linear equations and applications, inequalities and absolute values

8/11/2019 LINEAR EQUATIONS AND APPLICATIONS, INEQUALITIES AND ABSOLUTE VALUES

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CHAPTER VI - LINEAR EQUATIONS AND APPLICATIONS

Objectives:

1. Solve linear equations in one variable;

2. Translate verbal phrases into algebraic expressions or mathematical

phrases;

3. Solve different types of word problems.

6.1 SOLVING EQUATIONS

An equation is a statement that expresses the equality of two

mathematical expressions. Examples include:

x+2=7, 2y2+ 3y = 2, x2= -3 and (a+b)2=a2+ 2ab + b2

In particular, note that each equation has an equality sign. A solutionof

an equation that involves a single variable is any value of the variable that

makes the equation true. Thus x = 5 is a solution of x + 2 = 7 and y = -2 is

a solution of 2y2 + 3y = 2. On the other hand, the equation x2 = -3 has no

real solution.

To solve an equation means to find all its solutions. In this chapter we only

consider equations that can be expressed in the form of ax + b = 0 where

a and b are constant with a 0. Such equations are called linear

equations in x. The properties of the equality guarantee that there isalways a unique solution of such equations.

Example 1: Solve for x + 7 = 3

Solution:

x + 7 = 3 By Addition Property of Equality, this equation is equivalent to

(x +7) + (-7) = 3 + (-7), or

x = -4

Example 2: Solve for x in 3x + 17 = 4x

Solution:

Again, by Addition Property of Equality, the following is obtained:

3x + 17 + (-4x) = 4x + (-4x), or

-x + 17 = 0


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-x + 17 + (-17) = 0 + (-17)

-x = -17

Then by Multiplication Property of Equality, the only solution x = 17, is

obtained.

Example 3: Solve for y: 8y + 3 = 7

Solution:

By the Addition Property of Equality

(8y = 3) + (-3) = 7 + (-3), or

The Multiplication Property of Equality shows that the previous equation is

equivalent to

Therefore, the only solution is y =

Note that the solution obtained can be always be checked by substituting

in the original equation. For example, y = is indeed a solution of 8y + 3

= 7 since 8 +3 = 7

Instead of applying the addition and multiplication properties of the

equality, one can use the process of transposition that is illustrated in thenext example.

Example 4: Solve for p in 12p + 6 = 10p4

Solution:

To solve for p, 6 on the left must be eliminated. Thus, it is transposed to

the right by becoming -6. So the new equation is now 12 p = 10p10.

Now, the 10p on the right side must go with 12 p; so 10 is transposed to

the left, becoming10;

12p10p = -10 or2p = -10

Finally the 2 in 2p must be eliminated. It can be transposed to the right,

changing into :


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p = -10 p = -5

6.1.1 EQUATIONS CONTAINING FRACTIONS

One sure way of solving equations containing fractions is to get rid of the

denominators by multiplying the whole equation by the least common

denominator.

Example 1: Solve for x:

Solution:

The LCD is 12, so we can solve the equation as follows:

2x - 3x = 240

-x = 240

x = -240

Therefore, x = -240

Example 2: Solve for x

Solution:

The LCD is 4x, thus

75x = 6x4-5x 6x = -47

-11x = -11, or


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x = 11

6.1.2 LITERAL EQUATIONS

The procedure for solving a linear equation in one variable can also be

used to solve for a particular variable in terms of others in literal

equations as illustrated in the next examples:

Example 1: Solve for a: a + b = y

Solution:

a = y-b

Example 2: Solve for x: bx + c = a

Solution:

bx = ac

Example 3: Solve for l: P = 2l + 2w

Solution:

P2w = 2l or 2l = P2wSo, l = (P2w)

6.2 TRANSLATING VERBAL PHRASES OR SENTENCES IN

MATHEMATICAL PHRASES OR EQUATIONS

Before going into word problems, it is important to see how certain

phrases and statements can be translated mathematically. The examples

that follow illustrate some commonly used expressions.

Illustrations:

1. Five increase by a number:


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If y = the number, then the phrase is the same as 5 + y

2. The sum of m and n:

Therefore: m + n

3. The product of five and the sum of x and y:

Therefore: 5 (x + y)

4. Thrice a number diminished by five is 14.

If x = the number, the equation is 3x5 = 14

5. A number decreased by 10 is 20.

If m = the number, then m10 = 20

6. Twice the sum of x and y increased by 6 is 25.In this case the equation is 2 (x + y) + 6 = 25

6.3 WORD PROBLEMS

The following steps may be helpful in solving a word problem:

1. Choose a variable to represent an unknown quantity then write the other

unknowns (if there are any) in terms of the variable.

