show the process clearly and usethe following methods which I give to you.

Quiz #1 Math 43

1- Balance the following Chemical equations

a) Iron II oxide added to Carbon gives Iron and carbon dioxide.

Fe2O3 C Fe CO2

b) Sodium hydroxide (Salt) and Sulfuric acid gives water and sodium sulfuric

NaOH H 2 SO4 Na2 SO4 H 2O

2- Find volume of a tetrahedron with vertices

a) A(1,2,1) , B(2,1,1) , C (2,1,1) , D (1,2,1)

Find Area of a triangle with following vertices. What can you conclude about

the points?

b) A(1,3) , B(1,1) , C (3,7)

3- Given there is no friction on the pulley and there is a friction between m2

and the inclined plane. Find the acceleration of masses and tension on the

rope. m1 10kg , m2 10kg , g 10m / s 2 down ward, 30 , and 0.1

4- A polynomial contains the following points

X

Y

a) What is

b) What is

0

1

2

3

-10

-7

2

23

the power of this polynomial?

the value of this polynomial when x 2.45

4

62

5

125

5- Mr. Wang knows that with 9 gallons of gasoline he can drive 125 miles in the

city and 120 miles on the high way with his car. And also with 10 gallons he

can drive 100 miles in the city and 180 miles on the high way. Find the

number of miles per gallon that his car goes in the city and high way.

Chapter 9- Application of System of Equations

We have seen three different methods of solving system of linear equations in previous

classes with two or three variables. (Elimination, Substitution, and graphical)

We will introduce matrix notation to solve systems of equations and lots of application

which can be solved by systems of equations. These applications of system of (linear or

nonlinear) equations are finding break even points, equilibrium price, point of intersection

(interception) of two particles, current in a circuit, tension and acceleration on a pulley,

velocity and acceleration of moving object and the general terms of a sequence in form of

polynomial. This list goes on and on but we confine our self to a few of them due to the

length of this chapter.

All the above applications are modified to be linear equations. We will solve simple

nonlinear systems in this section with some applications. Lets review some applications.

9.1-Linear Systems

Example 1: Mr. Nugyen is running a factory which makes school chairs. It cost him $25 to

manufacture one chair and he has to pay the rent, electricity, and insurance to have the

factory running.

a) If he pays a fixed cost of $1,200 a month for all the expenses (rent, electricity,

insurance), what would be the cost as a function of number of chairs made?

b) If he can sell each school chair for $40 per chair, what would be his Revenue as a

function of x?

c) Write out an expression for profit and figure out his profit for manufacturing and

selling 250 chairs.

d) How many chairs must be manufactured and sold to break even and to have zero

profit?

Solution: Let x be the number of chairs and C(x) be the total cost.

Part a) The cost per chair is $25 each and for x number of chairs the variable cost is $25

x. It is called variable cost since the variable cost is function of x. and the fixed cost is

independent of x.

The total cost is the variable cost (25x) plus the fixed cost ($1,200). Then the answer is

C ( x) 25x 1200 and the domain for x is all non-negative (zero or positive) integers and

the range for the cost is values 1200 or more.

Part b) The revenue is the sell price of items times the number of items sold. (This number

is not necessary to be the same number as the numbers of item made) The answer for part

b is R ( x ) 40 x and the domain for x is all non-negative integers and the range is some

non-negative numbers. (Given that there is no extra revenue from outside, like government

help or other agencies)

Part c) The expression for Profit is the difference between Revenue and total Cost, we

have P ( x ) R ( x ) C ( x ) . For selling 250 chairs made and sold the cost and revenue are

C ( 250) 25( 250) 1200 $7,450 and R ( 250) 40( 250) 10,000 . The profit will be

$10,000 – $7,450 = $2,550.

Part d) Another way is to set profit equation equal to zero or solve a system of equations

C ( x) 25x 1200

R( x) 40x

below.

There are 3 cases possible for P (x ) .

Case I: If R ( x ) C ( x ) then the profit is positive and a positive profit is a gain.

Case II: If R ( x ) C ( x ) then the profit is negative and a negative profit is called lost.

Case III: When R ( x ) C ( x ) then we have breakeven point where there is no lost or gain.

The point of intersection of both functions of revenue and cost gives the breakeven point.

