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:

L

_{1}

L

_{2}

L

_{3}

L

_{4}=L

_{1}+L

_{3}

L

_{5}=2L

_{1}+L

_{2}

L

_{6}=L

_{5}-L

_{4}

L

_{7}Solve L

_{6}

L

_{8}Solve L

_{4}w/ a=4

L

_{9}Solve L

_{8}

L

_{10}Solve L

_{1}w/ a=4, b=-3

L

_{11}Solve L

_{10}

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:

L

_{1}

L

_{2}

L

_{3}

L

_{4}=L

_{2}-L

_{1}

L

_{5}+L

_{4}

L

_{6}Solve L

_{5}

L

_{7}Solve L

_{3}w/ c=31

L

_{8}Solve L

_{2}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:

L

_{1}

L

_{2}

L

_{3}

L

_{4}=L

_{3}-

_{2}

L

_{5}=L

_{2}-L

_{1}

L

_{6}=L

_{4}-2L

_{5}

L

_{7}Solve L

_{6}

L

_{8}Solve L

_{5}w/ a=-1

L

_{9}Solve L

_{1}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.

ben, I was just the scribe a couple of days ago.

ReplyDeleteSorry It's going to be Natnael

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