2. Form a table or draw a figure if appropriate. Then form the given

information, write an equation that involves the variable.

3. Solve the equation.4. Check the answer you obtained in the original words of the problem.

6.3.1 NUMBER RELATION PROBLEM

Example 1: Find three consecutive even integers whose sum is 72.

Solution:

Let x be the first number.

Then the next two given integers will be x + 2 and x + 4 respectively. Their

sum is therefore x + (x + 2) + x + 4. But it is also given that their sum is 72.

The following linear equation is then obtained.

x + (x + 2) + x + 4 = 72

This equation is solved as follows:

3x + 6 = 72


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3x = 7263x = 66

3x = 22

Therefore, x + 2 = 24 and x + 4 = 26. Now, these values clearly solve the

original problemthat is 22, 24, and 26 are consecutive even integersand their sum is 72. Therefore the answers are 22, 24, and 26.

Example 2: When the smaller of the two consecutive integers is added to three

times the larger, the result is 43. Find the smaller integer.

Solution:

Let x be the smaller integer.

Then the larger integer is x + 1. The first sentence of the problem is now

translated into the equation x + 3 (x + 1) = 43 which can be solved:

x + 3x + 3 = 43

4x = 4334x = 40

x = 10

Therefore, x + 1 = 11. Now, substituting in the original problem shows that

the solutions are correct. So the integers are 10 and 11.

Note: When you work with consecutive integers, the following may be helpful.

Three consecutive Variables

integers x, x + 1, x + 2

odd integers x, x + 2, x + 4

even integers x, x + 2, x + 4

Example 3: In an algebra test, the highest score was 42 points more than the

lowest. If the sum of the two scores was 138, find the highest score.

Solution:

Let x be the highest score.

Then the lowest score is x42.Since the sum of the two scores is 138,

x (x42) = 1382x = 138 + 42


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2x = 180

x = 90

Thus, x42 = 28. It can be checked that the values obtained solve theoriginal word problem. Therefore, 90 is the highest score.

6.3.2 MIXTURE PROBLEM

In solving mixture problems it is usually helpful to form a table. It is also

important to note the following fact:

When two or more solutions are mixed, the amount of a certain

ingredient in the final mixture is equal to the sum of the amounts of

the same ingredient in the different parts.

Example 1: How many liters of pure alcohol must be added to 10 liters of water

to be able to produce a solution that is 25 % alcohol?Solution:

Let x = the number of liters of pure alcohol to be added. The

following table can be formed.

100% water 100% alcohol 75% water25% alcohol

Total volume(in liters)

100 x (10 + x)

Amount of

alcohol(in liters)

0 x .25 (10 + x)

The equation suggested by the comment at the start of 6.3.2, is:

0 + x = 0.25 (10 + x)

which can solved

x = 2.5 + 0.25x

x - 0.25 x = 2.5

0.75 x = 2.5

x = 3 1/3

Therefore, 3 1/3 liters of pure alcohol is needed.

Example: A chemist would like to produce 700 gallons of a solution that is

28% acid, but only solutions that have 30% acid and 20% acid,

respectively are available. How many gallons of the 30% and 20%


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acid solutions must be mixed to be able to obtain the desired

solution?

Solution:

Let x be the number of gallons of the 30% solution. Since the final solution

is to contain 700 gallons, the 20% solution should have a total volume of

700x gallons. In table form:30% acid 20% acid 28% acid

Total volume x 700 - x 700

Amount of acid(in gallons)

0.3x 0.2 (700x) 0.28 (700)

The equation is therefore:

0.3x + 0.2 (700x) = 0.28 (700)which is solved below:

0.3x + 1400.2x = 0.28 (700)0.3x + 1400.2x = 196

0.3x0.2x = 1961400.1x = 56

x = 560

700- 560 = 140

It can be shown that the answers check if substituted in the word problem.

Therefore 560 gallons of 30% acid solution and 140 gallons of 20% acidsolution must be mixed.

6.3.3 WORK PROBLEMS

Example 1: A farmer can plow a field in 2 days using a hand tractor. His

brother, by using hand plow and a carabao can plow the same fieldin 8 days. How many days will be required for the plowing if they

work together?