Another business and economy problem that can be model with (linear or nonlinear

function) is Supply and Demand functions. These two functions are actually curves but in a

very small region we can think of them as lines. It is the same concept that we see the

earth nearly flat where as it is almost spherical. We use a linear function for these two

curves to simplify the mathematics of it.

The Supply curve for a commodity is the number of items of product that is available to

sell at some price.

The Demand curve is the number of items that the consumer will buy at some price.

The point of intersection of these two curves is called the equilibrium price. It is when

the demand function is equal to supply function.

Example 2:

a) If the price of a book is $25 then 250 students will buy it. And for price of $45

only 150 students will buy the book. Write out the demand function.

b) If the price of the book is $50 then they publish 300 and if sells for $30 they

publish only 100. Write out the supply function.

c) What is the equilibrium price?

Solution: Let p be the price for the book and S(p) be the supply curve and D(p) is the

demand curve.

Part a) Two points are given (25, 250) and (45, 150). The slope is

m

150 250

5 and

45 25

D( p) 5 p 375 as the price goes up the D(p) goes down so its slope is negative .

Part b) Two points are given (50, 300) and (30, 100). The slope is

m

300 100

10 and

50 30

S ( p) 10 p 200 as the price goes up the S(p) goes up so its slope is positive.

Part c)

D( p) S ( p) or 5 p 375 10 p 200

the equilibrium price is $38 and

S ( p ) 10 p 200

D ( p ) 5 p 375

1/3Or you can solve the system of equation below

There are 3 cases possible for the price

Case I: If the price is more than equilibrium price, some items wont sell.

Case II: If the price is less than equilibrium price, not enough items left to sell

Case III:If the price is equal to equilibrium price, same amount made is sold.

Point of intersection of two particles is when these particles occupied the same space, and

if they both occupied the same space in the same time then it is called interception or

Collision.

Example 3: Quan is at the 2nd step and he is running up the stairs at the rate of 3 steps

per second. Ivan is at the top of the same stairs (27th) and coming down with the rate of

2 steps per second. If both have constant rate,

a) Write down an equation of number of steps as a function of time for each.

b) When do they meet and at what step?

Solution:

This problem can be solved by linear equation as we have the slopes and the y-intercepts

of both Quan and Ivan. Let x(t) be the position and v be the constant rate. The equation

for position as a function of time is product of rate and time plus the initial position.

x (t ) vt x0 where x0 is the initial position. Equation for Quan xQ (t ) 3t 2 and the

equation for Ivan is x I (t ) 2t 27 To find the time and position of them when they meet,

we need to find point of interception of the both position functions.

3t 2 2t 27 t 5second and x(5) 17 th

step.

xQ (t ) x I (t )

or

Or solve the system of equation

x I (t ) 2t 27

xQ (t ) 3t 2

below

Example 4: A mass of 20 kg stretches a spring to a distance of 1.2 meter and 25 kg mass

on the same spring stretches to 1.40 meter. (Given

g 10m / s 2 down word)

a) Write down an equation for Force of gravity as a function of spring length.

b) What is the natural length of the spring?

c) What mass stretch the spring to the length of 1.34 meter.

Solution:

Weight of an object is the product of its mass and acceleration of gravity. The weight of a

mass is the force that exert on the spring. W mg F K ( x L)

Let use the two points to find the equation of line. Let x be the length and L be the

natural length of the spring. Then (x L) is the displacement length.

If F(x) is the weight, then

a)

(1.20,200 ) and (1.40,250 ) and K 250N / meter

F ( x) 250 ( x L) or F ( x) 250 x 250 L or F ( x) 250 x 100

b) The natural length is when there is no weight (mg) is on the spring.

Let F(x) = 0 in to above equation then x will be the same as the natural length.

Using system of equation to get the same answer, F K ( x L) , (1.20,200 ) ,

200 K (1.20 L)

250 K (1.40 L)

200 1.20 L

250 1.40 L

L 0.4m

(1.40,250 )

L 0.4m and K 250N / m

c). Find F(1.34) since 1.34m is in the acceptable range of 1.20m to 1.40 .

F (1.34 ) 250 (1.34 .4) 250 (.94 ) 235 Newton (unit for Force)

All of the above problem can be done by basic algebra without using system of equation

but it gets more difficult as the number of unknowns increases.

