February 8, 2013
Elio Godoy Vilches
Contents
1. Denitions 1
2. Solving simple equations: 1
3. Solving more complex equations 2
4. Equations with brackets 3
5. Equations with fractions 4
6. Solving problems with equations 5
7. Solving an equation for a specic variable 7
1. Denitions
An equation is a mathematical sentence containing an equal sign. The expression on the left side of the equal sign has the same value as the expression on the right side. It tells us that two expressions mean the same thing, or represent the same number.
An equation can contain variables and constants. Using equations, we can express math facts in short, easy-to-remember forms and solve problems quickly.
In an equation, letters stand for a missing number: one or both of the expressions may contain variables. Solving an equation means manipulating the expressions and nding the value of the variables.
An example might be: Solve the equation2a+ 3 = 7.
This means we need to nd the value of a. The answer isa= 2
2. Solving simple equations:
The rule we use to solve equations is change side change sign, that is to say, by using 'inverses', or undoing what the equation is doing.
Adding and subtracting are the inverse (or opposite) of each other. Multiplying and dividing are the inverse of each other.
Example 1:
x−5 = 11 (-5 opposite is +5)
So, undo the -5 by adding 5. Remember, you need to do it to BOTH sides!
x−5 + 5 = 11 + 5
So x= 16
Example 2:
3y= 18(times by 3 opposite is divide by 3)
So, undo times by 3 by dividing by 3. So, y= 6
Example 3:
4x−6 = 30 (-6 opposite is +6)
4x−6 + 6 = 30 + 6
4x= 36(times by 4 opposite is divide by 4)
Exercise 1:
Solve the following one-step equations:
1. x+ 12 = 3
2. 5 =x+ 2
3. 2x= 10
4. 3x= 12
5. 2 =x−1
6. x+ 5 = 7
7. 3 = 2−x
8. 5x= 4
9. 4−x= 6
10. 2x= 18
11. x
3 =−4
12. 2 = −x4
3. Solving more complex equations
Example 1
When a one step equation is solved, essentially the constants are sorted to one side of the equation and the variable to the other.
x−8 = 12
x−8 + 8 = 12 + 8
x = 16
Think of this as moving the constant 8 to the other side of the equation.
x−8 = 12
x = 12 + 8
Example 2
Solve3x−8 = 5x+ 2
Move terms involving variables the same way. Either the 3x or the 5x could be moved. Choose the term containing the variable with the smaller coecient. This doesn't matter now, but will make things easier in other equations.
3x−8 = 5x+ 2
3x−8−3x= 5x+ 2−3x
Now there is only a two step equation to solve.
−8 = 2x+ 2
Exercise 2:
Find the value of x in the following equations
1. 3x+ 6 = 12
2. 2 = 3x=1
3. 2x+ 1 =x+ 2
4. 3x+ 2 = 5x=3
5. 2x=3 =x
6. 6x=5 = 3x=1
7. 6x+ 2 = 2=3x
8. x+ 5 = 7
9. 3 = 2−x
10. −2x+ 1 =−3x+ 2
11. 5x−3 = 2−4x
12. 6x−4 = 15−3
13. 3x−2 + 4x= 5x−2
14. x+ 1−3 = 2x+ 4−x
4. Equations with brackets
To solve equations with brackets follow the steps: 1. Distribute.
2. Combine like terms.
3. Move all the variables to the same side. 4. Solve the two step equation.
a) Subtract or add b) Divide.
Every step isn't needed in every equation.
Example 1
To solve this equation 3(2x−8) = 12 First distribute 6x−24 = 12 Now it is a two step equation
6x−24 + 24 = 12 + 24
6x= 36
x= 6
Example 2
45 = 7(2x−8)−(7 + 5x) Distribute −8 + 7x−(4x−1)−3(−x−8) =−70 45 = 14x−56−7−5x Combine like terms −8 + 7x−4x+ 1 + 3x+ 24 =−70
45 = 9x−63 Add or subtract 6x+ 17 =−70
108 = 9x 6x=−87
108 9 =
9x
9 Divide
6x
6 =
−87 6
Example 3
If the equation contains nested parenthesis, work the distribution from the inside out.
