Use Cramer's Rule to find D_x, D_y, x, and y in the following system of equations.

3x – 4y = 8
2x + 2y = 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. To find x's determinant, we delete the x-values from the matrix and substitute in the constant 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 from the ends of both equations.

Now we use Cramer's Rule:

D_y = (3)(6) – (2)(8) = 18 – 16 = 2

Now we find y:

So:

Use Cramer's Rule to find D_x, D_y, x, and y in the following system of equations.

-2x + 3y = 7
4x – 3y = 6

First we hunt down 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 constant 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:

Use Cramer's Rule to find D_x, D_y, x, and y in the following system of equations.

2x + y = 4
4x + 2y = 6

First, we're going to find the coefficient determinant; we plug our diagonal values into the formula:

D = (2)(2) – (4)(1) = 4 – 4 = 0

Because we have a D of 0, we don't need to go any further; we know we can't get set values when D is 0 because the values will have 0 in the denominator. Boo-hoo.