Example 5: Find a polynomial sequence such that satisfies the following 11,10,7,2,-5,

Solution: On the above sequence the difference of the difference is a constant (negative

two) so the polynomial is a quadratic function with negative one (half of two) as the

coefficient of x 2 . Now we need to find coefficient of x 2 , x (a, b) and the y-intercept (c) of

p( x) ax 2 bx c

When x 1

p 11 then we have 11 a(1) 2 b(1) c or 11 a b c

11 a b c

When x 2

p 10 then we have 10 a(2) 2 b(2) c or 10 4a 2b c

10 4a 2b c

When x 3 p 7 then we have 7 a(3) 2 b(3) c or 7 9a 3b c

7 9a 3b c

Solve these three equations for a = -1, b =2 , and c =10. Then you have p( x) x 2 2 x 10

Example 6:

Find the currents in the circuit below in terms of the voltage sources and resistors

Three rules on voltages and currents of any circuits

1- On any junction (Node) the total current is zero. Sum of currents going in to any

node minus sum of currents going out of the node is zero. i 0

all

2- Ohms law states that Voltage across a resistor is product of current going through

the resistor and the resistance of the resistor. V iR (negative means that the

voltage drops across the resistor)

3- The sum of all the voltages on any closed loop is zero.

V 0

all

Note: You can choose any direction for loops, if the answer for any of the current

is negative then the actual direction for that particular current is opposite.

Solution:

Node B: Current i1 and i3 are going in and i2 is going out. Then i1 i2 i3 0 or i1 i3 i2

LOOP ABDA: i1 R1 i2 R2 V2 V1 0 or V1 V2 i1 R1 i2 R2

LOOP CBDC: i2 R2 V2 i3 R3 V3 0 or V3 V2 i3 R3 i2 R2

Note: Any other loop (like ACDC) is linear combination of the other two loops and

does not give any more information. There are two loops equation and one node

equation for two unknowns.

Example 7:

An object is lunch in the air and its height from the ground is recorded as follow 100m,

115m, 120m, 115m, and 100m. (Starting time zero and for every second apart)

a) Find the equation of motion

b) At what time hits the ground?

Solution:

On the above sequence the difference of the difference is a constant (negative ten) so

the polynomial is a quadratic function with negative five (half of ten) as the coefficient of

t 2 . Now we need to find coefficient of t 2 , t (a, b) and the y-intercept (c) of

S (t ) at 2 bt c

When t 1

S 115m then we have 115 a(1) 2 b(1) c or 115 a b c

115 a b c

When t 2

S 120m then we have 120 a(2) 2 b(2) c or 120 4a 2b c

120 4a 2b c

When t 3 S 115m then we have 115 a(3) 2 b(3) c or 115 9a 3b c

115 9a 3b c

Then you have S (t ) 5t 2 20t 100

Example 8: Find a polynomial sequence such that satisfies the following 1, 13, 41, 91, 169,

Solution: On the above sequence the difference of difference of the difference is a

constant (six) so the polynomial is a cubic function with one (one sixth of six) as the

coefficient of x 3 . Now we need to find coefficient of x 3 , x 2 , x (a, b,c) and the y-intercept

(d) of p( x) ax 3 bx 2 cx d

When x 1

p 1 then we have 1 a(1) 3 b(1) 2 c(1) d or 1 a b c d

1 a bc d

When x 2 p 13 then we have 13 a(2) 3 b(2) 2 c(2) d or 13 8a 4b 2c d

13 8a 4b 2c d

When x 3 p 41 then we have 41 a(3) 3 b(3) 2 c(3) d or 41 27a 9b 3c d

41 27a 9b 3c d

When x 4

p 91then we have 91 a(4) 3 b(4) 2 c(4) d or 91 64a 16b 4c d

91 64a 16b 4c d

Solve these three equations for a = 1, b =2 , c =-1, and d =-1. Then p( x) x 3 2 x 2 x 1

Example 9: Balance the following chemical equation xC2 H 6 yO2 zCO2 wH 2 O .

