Archive for the ‘Lesson 7: Linear Equations and Quadratic Equations’ Category


Lesson 7: Solution to Diophantus Riddle (VI)

March 17, 2012

This is the translation of the epitaph into Spanish:

” Aquí fueron sepultados los restos de Diofanto. La infancia de Diofanto duró 1/6 de su vida, 1/12 en la adolescencia, cuando la barba cubrió su cara, Después de 1/7 de su vida contrajo nupcias. Luego de cinco años de casado nació su primer hijo. El hijo vivió ½ de la vida de su padre, su padre buscó consuelo en los números pero no lo logró y murió cuatro años después que él.”

Let x be the age of Diophantus. If we translate the epitaph into Algebra:

x/6 +x/12 +x/7 +5 +x/2 +4 = x
14x/84 +7x/84 +12x/84 +420/84 +42x/84 +336/84= 84x
14x +7x +12x +420 +42x +336= 84x
14x +7x +12x +42x -84x= -420 -336
-9x = -756
x = -756/-9 =  84  years old.


Then we have that Diophantus was born about 200 and died about 284.

Thanks  Lidia for your effort!


Lesson 7: Linear and Quadratic Equations. Maths Are Magic (V)

March 5, 2012

How old are you? 

If you ask this to your great-aunt it would be indiscreet. So we are going to do it in other way …Play the following game

Tell her:Think of a number between 1 and 9.
Multiply it by 9.
Take away this result from 10 times your age and tell me the answer. 

Now you could guess your great-aunt’s age.


You will always get your great-aunt’s age. Is it Magic or Maths?

Why are we always right? Could you give a logic reason?

Freely translated from Matemágicas. Descartes Project

Lesson 7: Who Was Diophantus? Diophantus’s Riddle (IV)

February 25, 2012

Diophantus of Alexandria


Born: about 200   Died: about 284

Diophantus is often known as the ‘father of algebra’,but there is no doubt that many of the methods for solving linear and quadratic equations go back to Babylonian mathematics. Nevertheless, his remarkable, collection of problems is a singular achievement that was not fully appreciated and further developed until much later.

 He is best known for his Arithmetica, a work on the solution of algebraic equations and on the theory of numbers. The Arithmetica is a collection of 130 problems giving numerical solutions of determinate equations (those with a unique solution), and indeterminate equations. The method for solving the latter is now known as Diophantine analysis. Only six of the original 13 books were thought to have survived and it was also thought that the others must have been lost quite soon after they were written. Diophantus was the first Greek mathematician who recognized fractions as numbers; thus he allowed positive rational numbers for the coefficients and solutions.

 Diophantus did not use sophisticated algebraic notation, he did introduce an algebraic symbolism that used an abbreviation for the unknown and for the powers of the unknown.

 However, essentially nothing is known of his life and there has been much debate regarding the date at which he lived.

You can read a good biography on,

where I found this information. There are many biographies of famous mathematicians.

 Diophantus Riddle

 Much of our knowledge of the life of Diophantus is derived from a 5th century Greek anthology of number games and strategy puzzles. One of the problems (sometimes called his epitaph) states:

‘Here lies Diophantus,’ the wonder behold.

Through art algebraic, the stone tells how old:

‘God gave him his boyhood one-sixth of his life,

One twelfth more as youth while whiskers grew rife;

And then yet one-seventh ere marriage begun;

In five years there came a bouncing new son.

Alas, the dear child of master and sage

After attaining half the measure of his father’s life chill fate took him.

After consoling his fate by the science of numbers for four years, he ended his life.’


This puzzle implies that Diophantus lived …..Could  you tell it to us?

You will publish a post with your solution on this blog and there is an interesting reward for you!

 However, the accuracy of the information cannot be independently confirmed.


Lesson 7: Quadratic Equations (III)

February 11, 2011

6. Quadratic equations

 An equation is a quadratic equation in one variable (or unknown) when:

  • It has only one variable or unknown
  • The unknown is squared x2 at least once in the equation.

 For instance: 3x23x=x–1.

 Standard form

 If we bring all the terms over to the RHS (right hand side) and the LHS (left hand side) is equal to 0 we get:

3x2 – 4x + 1 = 0

which is the form we should always use to express quadratic equations to be able to solve them?

