How do you balance chemical equations?

  

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.

Don't use plagiarized sources. Get Your Custom Essay on
How do you balance chemical equations?
Just from $13/Page
Order Essay

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?

Basic features
  • Free title page and bibliography
  • Unlimited revisions
  • Plagiarism-free guarantee
  • Money-back guarantee
  • 24/7 support
On-demand options
  • Writer’s samples
  • Part-by-part delivery
  • Overnight delivery
  • Copies of used sources
  • Expert Proofreading
Paper format
  • 275 words per page
  • 12 pt Arial/Times New Roman
  • Double line spacing
  • Any citation style (APA, MLA, Chicago/Turabian, Harvard)

Our guarantees

Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.

Money-back guarantee

You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.

Read more

Zero-plagiarism guarantee

Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

Read more

Free-revision policy

Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.

Read more

Privacy policy

Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.

Read more

Fair-cooperation guarantee

By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.

Read more
× How can I help you?