Solution:

Let x = the number of days it will take to plow the field if they work

together. Then the following table can be formed:


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Farmer withhand tractor

Brother withcarabao

Farmer andbrother

No. of days 2 8 x

Part of work

done in one day

1/8 1/x

The values on the third row are obtained by taking the reciprocal of those

on the second. Such should be the case, for example the farmer takes 2

days to finish the work and so after 1 day, he has finished only of the

work. In the same way, his brother can finish 1/8 of the work in the day.

Therefore, if they work together, their combined work in one day is + 1/8

of the entire work. But this should be 1/x as the entry on the lower right

shows.

Thus:

+ 1/8 = 1/x

and if 8x is multiplied,

4x + x = 8

So, 5x = 8 and x= 8/5 or 1 3/5 which can be checked in the original

problem. Therefore, it will take 1 3/5 days to finish plowing the field

if the two work together.

6.3.4 MOTION PROBLEMS

When a body is moving in a path at constant speed, the body travels

equal distances in equals intervals of time. This motion is referred to as

uniform motion. The rate or speedof the body in its path is defined as

the distance traveled in one unit of time.

Thus, if

r = the rate or speed

d = the distance traveled, and

t = the amount of time that elapsed

Then,d = r t

Example 1:

Brain-Brain walked from point A to another point B along JP Laurel St. at

the rate of 3 km per hour. As soon as he reached B, he went back to A


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this time on a jeep that averaged 57 km per hour. How far is A from B if

the entire trip took him 2/3 of an hour?

Solution:

It is important to see that there are two sets of d, r and t referred to in the

problem the first being the walk from A to B and the second the rideback. But these two are related in the following way:

The time spent on the two adds up to 2/3 hours. And the distances

covered are equal.

Let x = the distance (in km) between A and B

Since d = r t, it follows that

A to B B to A

Distance x x

Rate 3 57

Time x/3 x/57

Since the total time spent is 2/3 of an hour, the following equation holds:

Which is solved as follows

19x + x = 38

20x = 38

x = 1.9

After checking on the original problem, it can be concluded that the

distance A and B is 1.19 kms.

6.3.5 PROBLEMS ABOUT INVESTMENTS, PROFITS AND DISCOUNTS

In solving investment problems, the following formula is used:

I = P r


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This formula computes the annual simple interest (I) that is earned by an

investment of amount P (for principal) at the interest rate r

Example 1: A Davao businessman invested part of 110 at 12% interest rate and

the rest at 10%. If the two investments earned the same interest,

determine how much was invested in each.

Solution:

Let x = the amount (in pesos) invested at 12%. Then the following table

can be formed:

12% interest 10% interest

Principal X 110,000xRate 0.12 0.10

Interest 0.12x 0.10 (110,000x)

As the 2 investments earned the same interest,

0.12x = 0.10 (110,000x)The equation can be solved:

0.12x = 0.10 (110,000x)0.12x = 11,0000.10x0.22x = 11,000

x = 50,000 and

110,000x = 60,000

It can be checked that these values solve the original problem.

6.3.6 AGE PROBLEMS

Example 1: The sum of the present ages of two children is 18. Six years from

now the age of the older child will be twice the age of the youngerbrother. Find the present ages.

Solution:

Let x = present age (in years) of the younger child.

Then the following table can be formed.

Present age Age 6 years from


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now

Younger child x x + 6

Older child 18x (18x) + 6

From the given information,

(18x) + 6 = 2 (x + 6)Thus: 24x = 2x + 12

-3x = -12

x = 4 and

18x = 14

It can be verified that these values solve the original problem. Therefore

the younger child is now 4 years old while the other child is 14.

Example 2: An antique jar is 45 years older than an exact copy. In 13 years, the

older jar will be four times as old as the copy will be then. Find the

present ages of the jars.

Solution:

Let x = present age (in years) of the antique jar.

Then the table below is formed from the information given.

Present age Age 13 years hence

The exact copy X X + 13

Antique jar 45 + x (45 + x) + 13

Also, (45 + x) + 13 = 4 (x + 13)

And so: 58 + x = 4x + 52

-3x = -6

x = 2 and 45 + x = 47

Therefore, the antique jar is 47 years old and the copy is 2 years old.

6.3.7 GEOMETRIC PROBLEMS

Example 1: The perimeter of a certain rectangle is 16 times the width. The

length is 12 centimeters more than the width. Find the width of the

rectangle.Solution:

Recall the formula for the perimeter of a rectangle (in centimeters).