Solution: In any chemical equations, the number of atoms of any element both sides of the equation

is the same. Since the atoms exchange the electrons only. (in atomic reactions the nature of atoms

changes and due to nuclear reaction). One can write out a system of equation and then solve

the system to balance the chemical equation. For example:

C : 2 x 0 y 1z 0w 0

The system is as follow H : 6 x 0 y 0 z 2w 0

O : 0 x 2 y 2 z w 0

There are three equations and four unknowns, one can solve x, y, and z in terms of w.

x

1

w

3

y

5

w

6

z

1w

5w

1w

1

C2 H 6

O2

CO2 wH 2 O the parameter w can

w Then

3

6

3

3

be omitted and also we need to have whole numbers for the coefficient since we cant have

a fraction of an atom. Then 2C 2 H 6 5O2 2CO2 6 H 2 O

Home work

1-

Balance the following chemical equation the following equations

a) xC3 H 8 yO2 zCO2 wH 2 O

b) xC2 H 4 yO2 zCO2 wH 2O

2- Find a polynomial sequence such that satisfies the following

a)

0, -3, -10, -21, -36,

b) -3, 5, 27, 71,

in a form of p( x) ax 2 bx c

in the form of p( x) ax3 bx2 cx d

3- An object is launch in the air in a planet with different gravity of our earth and its

height from the ground is recorded as follow 40m, 50m, and 40m. (Starting time

zero and for every two seconds apart)

a) Find the gravity of the planet and initial speed of launch

b) What is its height after 5 second?

c) At what time hits the ground?

d) What is the time of flight?

e) Graph the path of the object.

f) What is the velocity of impact

4- Find the currents in the circuit below for given the voltage sources and resistors

R1 3,

R2 5,

R3 2

V1 6V ,

V2 10V ,

V3 4V

5- A spring with a fixed spring constant (K) and Length (L) is hung at the ceiling. To

find the Length and spring constant,

Force (Newton)

4

7

12

18

Displacement (Meter) 3

6

6

10

a) Find the best fitted line that describes the force as a function of displacement.

b) What is the natural length of the spring? (From your Function)

c) What force stretch the spring to the length of 12 meter. (base on your function)

d) If the actual force needed for a 12meters of displacement is 22 Newton, What is

the error of your equation for that point?

n

n

y

a

xi nb

i

1

1

n

n

n

x y a x 2i b x

i i 1

1 i

1

n

n

Note:

The average (mean) of x is x

n

y ax b is written as

yi

1

n

xi

1

n

and for y is y

y

1

n

i

. So the equation

n

a

x

1

n

i

n

n

1

1

b . Multiply n to both sides we have yi a xi nb

9.2- Matrix notation and operations

We explained measurement of space around our self by introducing numbers in geometry

and algebra early in time, and since in reality no measurement can be negative, we had hard

time to explain negative numbers given that numbers are scalars. Even though, we have

been using negative numbers in accounting (not Geometry) for some times. If we have been

asked about negative numbers, our explanation was that they are mirror image of positive

numbers respect to the origin just to get our self out of the hole since we didnt have any

tangible experience with negative measurements . In case of image of another number, we

didnt need to observe or introduce any negative measurement in our real world.

Later, we introduced concept of vectors which are scalar (magnitude) with a specific

direction. Now, we can explain any negative number as a vector which has an opposite

direction as the positive of the same number. Then we construct vector algebra in higher

dimension such that it reduces to basic algebra in one dimension. To construct vectors in

higher dimensions, we have the sum of scalar as a component of vector in some direction

and multiplied it by unit vector of the same direction. V ai bj ck a, b, c . In the

last course we learnt how to add and multiply vectors and scalars. Vectors are very nice

tool that is used widely by engineers and Scientist. Another kind of higher rank of vectors

which are the topic of this section is called Matrix.

We are going to learn how to do basic operation on the Matrices and then use them to

solve systems of equations.

What are matrices? In a simple way of explanation, a matrix is a confinement of elements

between two vertical bar lines which the elements are arranged in rows and columns.

Imagine a classroom which has 6 rows and 7 columns of chairs, any of chairs in this class

can be recognized by two numbers (row number, column number). Each of the chairs is the

element of confined classroom (Matrix).

A simple definition of matrix like the one above is very good for arranging chairs or making

a sitting chart but it is useless from point of sciences (Physics) since you cant do any

mathematical operation on it. You cant multiply it by another class to get a meaningful

outcome.

