations, except solve for xx size 12{x} {} instead of yy size 12{y} {}.
  2. Use your formula to find xx size 12{x} {} in the two equations I did at the bottom (under “So what did we do?”)
  3. Test your answer by plugging your xx size 12{x} {} and my y=3y=3 size 12{y=3} {} into those equations to see if they work!

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