 Any quadratic equation in one variable can be expressed using its standard form:

 ax2+bx+c = 0

 where a, b  and c are known numbers and a≠0.


 1) If a quadratic equation has a = 0, then the equation is a linear equation; this is a first degree equation

 2) If you look at the LHS you find a second degree polynomial in one variable. Because of it, these equations are also named second degree polynomial equations in one variable or second degree equations for shorting (it isn’t very precise).

 3) In many cases once the equation has this form it can be simplified which is very useful.

 7. Solving quadratic equations

 We saw that any quadratic equation can be expressed in its general form ax2+bx+c = 0

 where a, b  and c are known numbers and a≠0.

 We name incomplete quadratic equations to that quadratic equations that have  or . They have the forms:

ax2+c = 0    where    b = 0 

ax2+bx = 0    where    c = 0. 

There are several methods you can use to solve a quadratic equation: 

  • Factoring
  • Completing the Square
  • Quadratic Formula
  • Graphing

 This course, we will practice the Quadratic Formula and a very elemental case of factoring for incomplete quadratic equations.

 First of all, we are going to study the easy cases, when b = 0 or c = 0.

 7.1. Solving quadratic equations of the form ax2+c = 0    where    b = 0 

 In this case you have to follow the same method as in linear equations:

  • Transpose the number to the LHS

 ax2 = c

  • Transpose the coefficient of the quadratic term to the RHS

 x2 = c/a

  • Isolate the unknown by squaring the RHS.

 x  = ±√¯c/a

Take into account that in a square root:

  • If the radicand is positive there are two roots of the same absolute value and opposite signs.
  • If the radicand is zero there one only root, zero.
  • If the radicand is negative there are no real roots.

 When solving quadratic equations of the form  ax2+c = 0    where    b = 0 

  • If –c/a > 0 , there are two solutions x1  = +√¯c/a  and x2  = √¯c/a
  • If –c/a = 0 , there are two solutions x1  = x2  =0
  • If –c/a < 0 , there are no solutions

 7.2. Solving quadratic equations of the form ax2+bx = 0    where    c = 0. 

In this case you have to follow the method of factoring the equation:

In general, if we are solving an equation of the form

  • Factor out the common factor .

 x·(ax + b) = 0

  • Apply the following property: “If the product of two numbers is zero then, one of the factors is zero”.

 x·(ax + b) = 0  →  x = 0    or   ax + b = 0

  • Solve the resulting equations.

 x = 0                    and           ax + b = 0   →   x= –b/a

  When solving quadratic equations of the form  ax2+bx = 0    where    c = 0.the solutions are x = 0 and x= –b/a 

 7.3. Solving equations in the standard form 

 The quadratic formula uses the “a“, “b“, and “c” from “ax2 + bx + c“, where “a“, “b“, and “c” are just numbers; they are the “numerical coefficients”.

 The solutions of an equation of the form  can be obtained by applying the quadratic formula:

 This is there are two solutions corresponding to each sign preceding the square root. We will denote them x1 and x2 .

Practise plugging into the formula on this web page.

Warning: Don ‘t panic if you find “complex numbers” roots with i. These number belong to a set of number you don’t know. If the radicand is negative there are no real solution. This is enough for you.

 The Discriminant

Notice the expression under the radical b2 – 4ac in the quadratic formula is called the discriminant.  From this number we can determine the nature of the solutions of a quadratic equation.

Three possible cases:
1. Δ = b2 – 4ac = 0 Exactly one real number solution exists.
2. Δ = b2 – 4ac > 0 Two real number solutions exist.
3. Δ = b2 – 4ac < 0 There isn’t any real solution.


 The argument of the square root, the expression b2 – 4ac, is called the “discriminant” because, by using its value, you can discriminate between (tell the differences between) the number of solutions of a quadratic equation.

 No real number has a negative square.

Practise finding the of a quadratic equation and deducing from it the number of solutions of the equation in this web page


 1. When you are using the quadratic formula you’re just plugging into a formula. There are no “steps” to remember, and there are fewer opportunities for mistakes, but if you have an incomplete equations use the other techniques, are simpler. You can also use the formula, of course!