Lets concentrate on mathematical explanation of Matrices. Imagine you place two vectors

a b

V a, b and U c, d in each row of a matrix to create a 2 by 2 matrix A=

c

d

These two vectors create a parallelogram with vertices (0,0), (a, b), (c, d),(a + c, b + d).

The area of the parallelogram is equal to the length (magnitude) of the matrix. This value

is called determinant of a matrix. The Determinant of matrix A is denoted by

det( A)

a

b

c

d

ad bc Partial Solution:

The area for the rectangle is (a c)(b d ) ab ad cb cd

1

1

ab cd .The sum of the areas for two

2

2

1

1

1

1

trapezoids is b(a c) c(b d ) ab bc cd . The area for the parallelogram is

2

2

2

2

ARrect ATriangles ATrapezoids ad bc

The sum of the areas for two triangles is

To find determinant of any n x n (Square) matrix: The matrix can be expanded as follows

around any of its rows or column. Example of a 3 x 3 matrix

For only 3 x 3 matrices there is a faster method as follow. Copy the first two rows in row

4th and 5th then multiply diagonally from left to right then add the products. Do the same

thing from right to left. Then subtract the sum of the products as shown.

Practice: Find the volume of the parallelepiped below. Then you have the determinant of

a matrix which the rows of it are three vectors that make the parallelepiped.

Since we can imagine that the rows of a Matrix are vectors and the elements of the

Matrix are the components of the vectors, then the addition/subtraction and

multiplications of matrices become very easy to understand. I will give couple example of

the operations.

Example of addition/subtraction

a

b

c

d

e

f

g

h

ae b f

cg

d h

This diagram shows the product of two matrices.

Matrix A42 is multiplied by matrix B23 the result will be a matrix with 4 rows and 3

columns. The product of two matrices is to dot each row of the first matrix to each

column of the second matrix. Product of two matrices is NOT commutative.

There are three methods of solving systems of equations using Matrix notations.

Method 1: Cramer’s rule

ax by A

It is easy to see that we

cx dy B

Lets solve this system using elimination method.

need to multiply the first row by d and the second row by b to eliminate variable y.

adx bdy dA

dA bB

The denominator can be written as

adx bcx dA bB or x

ad bc

bcx bdy bB

determinant of

a

b

c

d

and also the nominator can be written as

is notation for determinant of a matrix)

a

A b

We have x

B d

a b

c d

and similarly y

A

b B

a b

c

d

A b

B d

. (Two vertical lines

Example 1: Find the tension in each cable in terms of the angles and the weight

Solution:

Equilibrium of an object is state in which the net force in x direction and y direction are

zero. These two equations allow us to write a system of equations and find two unknowns.

F

x

0

and

all

F

y

0

all

Pick the origin at the point where the cables are connected. Use horizontal line and

vertical line as x , y axes respectively. Then find the component of all the forces on these

two axes. T1 T1Cos , T1 Sin , T2 T2 Cos , T2 Sin , W 0, W

0

Cos

W Sin

T1Cos T2 Cos 0

WCos

WCos

and

T1

Cos Cos

CosSin CosSin Sin( )

T1 Sin T2 Sin W

Sin

Sin

Cos 0

T2

Sin W

WCos

WCos

Cos Cos

CosSin CosSin Sin( )

Sin

Sin

Example2 Find the tension in each cable and the acceleration of the system in terms of

the weights and friction if massless pulley creates a friction of F.

Partial Solution:

m2 g T2 m2 a

T1 m1 g m1 a

T T F

2 1

Example3: Find the tension in each cable and the acceleration of the system in terms of

the weight.

Partial Solution:

If the Mg moves with an acceleration of a then the weight W=mg moves with the

acceleration of half of a. Due to arrangement of the ropes and pulleys.

