Class notes from 8/29

Try same ideas with

3x + 7 y = 9

81 x – 35 y = 17

and with

2 x + 32 y = 7

3 x – 5 y = 8.

Solutions

(291/106, 5/106)

(1.55, .62)

(2.745,.047)

3 * [ 2x + 32 y = 7 ]

-2 * [ 3 x – 5 y = 8 ]

6 x + 96 y = 21

-6 x + 10 y = - 16

Sum of equations is

0 x + 106 y = 5 so y = 5 / 106

Plugging y into 2x + 32 y =7 we get

2x + 32 ( 5/106) = 7

Multiplying both sides by 106 we get

106 ( 2 x ) + 32 ( 5 / 106) * 106 = 106 * 7

212 x + 160 = 742

212 x = 742 – 170 = 582

x = 582 / 212 = 291 / 106

Should also verify in both equations to be sure.

Solution is (291 / 106, 5 / 106) or approx. (2.75, .047).

Next: get model by simul eq, compare, etc..

The form of a quadratic function is

If we use the data points (10,80), (46, 50) and (99,20) we can get three simultaneous linear equations.

(10, 80) means y = 80 when t = 10. In this case our equation will be

The point (46, 50) gives us the equation

The point (99, 20) gives the equation

Squaring and rearranging we see that we end up with the equations

100 a + 10 b + c = 80

2116 a + 46 b + c = 50

9801 a + 99 b + c = 20.

How we gonna solve this system?

Eliminate one of the variables to get 2 equations in 2 unknowns. Pick on c:

1st eqn + (–2d) eqn gives:

100 a + 10 b + c = 80

-2116 a –46 b – c = -50

sum is

-2016 a – 36 b = 30

Do the same with the first and third equations:

100 a + 10 b + c = 80

-9801 a - 99 b - c = -20.

Sum is –9701 a – 89 b = 60.

Now we have the system

-2016 a – 36 b = 30

–9701 a – 89 b = 60.

To solve eliminate b. Multiply 1st by 89 and 2d by –36 to get

89 [ -2016 a – 36 b = 30 ]

-36 [ –9701 a – 89 b = 60 ]

so

-179,424 a – 3204 b = 2670

349,232 a + 3204 b = -2160.

Sum is

169,812 a = 510 so

a = 510 / 169,812 = .003003 (approx.).

Now plug a into -2016 a – 36 b = 30 to get

-2016 (.003003) – 36 b = 30

-6.054 - 36 b = 30

-36 b = 36.054

b = -1.0015.

Now plug a and b into the first equation:

100 a + 10 b + c = 80 so

100 ( .003003) + 10 (-1.0015) + c = 80

.3003 – 10.015 + c = 80

-9.712 + c = 80

c = 89.71.

So our y = a t^2 + b t + c model is now

The Excel solution, which used ALL the data points and not just the three we chose, was

so we're pretty close.

According to the rounded-off model y = .003 t^2 – t + 90:

What should be the depth at t = 60?

When should the depth be 25?

25 = .003 t^2 – t + 90.

Solve this for t. This is a quadratic equation in t. Rearrange to get

.003 t^2 – t + 65 = 0 and use the quadratic formula.

When should the depth reach 0?

What should be the minimum depth?