**Cramer's Rule** has nothing to do with Seinfeld, guitars, or horrible movie divorces. No, this has everything to do with solving square systems using our uptight friends, **determinants**. This is how Cramer rolls. Here are two random, innocent equations, ready to be figured out by us. We want to learn the values of *x* and *y*, so:

3*x* – *y* = 5

-2*x* + 4*y* = 7

First, we are going to go with what we know and find the coefficient determinant:

D = (3)(4) – (-2)(-1) = 10

Next we will find the D of the *x* set, Dₓ. You take D as you know it and go from there:

and delete those *x* values out of there:

add in the = values and you've got D_{ₓ}:

D_{x} = (5)(4) – (7)(-1) = 27

Finally, you can find *x*. That's because

At last, our purpose becomes clear, right? Just like Hannibal, we love it when a plan comes together. Now, to find *y*, same bat procedure, same bat equations:

This time you delete the *y* values out and—you guessed it—substitute in the *=* values to get *y*.

And of course you know how to find *y*:

Looking for this?

(_{x}, _{y})

Now we know that:

There's only one *leetle* thing to remember. Remember how you learned that a fraction with a zero as a denominator is a sign of the apocalypse? In this case, since we're ultimately talking about lines (remember graphing a bajillion equations into lines?), in this case you can't get convenient *(x, y)* values because either the two lines represented by your two equations are parallel or *the same freakin line*. What a mind bender.

If we figure out our coefficient determinant is zero, that's the end of the road. We know that 0 will be our denominator for both *x* and *y*, so we can't get nice neat values.

Just when we thought we knew all there is to know about determinants, it turns out that those cute little two by two matrices are just the tip of the iceberg.

## Practice:

Use Cramer's Rule to find: D_{x}, D_{y}, *x*, and *y* 2*x* + *y* = 4 3*x* + 2*y* = 6 | |

First, we are going to find the coefficient determinant; we plug our diagonal values into the formula: D = (2)(2) – (3)(1) = 4 – 4 = 0 Because we have a D of 0 we don't need to go any further; we know we cannot get set values when D is 0 because the values will have 0 in the denominator. | |

Use Cramer's Rule to find: D_{x}, D_{y}, *x*, and *y* 3*x* – 4*y* = 8 2*x* + 2*y* = 6 | |

We find D, the coefficient determinant: We multiply down the diagonal from left to right and then subtract the value we get by multiplying up the diagonal from left to right: D = (3)(2) – (2)(-4) = 6 – (-8) = 14 Next we find our variable-specific determinants. This is because to find *x*'s determinant we delete the *x* values from the matrix and substitute in the *=* values. Then we use Cramer's Rule as usual with the new values: D_{x} = (8)(2) – (6)(-4) = 16 – (-24) = 40 Now we can find* x*: We work out *y*'s determinant the way we did *x*'s; we remove the *y* values and substitute in the *=* values. Now we use Cramer's Rule: D_{y} = (3)(6) – (2)(8) = 18 – (16) = 2 Now we find *y*: So:
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Use Cramer's Rule to find: D_{x}, D_{y}, *x*, and *y* -2*x* + 3*y* = 7 4*x* – 3*y* = 6 | |

We find D, the coefficient determinant: We multiply down the diagonal from left to right and then subtract the value we get by multiplying up the diagonal from left to right: D = (-2)(-3) – (4)(3) = 6 – 12 = -6 To find *x*'s determinant we delete the *x* values from the matrix and substitute in the *=* values: Then we use Cramer's Rule as usual with the new values: D_{x} = (7)(-3) – (6)(3) = -21 – 18 = -39 We do the same thing to find the y determinant: Follow up with Cramer's Rule: D_{y} = (-2)(6) – (4)(7) = -12 – 28 = -40 Now we can find *x* and *y*:
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Solve for *x* and *y* using Cramer's Rule:

2*x* – 2*y* = 5-*x* – 3*y* = 0

Solve for *x* and *y* using Cramer's Rule:

3*x* + *y* = 95*x* – 2*y* = 3

Answer

3. Solve for *x* and *y* using Cramer's Rule:

4*x* + *y* = -2-2*x* – (-5*y*) = -3

Answer