 2. When using the formula, make sure you are careful not to omit the “±” sign, and be careful with the fraction line (don’t draw it as being only under the square root; it’s under the initial “–b” part, too). And, though many of your quadratic equations will start with “x2” so a = 1, don’t forget that the denominator of the Formula is “2a“, not just “2”; that is, when the leading term is something like “3x2“, you will need to remember to put the “a = 3″ value in the denominator.

 Finally here you are a video on how to use the discriminant to find out the number of solutions of a quadratic equation.


Lesson 7: More Equations. Word Problems (II)

February 1, 2011

4.2. Solving equations with brackets

 When equations involve brackets then we have to:

1º.  Eliminate brackets

2º.  Group terms with  in the left-hand side and numbers in the right-hand side.

3º.  Collect like terms

4º.  Isolate  in the left-hand side, that results in  equals one number. This number is the solution.

5º.  Check the solution.


 Remember that if a – sign precede brackets when you eliminate brackets, all the terms inside changes their signs. For instance;  –2(x – 3) = –2x + 3

View the following video on solving equations with brackets

4.3. Solving equations with denominators

When equations have denominators you have to:

1º.  Eliminate denominator by multiplying both sides by the HCF of all the denominators that appear

2º.  Eliminate brackets

3º.  Group terms with  in the left-hand side and numbers in the right-hand side.

4º.  Collect like terms

5º.  Isolate  in the left-hand side, that results in  equals one number. This number is the solution.

6º. Check the solution.

Practise on this web pages equations with brackets and denominators, easy, I believe

In this other link you can find harder equations. Solve only linear equations. The others have radicals and algebraic fractions. When you click on solution you will find the example solved. If you go at the bottom of the page where the solution is you will find other example (click on it). If you would like to work with some similar problems click on problem.

Finally a video. Try to understand. If you are in a hurry write to me.

4.4. Equations that are identities

In some exercises, you have to solve an equation that results to be an identity. Let’s see the following example.


Solve 2x – 2 =2(x +2)  – 6

 First we eliminate brackets: 2x – 2 =2x +4  – 6

 Transposing terms: 2x – 2x = 4  – 6 + 2

 Collecting like terms: (2 – 2)x = 0 →0·x = 0

 But any number (known or unknown) multiplied by zero gives zero, so:

 We deduce that the equality is always true. This equation is true, regardless of the value of which it indicates it is an identity. 

 We can say that an equation of the form , is an identity.

4.5. Equations without solutions

Some equations have no solutions. They are named inconsistent equations. Here it is an example.


Solve x – 3 = 2 + x 

 Transposing terms: x – x = 2 + 3

 Collecting like terms (1 – 1)·x = 5  → 0·x = 0

 But any number (known or unknown) multiplied by zero gives zero, so: 0 = 5

 In the example you found that . What does this mean? Obviously, this equations cannot be true, regardless of the value of which it would indicate it is an identity. The equality is always false.

 We can say that an equation of  the form   0·x = number, doesn’t have a solution.

 5. Solving words problems with linear equations

 Most of the time when someone says “word problems” there is automatic panic. By setting up you can be successful with word problems. So what should you do? Here are some recommended steps:

  1.  Read the problem carefully and understand what it is asking you to find. Usually, but not always, you can find this information at the end of the problem.
  2. Assign a variable to the quantity you are trying to find. Most people choose to use x, but feel free to use any variable you like. For example, if you are being asked to find a number, some students like to use the variable n. It is your choice.
  3. Write down what the variable represents. At the time you decide what the variable will represent, you may think there is no need to write that down in words. However, by the time you read the problem several more times and solve the equation, it is easy to forget where you started.
  4. Re-read the problem and write an equation for the quantities given in the problem. Make yourself the following questions: What do we know? (DATA) How can I translate the information into algebra?
  5. Solve the equation.
  6. Answer the question in the problem. Just because you found an answer to your equation does not necessarily mean you are finished with the problem. Many times you will need to take the answer you get from the equation and use it in some other way to answer the question originally given in the problem.
  7. Check your solution. Your answer should not only make sense logically, but it should also make the equation true. If you are asked for a time value and end up with a negative number, this should indicate that you have made an error somewhere. If you substitute these unreasonable answers into the equation you used in step 4 and it makes the equation true, then you should re-think the validity of your equation.

