When F = 40, C = 5/9(40-32) = 5/9(8) = 4.44 degrees

When F = 60, C = 5/9(60-32) = 5/9(28) = 15.56 degrees

When C = 30, F = 9/5(30) + 32 = 9*6 + 32 = 54 + 32 = 86 degrees

When C = 40, F = 9/5(40) + 32 = 9*8 + 32 = 72 + 32 = 104 degrees

Find where y = 7x + 4 and y = 10x - 5 intersect.

Step 1: Set the equations equal to each other and solve for x.

7x + 4 = 10x - 5_____Subtract 7x from both sides.

4 = 3x - 5_________Add 5 to both sides.

9 = 3x___________Divide both sides by 3.

3 = x

Step 2: Plug x=3 into either equation to find y.

y = 7(3) + 4 = 25

The point of intersection is (3, 25).

Kramer's rule problems.

3x + 4y = 25

4x - 2y = 4

Matrix version

[3 _4 | 25]

[4 -2 | _4]

Three matrices.

M

[3 _4 ]

[4 -2 ]

determinant = 3x(-2) - 4x4 = -6 - 16 = -22

Mx

[25 _4 ]

[ 4 -2 ]

determinant = 25x(-2) - 4x4 = -50 - 16 = -66

My

[3 _25 ]

[4 __4 ]

determinant = 3x(4) - 4x25 = 12 - 100 = -88

x = -66/-22 = 3

y = -88/-22 = 4

Answer: (3, 4)

2x + 4y = 10

4x + 8y = 30

Matrix version

[2 4 | 10]

[4 8 | 30]

Three matrices.

M

[2 4 ]

[4 8 ]

determinant = 2x8 - 4x4 = 16 - 16 = 0. This means no single solution.

Mx

[10 4 ]

[30 8 ]

determinant = 10x8 - 30x4 = 80 - 120 = -40. Since this isn't 0, the two lines are parallel.

We don't have to do the third determinant. Yay!

Answer: No solution, two parallel lines.

_x + _y = 13

3x - 3y = 9

Matrix version

[1 _1 | 13]

[3 -3 | _9]

Three matrices.

M

[1 _1 ]

[3 -3 ]

determinant = 1x(-3) - 3x1 = -3 - 3 = -6

Mx

[13 1 ]

[ 9 -3 ]

determinant = 13x(-3) - 9x1 = -39 - 9 = -48

My

[1 _13 ]

[3 __9 ]

determinant = 1x9 - 3x13 = 9 - 39 = -30

x = -48/-6 = 8

y = -30/-6 = 5

Answer: (8, 5)

## Wednesday, July 29, 2009

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