Simple equations to an unknown factor     
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Reception

SECOND PART : ALGEBRA

4. - EQUATIONS - IDENTITIES

4. 1. - INTRODUCTION TO THE CONCEPT OF EQUATION

1 - When you write 3 + 7 = 7 + 3, you obtain a numerical equality. You can check it quickly besides, by carrying out calculations indicated on both sides sign = and by you ensuring that the found results are the same ones.

2 - If you write 3x now + 7 = 7 + 3x, you obtain a literal identity (notice that while replacing (x) by an unspecified numerical value, you find a numerical equality).

Definition of the identity: An identity is a literal equality which is checked for all the numerical values that one can give to the letters that it contains.

Example : Let us take again the identity 3x + 7 = 7 + 3x and give to (x) the following values :

x = 1 gives us 3 + 7 = 7 + 3

x = - 1 gives us - 3 + 7 = 7 - 3

x = 10 gives us 30 + 7 = 7 + 30 ….

Also let us quote : (a + b)2 = a2 + 2ab + b2 ....

3 - If you now write 2 (x + 6) = 4 x + 6, you can believe at first sight that they is false, and they would be false indeed, if you had wanted to write an identity. The equality above, is checked only for one particular value of X. It is an equation. If you give value 3 for x, the equation is solved and you found a numerical equality. No other value of X solves this equation.

Definition: An equation is a literal equality which is checked only for certain values allotted to the letters that it contains.

These values are the solutions or roots of the equation.

To solve an equation is to find all these roots (or solutions). In complex equations, several solutions can answer the problem and solve the equation. Us will study not equations known as of second or third degree, which would leave the framework of our lesson, the purpose of which is not, as we said, to facilitate the comprehension of the various theories to you.

4. 2. - RESOLUTION OF THE EQUATIONS

To solve an equation, it is necessary to try to replace it by a simpler equivalent equation, of which one knows the roots.

Regulate 1 : When one adds or cuts off the same algebraical expression with the two members from an equation, one obtains an equation equivalent to the first.

That is to say    3 (x + 2) = 2x + 3

One can add (or cut off) with the two members, an algebraic number “A” unspecified without modifying the equation.

One can thus write : 3 (x + 2) + A = 2x + 3 + A

A” can be a pure number, or contain unknown factor x.

Note : One can transpose a term of a member of an equation in the other with the proviso of changing his sign.

Example :

3 + 2x = 3x + 4

Let us cut off 3 in each member :

3 + 2x - 3 = 3x + 4 - 3

Let us carry out the possible operations in the first member; it remains :

2x = 3x + 4 - 3

We observe that the term 3, which had the sign + in the first member, has the minus sign in the second, which confirms the stated rule well.

Thus let us observe this rule for the 3x term :

2x - 3x = 4 - 3

                                                                                           - x = 1

                                                                                or         x = - 1

Note : It is always necessary, in the final result, to express “x positive”. If one obtains “x negative”, it is enough to multiply the two terms of the equation by - 1 to obtain “x positive”.

Regulate 2 : When one multiplies or divides the two members of an equation by the same nonnull algebraical expression, one obtains an equation equivalent to the first.

  P7 

A, B and C being algebraical expressions.

APPLICATIONS :

1 - One can “drive out the denominators” of an equation, by multiplying one of his members by the denominator of the other and conversely.

Example :

5x - 11 / 4 = 3x + 20 / 15

“To drive out” the denominators, one multiplies by 15 the member of left and 4 the member of right-hand side :

(5x - 11) . 15 = (3x + 20) . 4

Foot-note: That points out the property of the proportions to us “produces extremes = produced means”.

The solution is :

75x - 165 = 12x + 80

63x = 245

x = 245 / 63

Another example :

That is to say : 5x - 11 / 4 = 3x + 20

Let us multiply by 4 the member of right-hand side. 4 “is driven out” of the left member and comes to multiply the right member.

5x - 11 = 4 (3x + 20)

The resolution of the equation continues, then, as we saw above:

5 x - 11 = 12x + 80

- 91 = 7x

from where : x = - 91 / 7 = - 13

2 - One can simplify an equation by dividing the two members by the same number not no one.