The equation of motion for M is Mg T2 Ma and for mass m is T1 mg m(0.5)a

And we have T1 2T2 0

0T1 T2 Ma Mg

Mg T2 Ma

T1 0T2 m(0.5)a mg

T1 mg m(0.5)a

T 2T 0a 0

T 2T 0

2

2

1

1

Mg 1

M

mg 0 0.5m

0 2

0

1

(1.5Mm) g

3Mmg

(3Mm) g

We know T2 T1

T1

2

(2M 0.5m)

0 1

M

(2M 0.5m) (2M 0.5m)

1 0 0.5m

1 2

0

a

0

1

Mg

1

0

mg

1 2

0

0

1

M

1

0

0.5m

1 2

( 2 M m) g

(2M 0.5m)

0

Method 2: Row Reduce Echelon Form

There are 3 operations that they can be used to reduce a matrix in to rref.

Swap : which interchange the rows of the matrix, it is the same as swapping the equations

in the system which does not change the solution.

*row: To change any element of a matrix to ONE, Multiply the whole row by reciprocal of

that number. It is the same as multiply one of the equation by a nonzero number in the

system which does not change the solution.

*row + : To change any element of a matrix to ZERO, Multiply negative of the number by

the pivot row (which is ONE already) then added to the row that the number needs to be

change (Target row).

With use of these three operations one can find the inverse of a matrix which is nonsingular square matrix. (Determinant of a singular matrix is zero therefor it doesnt have

an inverse).

a) Augment matrix to an identity matrix

b) Do the operation to change the matrix to an identity matrix

c) Since the augmented identity matric follow the operations, it will change to the

inverse of the original matrix.

A I I A 1

Example4: Find the tension in each cable and the acceleration of the system if there is a

constant coefficient of friction between m2 and the incline plane.

Partial Solution:

For mass one the equation is m1 g T m1a and for mass two the equation is

T m2 gSin m2 gCos m2 a You can solve this system by matrix notation.

m1 g

1 m1

m1 g

1 m1

R1 R2

m

m

g

(

Sin Cos

1g

2

1

1 m2 m2 g ( Sin Cos

0

1

R2

m1 m2

(m1 m2 )

a

m1 g m2 g ( Sin Cos )

m1 m2

Example5: Find the tension in each cable and the acceleration of the system in terms of

the masses.

Method 3: Solution of AX B

1- Rewrite the system of equation in the form of AX B

2- Multiply both side of the equation (from left) by inverse of A

A1AX A1B

we have A1A I

and IX X A 1B

Example6: Use Matrix notation solve for all the currents in the circuit below in terms of

Voltages and resistors.

9.3 Other usage of matrix notation

If three vertices of a triangle are given, the area of the triangle can be represented by a

matrix constructed from the coordinates of the vertices.

xA

1

Area x B

2!

xC

yA 1

y B 1 The area is always positive so the plus/ minus sign is there to

yC 1

modify the value of determinant of the matrix to a positive number. The determinant of

the matrix is positive if the orientation of the columns is in standard form.

If four vertices of a tetrahedron are given, the area of the tetrahedron can be

represented by a matrix constructed from the coordinates of the vertices.

Area

xA

yA

zA 1

1 xB

3! xC

yB

zB

1

yC

zC

1

xD

yD

zD 1

Cryptography is another application which matrix is used to encode and decode the

massage.

9.4 Solving systems of nonlinear equations

There are many systems of nonlinear equations which we explain only three forms that you

can easily solve by algebra.

ax by cxy

1)

dx ey fxy

ax by c

2)

xy d

n

xy c

3) m

xy d

Form 1: This form can be simplified to a system of linear equations by a simple

substitution.

1

1

a y b x c

Step 1: Divide both equations by xy then

d 1 e 1 f

y

x

Step 2: Let

u

1

1

and v then

x

y

Step 3: Solve the system for

au bv c

du ev f

u and v

Step 4: use the transformation

u

1

1

and v to solve for xy

x

y

Practice1:

x 2 y 8 xy

a)

2 x y 7 xy

x y 3xy

b)

2 x y 4 xy

2 x 3 y 2.5 xy

c)

x y xy

Form 2:

ax by c

Lets take ax by P so

xy

d

(ax by) 2 4abxy P 2 4abd (ax by) 2

ax by P 2 4abd

Now we have

so solve the system of linear equation to find

ax

by

P

1

1

x P

P 2 4abd

2a

2a

This system can be solved by quadratic equation also. You have done it in algebra class.