4−3(2x+ 4(7−6x)) = 90 Distribute 3(4x−5(2x−6) + 7)−(x−(x−1)) = 200 4−3(2x+ 28−24x) = 90 Combine lilke terms 3(4x−10x+ 30 + 7)−(x−x+ 1) = 200
4−3(28−22x) = 90 Distribute 3(−6x+ 37)−1 = 200
4−84 + 66x= 90 Combine like terms −18x+ 111−1 = 200 66x−80 = 90 Add or subtract
66x= 170 Divide −18x= 90
66x
66 = 170
66
−18x
−18 = 90
−18
x= 17066 = 8533 x= −9018 =−5
Exercise 3
Practice with distribution
1. 3(2x+ 5 =−8
2. 9 = 4(4x−8) + 12
3. 15 = 3(4x−8)
4. 2(3f + 6) =−9
5. 7 = 5(2r−10) + 1
6. 60 = 60(7t−7)
7. 7(2v+ 5) = 0
8. 3 = 14(6b−2) + 20
9. 25 = 3(8a−4)
10. 3(2x−5) + 4(6−5x) = 23
11. 5(2x−9)−(2x−8) = 0
12. 9 = (7x−9)−8(2−2x)
5. Equations with fractions
To solve an equations with fractions, we transform it into an equation without fractions which we know how to solve. The technique is called clearing of fractions.
Example 1
Solve for x x
3 +
x−2 5 = 6
Multiply both sides of the equation every term by the LCM of denominators. Every denominator will then cancel. We will then have an equation without fractions. The LCM of 3 and 5 is 15. Therefore, multiply every term on both sides by 15:
15·x 3 + 15·
x−2
5 = 15·6
Each denominator will now cancel into 15, that is the point, and we have the following simple equation that has been "cleared" of fractions:
8x = 90 + 6
8x = 96
x = 96÷8
x = 12
We say "multiply" both sides of the equation, yet we take advantage of the fact that the order in which we multiply or divide does not matter. Therefore we divide by each denominator rst, and thus clear of fractions.
Notice, that in some problems, when a single fraction is equal to a single fraction, then the equation can be cleared by cross-multiplying
x−2
4 =
x
7 7(x−2) = 4x
7x−14 = 4x
7x−3x = 14
4x = 14
x = 14÷4
x = 7 2
Exercise 4 Practice
1. 2a−9
−3 = 3a+8
5
2. 2x−5 3 =
3x−8 5
3. m+10 7 =
m−4 5
4. 2a−9 3a+2 =
8 5
5. m+10 5m+2 =
4 3
6. 2−9f
5−f =
10 15
7. 9
−1 = 8s+9 5s−10
8. 3 4 =
t+3
t−8
9. 24 8 =
k
10k−8
Exercise 5
Solve the following equations:
1. x−1 6 −
x−3 2 =−1
2. x−1 4 −
x−5 36 =
x+5 9
3. 2(4x−7)
−3 + 8
−6 =
4(−4x+1) 6 +
16 6
4. −5(x−5)
−1 −
−7
−2 =
3(−8x−2) 6 +
−3 6
5. 2(5x+8) 9 +
−6 9 =
−6(x−1) 9 +
−12 9
6. 2(−7x−6) 2 +
−3
−1 =
3(−3x−2) 2 +
15 2
6. Solving problems with equations
So far you have concentrated on solving given equations. Making up your own equations helps you to solve dicult problems. There are four steps:
2. Write the problem in the form of an equation. Read the problem carefully. 3. Solve the equation and give the answer in words.
4. Check your solution using the problem and not your equation.
Exercise 6
Find the number in each question by forming an equation and solving it: 1. If you multiply the number by 2 and then subtract 5, the answer is 4 2. If you multiply the number by 10 and then add 19, the answer is 16
3. If you add 3 to the number and then multiply the result by 4, the answer is 10 4. If you subtract 11 from the number and then treble the result, the answer is 20 5. If you double the number, add 4 and then multiply the result by 3, the answer is 13 6. 6) If you treble the number, take away 6 and then multiply the result by 2, the answer is
18
7. If you double the number and subtract 7 you get the same answer as when you add 5 to the number.
8. If you multiply the number by 5 and subtract 4, you get the same answer as when you add 3 to the number and then double the result.
9. If you multiply the number by 6 and add 1, you get the same answer as when you add 5 to the number and then treble the result.
10. If you add 5 to the number and then multiply the result by 4, you get the same answer as when you add 1 to the number and then multiply the result by 2.
Exercise 7
Solve each problem by forming an equation. The rst questions are easy but should still be solved using an equation, in order to practise the method.
1. The length of a rectangle is twice the width. If the perimeter is 20 cm, nd the width. 2. The width of a rectangle is one-third of the length. If the perimeter is 96 cm, nd the
width.
