Applications with Systems of Linear Equations

Hi it's me ben again. Today in class Mr.K created a new rule to get people to participate in class and that's to pair up and let one person speak for the other person. Okay back to the math. Today we started with some Systems Of Linear Equations With Three Variable questions for practice.

a+2b+c=0
3a-b-2c=11
2a+b-c=3

3a+3b=3
5a+3b=11
2a=8
a=4

3(4)+3b=3
12+3b=3
3b=-9
b=-3

4+2(-3)+c=0
4-6+c=0
-2+c=0
c=2

EXPLANATION:
L₁
L₂
L₃
L₄=L₁+L₃
L₅=2L₁+L₂
L₆=L₅-L₄
L₇ Solve L₆
L₈ Solve L₄ w/ a=4
L₉ Solve L₈
L₁₀ Solve L₁ w/ a=4, b=-3
L₁₁ Solve L₁₀

Lines intersect @(4, -3, 2)

After a little work on that question we had to apply our Linear Equation skills we just learned and use them in word problems.

1.When 3 numbers are added in pairs the sums of the pairs are 28, 39, 45. What are the three number?
2Mark sells wildlife photos at the county fair, his revenue, R dollars is determined by the price, p dollars, that he charges for each picture as shown in the equation:
R(p)=ap²+bp+c
At a price of $40, his revenue is $2900
At a price of $50, his revenue is $3000
At a price of $70, his revenue is only $2600
Find the values of a,b, and c

Okay for #1 we need to find out the equation. So it states that the numbers are add in pairs so we need the possible combinations for the 3 equations and they are shown below then now solve.
a+b=28
a+c=39
b+c=45

-b+c=17
2c=62
c=31

b+31=45
b=14

a+31=39
a=8

EXPLANATION:
L₁
L₂
L₃
L₄=L₂-L₁
L₅+L₄
L₆ Solve L₅
L₇ Solve L₃ w/ c=31
L₈ Solve L₂ w/ c+31

Numbers are 8, 14, 31

For #2 we need to get another equation for each price and revenue. Notice that the price is p so just sub the price of each set of pictures and same with the revenues and you should get this:
a(40)²+40b+c=2900
a(50)²+50b+c=3000
a(70)²+70b+c=2600

Now just simplify and Solve
1600a+40b+c=2900
2500a+50b+c=3000
4900a+70b+c=2600

2400a+20b=-400
900a+10b=100
600a=-600
a=-1

900(-1)+10b=100
-900+10b=100
10b=1000
b=100

-1600+4000+c=2900
4000+c=4500
c=500

EXPLANATION:
L₁
L₂
L₃
L₄=L₃-₂
L₅=L₂-L₁
L₆=L₄-2L₅
L₇ Solve L₆
L₈ Solve L₅ w/ a=-1
L₉ Solve L₁ w/ a=-1, b=100

Therefore a=-1, b=100, and c=500

That's what we did today on Applications with Systems of Linear Equations and homework for tonight is EXERCISE #27 *But don't do #'s 1-5*
The next scribe will be M@RK and good night.