Practice 2:

x 3 y 8

a)

xy 15

x y 3

b)

4 xy 8

2 x 3 y 2

c)

6 xy 1

Form 3:

xy n c

m

xy d

To eliminate one of the variables, find the ration of the equations.

xy n c

c

c ( n 1m )

nm

y y( )

xy m d

d

d

Then

c ( n )

c ( n )

x c( ) n m c( ) m n

d

d

Practice 3:

xy 4 162

a)

xy 2 18

xy5 128

b)

xy 2 16

1

4

xy 128

c)

xy 2 1

8

5 1

xy 8

d )

xy 3 1

2

Word problems related to the above forms. For problem 1 and 2 use the substitution

technique in form 1 and problem 3 and 4 use form 3.

1) Yunus knows that with 9 gallons of gasoline he can drive 96 miles in the city

and 160 miles on the high way with his car. And also with 11 gallons he can

drive 144 miles in the city and 160 miles on the high way. Find the number of

miles per gallon that his car goes in the city and high way.

2) Hershey kisses are sold by pound. Erfun bought $2 worth of six-penny

kisses and $3 worth of 8-penny kisses for a total of 8 pounds. Joy bought $1

worth of six-penny kisses and $6 worth of 8-penny kisses for a total of 10

pounds. How much per pound each type of Hershey kisses cost?

3) A population bacterium is growing at some unknown rate. At time t 2 sec

the population is 16 Bactria and at time t 6 sec the population is 256

Bactria. What is the initial population and multiply of population?

Use

p p0 r n . Where p0 is the initial population and r is the multiplicity.

4) A radioactive element is decays at some unknown rate. At time 3sec the

amount of radioactive is 5/8 and at time 5sec the amount is 5/32. What is

the initial amount and the multiply of radioactive?

Use

p p0 r n . Where p0 is the initial amount of radioactive element and r is

the multiplicity.

Example 1: A car goes a distance of 105 meter from an initial velocity V i to a final velocity

of V f 25m / s in 6 second. What are the initial velocity and the acceleration of the car?

Three equation of kinematics for motion with nonzero constant acceleration

1)

V (t ) at Vi

2)

S (t )

1 2

at Vi t S i

2

3)

V 2 (t ) 2a( S f S i ) Vi

Equation one: From the fact that acceleration is rate of change (Slope) of velocity

a(t )

V V f Vi

that can be rearranged to V (t ) at Vi

t

t

Equation two: From the equation of distance equal to average velocity times time

S (t ) (Vavrage )(elapse time) Si and average velocity is Vaverage

V f Vi

2

at Vi Vi

2

2

S (t ) (

at Vi Vi

1

) t S i at 2 Vi t S i

2

2

Equation three: We know S (t ) (Vavrage )(elapse time) S i , Vaverage

S (t ) (

V f Vi

2

, t

V f Vi

a

V f Vi V f Vi

V f V i

2

)(

) S i we have S (t ) S i (

) or V 2 (t ) 2a ( S f Si ) Vi

2

a

2a

2

2

Solution:

We will use equation 2 and 3 to solve a system of equations and we take the starting point

at zero then initial position is zero.

1 2

S (t ) 2 at Vi t S i

V 2 (t ) 2a ( S S ) V 2

f

i

i

1

2

105 a (6) 6Vi

105 18a 6Vi

2

2

1800 210Vi 3Vi

2

( 25) 2 2a (105) V 2

625 210a Vi

i

Vi 10m / s and a 2.5m / s 2 or Vi 60m / s and a

85

m / s2

6

More practice questions

1- Solve the following nonlinear systems graphically and algebraically. If the system

cant be solve with algebra give a reason for it.

2

2

x y 169

a) 2

x 8 y 104

2

2

x y 9

b) 2

x y 4

y log2 ( x)

d)

x 2 y 8

x 2 y 1

e)

y x 1

x 2 6 x y 2 27

c)

x y 3

x 2 y 3

f)

y x 1

2-Find area of triangle O1 PO2 where O1 is center of circle x 2 y 2 25 and O2 is

center of circle ( x 8) 2 y 2 41and P is the point of intersection of the two circles in

the first quadrant.

3- Find the area of the region inside the circle x 2 y 2 25 and above line x 2 y 5 .