3. The sum of three consecutive numbers is 276. Find the numbers. Let the rst number be x.
9. The sum of three numbers is 28. The second number is three times the rst and the third is 7 less than the second. What are the numbers?
10. David weighs 5 kg less than John, who in turn is 8 kg lighter than Paul. If their total weight is 197 kg, how heavy is each person?
11. A man is 32 years older than his son. Ten years ago he was three times as old as his son was then. Find the present age of each one.
12. Mr Lee left his fortune to his 3 sons, 4 nieces and his wife. Each son received twice as much as each niece and his wife received ¿6000, which was a quarter of the money. How much did each son receive?
7. Solving an equation for a specic variable
Some equations contain more than one variable. To solve an equation or formula for a specic variable, you need to get that variable by itself on one side of the equation. When you divide by a variable in an equation, remember that division by 0 is undened.
Example
Solve the formulad=rt for t.
The variable t has been multiplied by r, so divide each side by r to isolate t. d r =
rt r or
d r =t.
So t= dr, wherer6= 0.
Exercise 8
1. Solve4a+b= 3afor a(Hint: begin by subtracting3afrom each side)
2. Solve c+d
3 = 2c for c(Hint: begin by multiplying each side by 3)
Exercise 9
Solve each equation for the variable specied:
1. f =epd, for e
2. 12g+ 31h=−8g, forh
3. y =mx+b, forb
4. v =r+at, forr
5. P V =nRT, for V
6. G=F−D, for D
7. 3c+ 5d= 7d−6c, ford
8. W =mgh, forg
9. 3x+y
c = 4, forc
10. 5xy+n
2 =−6, fory
8. Reading activity
AL-KHWARIZMI Muhammad Ibn-Musa (780-850)
He was born in the town of Khwarizm (now Khiva), in Khorasan province of Persia (now in Uzbekistan). The name al-Khwarizmi means the person from Khwarizm. His family moved soon afterwards, to a place near Baghdad, where he accomplished most of his work in the period between 813 and 833. All of Al-Khwarizimi's works were written in Arabic.
He developed the concept of the algorithm in mathematics (which is a reason for his being called the grandfather of computer science by some people), and the words "algorithm" and "al-gorism" come from Latin and English corruptions of his name. He also made major contributions to the elds of algebra, trigonometry, astronomy, geography and cartography. His systematic and logical approach to solving linear and quadratic equations gave shape to the discipline of algebra, a word that is derived from the name of his 830 book on the subject, Hisab al-jabr wa al-muqabala).
While his major contributions were the result of original research, he also did much to synthe-size the existing knowledge in these elds from Greek, Indian, and other sources. He appropriated the place-marker symbol of zero, which originated in India, and he is also responsible for the use of Arabic numerals in mathematics.
FERMAT, Pierre de (1601-1665)
Pierre de Fermat was born in Beaumont-de-Lomagne, France in August of 1601 and died in 1665 in Toulouse. He is considered to be one of the greatest mathematicians of the seventeenth century. Fermat's father was a leather merchant and his mother's family was in the legal profes-sion. Fermat attended a Franciscan monastery before moving on to obtain a Bachelor's Degree in civil law from the University of Orleans in 1631. He married, had ve children and practiced law. For the most part, Math was a hobby for Fermat. Fermat was a busy lawyer and did not let his love of math completely take over his time. It's been said that Fermat never wanted anything to be published as he considered math to be his hobby. The only one thing he did publish - he did so anonymously. He sent many of his papers by mail to some of the best mathematicians in France. It was his link with Marin Mersenne that gave Fermat his international reputation. Fermat loved to dabble in math and rarely provide his proofs (evidence or procedures for reaching conclusions), he would state theorems but neglected the proofs! In fact, his most Famous work 'Fermat's Last Theorem' remained without a proof until 1993 when Andrew J. Wiles provided the rst proof. During Fermat's lifetime, he received very little recognition as a mathematician, if not for the fact that others saved his papers and letters, he may not be the legacy that he is seen as today.
The Most Famous Question In Math History for 350 Years!. Fermat's most important work was done in the development of modern number theory which was one of his favorite areas in math. He is best remembered for his number theory, in particular for Fermat's Last Theorem. This theorem states that: xn+yn=zn has no non-zero integer solutions forx, y andz whenn is greater than 2.
Famous Quote: "I have a truly marvelous demonstration of this proposition which this margin is too small to contain."
Exercise 10
