## Lesson 13: Simultaneous Linear Equations (I)

February 28, 2011

Vocabulary on simultaneous linear equations

### 1. Linear equations

A linear equation in two unknowns is an equation of first degree with two unknowns.

Any linear equation can be expressed in its standard form

ax + by = c

where a, b and c are known numbers.

A solution of a linear equation in two unknowns is a pair of values that verify the equation (they make it true)

A linear equation in two unknowns has infinite solutions.

We call system (or pair) of linear equations or simultaneous equations in two unknowns to a collection of such equations.

ax + by = c

a’x + b’y = c’

A solution of the system is any pair of numbers that satisfies them simultaneously.

Solving a pair of simultaneous equations is to look for the values of the unknowns which make all the equations true simultaneously

### 2. Methods of solving simultaneous equations

We now focus on various methods of solving simultaneous equations. All these methods have the same philosophy: to obtain from the two equations another equation only in one unknown. Once we solve it is easy to determine the value of the other unknown.

### 2.1. Substitution method

Here it is the method of substitution:

• Solve one of the equations for one unknown in terms of the other.
• Then, substitute that in the other equation.
• That will yield one equation in one unknown, which we can solve.
• We obtain the other unknown substituting in any equation of the system

In this page you can introduce any system of linear equations, solve it by yourself and find a complete explanation about how to do it by using substitution method. Unfortunately, this algebra solver don’t use elimination method.

Here you are a video on substitution method.

### 2.2. Equalizing Method

This method is not explained in Anglo-Saxon countries, probably because it is a slight variation of substitution method.

Here it is the method of “Equalizing”:

• Solve both equations for the same unknown in terms of the other.
• Then, equalize the expressions in terms of the other.
• That will yield one equation in one unknown, which we can solve.
• We obtain the other unknown substituting in any equation of the system

Sorry, I couldn’t find a web page where the equalizing method were explained and that had interactive exercises.

### 2.3. Elimination method

This method is also named addition method.

As the name suggest, this method tries to eliminate variables until there’s only 1 variable left by adding the equations.

Here it is a general strategy for solving simultaneous equations:

• Look at the equations and try to find 2 equations which has the same coefficient (plus or minus) for the same variables.
• When the coefficients don’t have the same coefficients (plus or minus), multiply the equations so that the coefficients of that unknown are the same
• Add or subtract  the equations vertically, and that unknown will cancel(we eliminate the unknown)
• We will then have one equation in one unknown, which we can solve.
• To solve for y, the other unknown substitute the value of x in one of the original equations.

This page is highly recommendable because you can find several videos on solving simultaneous equations by using elimination method, but they have got a higher level. The video below is elemental but good to improve your listening

In the page I am going to recommend you, you have interactive exercises on solving simultaneous equations by using elimination method. It has a difficulty because they always subtract instead of add up as we do.

Take into account that if you subtract an equation from other you have to change every sign in the LHS and in the RHS to subtract correcly.

Go down the page to find the exercise. Get an example and try it by yourself. If you have got any difficulty, click on the buttons below. Finally, use only example 1. Example 2 is harder to understand for you.

## Lesson 7: Quadratic Equations (III)

February 11, 2011

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.

NOTES:

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.

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
• 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.

Remark.

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.

NOTE

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.

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.

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.