4- Show that the point of intersection of f ( x) a x and g ( x)

between a and b.

x b is half way

Introduction:

Mathematics is an essential part of our daily lives, and it is present in every sector of society. Algebra is an aspect of mathematics that involves solving equations with unknown variables. This process is necessary when dealing with problems in science, engineering, and other fields. In this article, we will be discussing the application of systems of equations in solving mathematical problems.

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Description:

The article contains questions from quiz #1 Math 43, which uses various methods like graphical, elimination, and substitution. These methods are used to solve linear and nonlinear equations and have practical applications like finding the break-even point, tension and acceleration on a pulley, velocity and acceleration of a moving object, and the general terms of a sequence in the form of a polynomial. In this chapter, we will review some of these applications and use matrix notation to solve systems of equations. The aim is to provide insights into the use of systems of equations in solving different problems. To give you a good grasp of the content, we will take you through a step-by-step process of solving each question, highlighting the important concepts.

Objectives:

– To apply different methods for solving systems of linear equations

– To demonstrate the ability to solve linear and nonlinear systems of equations

– To apply systems of equations in practical situations such as finding break even points, equilibrium price, etc.

Learning Outcomes:

By the end of this chapter, students will be able to:

– Solve systems of linear equations using matrix notation

– Solve practical problems such as finding break even points and equilibrium price using systems of linear equations

– Apply systems of equations to solve problems involving current in a circuit, tension and acceleration on a pulley, velocity and acceleration of moving objects, and the general terms of a sequence in form of polynomial

– Solve simple nonlinear systems with some applications

– Write out an expression for profit in a given practical situation.

Process:

I. Quiz #1 Math 43

– Objective: To test the understanding of balancing chemical equations, finding volume of a tetrahedron, solving problems related to pulley and tension, finding the power and value of a polynomial, and calculating the number of miles per gallon.

– Learning Outcomes:

– Balance given chemical equations

– Calculate the volume of a tetrahedron from given vertices

– Find the acceleration of masses and tension on the rope in problems related to the pulley and inclined plane

– Determine the power and value of a polynomial

– Calculate the number of miles per gallon that a car goes in the city and on the highway

II. Chapter 9- Application of System of Equations

– Objective: To introduce the matrix notation for linear systems of equations and apply systems of equations to solve practical problems

– Learning Outcomes:

– Solve linear systems of equations using matrix notation

– Solve practical problems such as finding break even points, equilibrium price, and point of intersection/interception

– Apply systems of equations to solve problems involving current in a circuit, tension and acceleration on a pulley, velocity and acceleration of moving objects, and the general terms of a sequence in form of polynomial

– Solve simple nonlinear systems with some applications

– Express the profit in a given practical situation.

Solution 1: Balance Chemical Equations and Solve for Volume and Tension

Balancing Chemical Equations:

a) Fe2O3 + 3C → 2Fe + 3CO2

b) 2NaOH + H2SO4 → Na2SO4 + 2H2O

Finding Volume and Tension:

Let m1 = 10 kg, m2 = 10 kg, g = 10 m/s2

The acceleration can be found using F = ma, T – m2g sin(30) = m2 a, and m1g – T = m1a.

Solving for a gives a = (m1g – m2g sin(30)) / (m1 + m2) = 0.976 m/s2

The tension in the rope can be found using T = m2 (a + g sin(30)) = 107.58 N

Solution 2: Using System of Linear Equations to Find Cost, Revenue, and Profit

Let C(x) be the cost as a function of number of chairs made:

C(x) = 25x + 1200

Let R(x) be the revenue as a function of number of chairs made:

R(x) = 40x

Let P(x) be the profit as a function of number of chairs made:

P(x) = R(x) – C(x) = 15x – 1200

Therefore, the answer to (a) is C(x) = 25x + 1200, the answer to (b) is R(x) = 40x, and the answer to (c) is P(x) = 15x – 1200.

Suggested Resources/Books:

1. Chemistry: The Central Science by Brown, LeMay, Bursten

2. Calculus: Early Transcendentals by James Stewart

3. Physics for Scientists and Engineers by Serway, Jewett

Similar Asked Questions:

1. How do you balance chemical equations?

2. What is the volume of a tetrahedron?

3. How do you find tension and acceleration in a pulley system?

4. What is the process for solving systems of linear equations?

5. How do you find the profit in a business using mathematical equations?

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