Then the length l and width w:

P = 2l + 2w


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Let w = width of the rectangle (in centimeters). The length is w + 12

cm,. This the perimeter is

P = 2 l + 2 w = 2 (w +12) + 2w

But the first sentence of the problem says that

P = 16w

Therefore:

16w = 2 (w + 12) + 2w

16w = 2w + 24 + 2w

16w4w = 24

12w = 24w = 2

Therefore, the width is 2 cm.

Example 2: One side of a triangle is 10 centimeters longer than the shortest

side. A third side is 20 centimeters longer than the third side. If the

perimeter of the triangle is 120 centimeters, find the length of the

shortest side.

Solution:Recall the formula for the perimeter of a triangle with sides having

lengths a, b, and c namely:

P = a + b + c

Let x = the length of the shortest side (in cm). Then the other two

sides have lengths x + 10 cm and x + 20 cm respectively.

Therefore:

(x) + (x+ 10) + (x + 20) = 120

or 3x + 30 = 120

3x = 90

x = 30

So, the shortest side is 30 cm long.


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6.3.8 RATIO

The quotient obtained by dividing two quantities is called their ratio. The

ratio of x to y is also denoted by the symbol x:y which read as x is to y. Itmay also be written in the form x/y where y0.

Example 1: Divide P72.00 between Jon and Pol so that the ratio of Jonsmoney to Pols money is 4:5

Solution:

There should be some number x for which Jons share = 4x and PolsShare = 5x. From the given problem, it follows that

4x + 5x = 72

or 9x = 72 and

x =8 Therefore, 4x = 32, and 5x = 40.

6.3.9 PROPORTION

A proportion is a statement of the equality of two ratios. This may be

written as a: b = c:d which is read as a is to b as c is to d, or a/b = c/d.Both are equivalent to ad = bc.

Example 1: A piece of steel chain 2 feet long weighed 4 pounds. What would

be the weight of a piece of the same chain 5 feet long?

Solution: Let x be the weight of the chain (in pounds). Then


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6.3.10 VARIATION

1. Direct variation

A variable y is said to vary directly as another variable x (or y is

directly proportional to x) if there is a nonzero constant k such

that y = kx. Here, k is called the constant of proportionality.

Example 1: Suppose y varies directly as x and y = 6 when x= 3

(a) Write y in terms of x

(b) Find y when x = 4

Solution:

It is desired to determine the value of k un y = kx.

Since y = 6 when x = 3, 6k 3 or k = 2Therefore, y = 2x

Finally, if x = 4, then y = 2 4 = 8

2. Inverse variation

When y = k/x, for some k 0, y is said to vary inversely as x (or yis inversely proportional to x)

Example 2: If y varies inversely as x and y = 4 when x = 3

(a) Write y in terms of x


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(b) Find y when x = 6

Solution:

This time y = k/x. Since y = 4 when x = 3,

4 = k/3 or k = 12

Therefore y = 12/x and y = 12/6 = 2 when x = 6.

3. Joint variation

If y, u, and v are variable and y = kuv, for some k 0, y is said tovaryjointly as u and v.

Example 3: The volume of a right circular cone varies jointly as the square of

the radius of its base and its height. Express this variation by an

equation.

Solution:

Let V be the volume, r the radius of the base and h the height of

the cone. Then, V= kr2h, for some k 0.

4. Combined variation

If y = kn/v for some k 0, y is said to vary directly as u andinversely as v.

Example 4: According to Newtons Law of Gravitation, force exerted by 2particles on each other varies jointly as their masses and inversely

as the square of the distance between them. Express this variation

by an equation.

Solution:

Let F be the force, m1and m2the masses of the d the distance.

Then F = km1m2/d for some k 0.CHAPTER VII INEQUALITIES AND ABSOLUTE VALUES

Objectives:

1. Identify and use the properties of inequalities and absolute values

2. Solve inequalities in one variable


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3. Simplify expressions involving values.

7.1 INEQUALITIES

One of the important properties of the real numbers is that they can be

ordered. Thus, 3 is smaller than 5 and 2 is bigger than1. Under thisordering the real numbers can be paired off with points on the real line

Given any 2 real numbers a and b, we write:

(i) a < b, read as a is less than b, if the point on the linecorresponding to a is on the left of the point corresponding to b:

(ii) a > b, read a is greater than b, if the point on the linecorresponding to a is on the right of the point corresponding to b:

Therefore a < b and b > a are one and the same inequality.