The links below contain different types of problems. Try the type you want to practise.


Lesson 7: Linear and Quadratic Equations (I)

January 27, 2011

Vocabulary on Linear and Quadratic Equations 

 Lesson 7: Linear and Quadratic Equations (Notes)

Equation Worksheets

Linear Word Problems Worksheet

I Have to Know by the End of this Lesson  


1. Identity, formula and equation 

An equality is formed by two expressions separated by equal sign =.

 A numeric equality is formed only for numbers.

 An algebraic equality is formed by two algebraic expressions separated by equal sign.

 An algebraic equality can be:

  • an identity when it is true for any value of the variables (letters).
  • an equation when it is true for some values of the variables (letters).
  • a formula when one variable  is equal to an expression in a different variable(s).


5x-2=7x , algebraic equality, equation

x+x=2x , algebraic equality, identity

P= 4l , algebraic equality, it is the perimeter of a square of side l

 2. Elements of an equation

 Remember that an equation is made up of two expressions separated by an equal sign. You have to learn the following terminology:

 In any equation there are two expressions separated by an equal sign, the expression on The Left Hand Side (LHS), in Spanish “primer miembro” and the expression on The Right Hand Side (RHS), in Spanish “segundo miembro”.

 Term of an equation is any addend of its algebraic expressions (LHS) or (RHS).

 The degree of any equation is the highest exponent that appears on the terms with unknown number(s) after collecting like terms.

 Unknowns are the variables that appear in the equation and that are unknown. They are typically called .

 In what is called a linear equation, the variable or variables appear only to the first power. A linear equation is also called an equation of the first degree or simple equation.

 Solutions are the values of the unknown(s) that make true the equation. We will say they satisfy the equation.

 Equivalent equations are equations that have the same solutions. 

 Solving an equation is finding its solution(s).

Learn to check if your solution is valid  on this link

Does x satisfy the equation?

 3. Transposing terms

 An equation is like a weighing scale – both sides should always be perfectly balanced. To solve the equation you need to find the value of missing numbers and perform the same operation to each side.

 We know that:

  • If we add or subtract the same number or algebraic expression in both sides of the equation we obtain an equivalent equation. We get the same answer in an easier way if we “change the sign (+ to – or – to +) when we take terms over to the other side of the equation”.
  • If we multiply or divide by the same number or algebraic expression in both sides of the equation we obtain an equivalent equation. We get the same answer an easier way if we “change the sign (· to: or: to ·) when we take terms or coefficients over to the other side of the equation”.

 This technique is named transposing terms. We will use it to solve equations.

 4. Solving linear equations

After transposing and collecting like terms, every linear equation is equivalent to an equation of the form


with a, b, known numbers.

This equation is called standard form of the linear equation.

 4.1. Solving easy equations

In general, to solve an equation for a given variable, you need to “undo” whatever has been done to the variable. You do this in order to get the variable by itself; in technical terms, you are “isolating” the variable. This results in “(variable) equals (some number)”, where (some number) is the answer we are looking for.

The steps are:

1º.  Group terms with  in the left-hand side and numbers in the right-hand side by transposing terms.

2º.  Collect like terms

3º.  Isolate  in the left-hand side, that results in  equals one number. This number is the solution. Recall  it must be positive.

.  Check the solution.


3x-3 = x+9

We want to isolate the variable in the LHS so we first transpose 3 that goes to the RHS adding

 3x = x+9+3

We collect like terms

 3x = x+12

We transpose  x , that goes to the RHS subtracting

 3x-x = 12

We collect like terms

 2x = 12

Finally, we transpose 2, that goes to the RHS dividing:

 x = 12/2

x = 6

Remember you can check your solution

 3· 6 -3 = 6+9

Then the solution is correct.

Here you are several links to practise solving easy equations.

Solve one-step linear equations

Solve two-step linear equations

Solve multi-step equations

Solve equations involving like terms

Finally,  here you are a couple of videos on solving linear equations. Improve your listening an pronountiation