Aside from < and >, we also use the symbols and . We write a b, read aus less than or equal to b if eithera < b or a = b. Similarly, we write a b, reada is greater than or equal to b if either a > b or a = b.

The next two properties of inequalities are very important in solving inequalities:

1. (The Addition Property of Inequality) If a and b are any real numbers with

a < b, then a + c < b + c for any real number c

2. (The Multiplication Property of Inequality) If a and b are any real numbers

a < b and


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(i) c > 0 then ac < bc,

(ii) c < 0 then ac > bc

Illustrations:

1. Clearly, 9 < 14 and 9 + 3 < 14 + 3 and even 93 < 143. Note thatsubtracting 3 is the same as adding3, so that the addition property alsoworks when the same number is subtracted from both sides of an equality.

2. Also, 12 < 20, 12 2 < 20 2, and 12/2 < 20/2. Since dividing 2 is thesame as multiplying 1/2, the multiplication property also works when the

two sides of an equality are divided by the same number (that is not 0).

3. Finally, -6 < 4, -6 3 4 (-3) and -6/2 > 4/-2.

From the definition of the other inequalities, (>, , ), the properties listed above

still hold after making the appropriate modifications. For example, if a b then a + c b + c for any c,ac bc if c > 0, andac bc if c > 0.

Any inequality that can be expressed in the form ax + b < 0, ax + b 0, ax + b 0 or ax + b 0, where a and b are constant and a 0, is called a linearinequalityin the variable x. To solve such inequality means to determine all the

values of x that make the inequality true. The set of these values is called the

solution set of the inequality.

Example 1: Solve the inequality 4 (2x - 5) < 12

Solution:

We can divide both sides of the inequality by 4 (which is > 0) to get 2x 5< 3

Adding 5 to both sides (or transposing -5 to the other side), we get 2x < 8

Finally, dividing both sides by 2 (which is > 0) will give us x


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Example 2: Solve the inequality 2 (x4)8 > 2x6

Solution:

2 (x4)8 > 2x62x88 > 2x6Subtracting 2x from both sides and simplifying, we get -16 > -6

which is never true. Therefore the solution is empty.

The notation a < b < c is used to denote two inequalities and a c orsomething similar, is undefined.

Example 3: Solve the inequality and graph the solution set on the real line: 3 < -2x + 3 7

Solution:

The inequality can be solved by splitting into two, then solving these two

separately. The first, 3 < -2x + 3 can be solved as follows:

Subtracting 3 from both sides, we get 0 < -2x and transposing -2x

to the other side, we have 2x < 0. Dividing by 2, we obtain x < 0.

For the second inequality, -2x + 3 7, we get:-2x 4 and dividing by2 (which is < 0), x -2.

The solution set of the original therefore is:{x | x < 0 and x -2} or {x | -2 x < 0}

In the real line, the graph will be:

A better solution can be obtained by working on the two inequalities

simultaneously:

3 < -2x + 3 7Adding -3, we get 0 < -2x 4Then dividing by -2 0 > x -2, or -2 x < 0Which gives the solution set.

7.2 ABSOLUTE VALUES


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The absolute valueof a real number a is defined as follows:

a if a > 0,

| a | = 0 if = 0

-a if a < 0

Thus | 4 | = 4, | -3 | = -3 and | 0 | = 0. In the real line, we have:

3 4

That is | a | can be interpreted geometrically as the distance of the point afrom 0 on the real line. So | a | is never negative.

By comparing their definitions, it can be verified that for any real number a,

| a | = a2

For example, (4)2 = 16 while (-3)2 = 9 = 3In simplifying expressions with absolute values, the following properties

are important. For any real numbers a and b,

| ab | = | a | | b |, anda a

b b (if b 0)

Example 1: Solve for x: | 2x | = 3

Solution:

From the definition of the absolute value of a number, there are exactly 2

numbers whose absolute value is equal to 3, name 3 and3. Therefore, 2- x = 3 or 2x = -3. Simplifying separately, we get:

x = -1 or x= 5 (It can be verified that both actually solve the original

problem).

Example 2: Solve


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Solution:

Since , the equation can be rewritten:

As in example 1, it follows that

or

Which simplifies to

x4 = 22x or x4 = -2 + 2x3x = 6 or -x = 2

x = 2 or x= -2

Therefore, x = 2, or -2

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linear equations and applications, inequalities and absolute values

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