P8

With C, not no one. (Another property of the proportions).

Note :

P9

Example of calculation :

We have the following equation :

(7x + 3) (x - 2) = (4x - 1) (x - 2)

One notices the presence of (x - 2) in the two members. We can thus simplify them by x - 2 according to rule 2. But this expression (x - 2) is cancelled for x = 2. The two members are then multiplied by 0 and the equation is checked. (One obtains 0 = 0).

Number 2 is a first root. We can now simplify by (x - 2). We have :

7x + 3 = 4x - 1

from where : 7x - 4x = - 3 - 1

3x = - 4

x = - 4 (4 / 3) which is the 2nd root

We thus found with this equation the two roots :

x = 2 and x = - (4 / 3)

You see according to this example, that it is necessary to take care not to simplify without precaution, by an expression which can be cancelled: you will run the risk to remove a solution of the equation.

HIGH OF PAGE 5. - SIMPLE EQUATIONS A AN UNKNOWN FACTOR

Example of calculation :

That is to say to solve: (2x + 1) / x = (2x - 5) / (X - 2)

1 - One “drives out” the denominators :

(2x + 1) (x - 2) = (2x - 5) x

2 - One carries out :

2x2 + x - 4x - 2 = 2x2 - 5x

3 - One makes pass the unknown factors on the left and the numerical values on the right.

2x2 - 2x2 + x - 4x + 5x = 2

2x = 2

x = 1

One makes sure now that the found square root does not cancel a denominator :

B1

One can thus confirm that 1 is the root of the equation.

Note : The presence of 2x2 indicates a quadratic equation.

In this case, the term in x2 disappears by calculation, because there exists in the two members. We return to an equation of the 1st degree which we know to solve. So on the other hand after simplification, a term in x2 had remained present, us could not have solved the equation not having studied the quadratic equations.

5. 2. - EQUATION A LITERAL COEFFICIENTS

That is to say to solve : (ax - c) / b = (c - ax) / a

This equation has direction only if the terms a, b and c are different from zero.

1 - Let us drive out the denominators :

(ax - c) a = (c - ax) b

a2x - ac = bc - abx

2 - Let us pass the terms in (x) in a member, the terms without (X) in the other.

a2x + abx = ac + bc
3 - Let us put (x) in common factor
x (a2 + ab) = c (a + b)

Let us put now “a” in common factor in the first member :

ax (a + b) = c (a + b)

4 - To obtain x alone, we should divide the first term by a (a + b). But to preserve the equality, let us divide the two members by this value :

 B2

5 - Discussion :

If a + b is different from 0, we can simplify by a + b and we obtain :

x = c / a

If a + b is equal to 0, the equation is form :

ax . 0 = c . 0

It is unspecified. Indeed, the two members are always null whatever the value which one gives to x.

These calculations can appear complicated to you. It of it is nothing. They are only elementary applications of rules which we studied until now. On the other hand, it is very important to attentively observe an equation before developing calculations.

5. 3. - SOLUTIONS Of EQUATIONS

5. 3. 1. - BY THE PRODUCTS OF FACTORS

(3 - x) (x + 4) = 0

It is obvious that so that a product of factors is null, it is necessary and it is enough that one of the factors is null.

Example :

107 x 109 x 0 = 0

From where two possibilities and two solutions :

1)

3 - 4 = 0

i.e.

x = 3

2)

x + 4 = 0

i.e.

x = - 4

5. 3. 2. - BY COMMON FACTORIZATION

That is to say the equation :

2x2 - 3x = 21x

2x2 - 3x - 21x = 0

2x2 - 24x = 0

Let us put x in common factor

x (2x - 24) = 0

So that a product of factors is null, it is necessary and it is enough that one of the factors is null.

From where two solutions :

1)

x = 0

x = 0

2)

2x - 24 = 0

x = 12

5. 3. 3. - EQUATION CONTAINING the UNKNOWN IN FACTOR IN the TWO MEMBERS

(x + 3) (2x - 2) = (x + 4) (x + 3)

One sees the term (x + 3) in the two members. We encountered such a problem a few moments ago.

If the term (x + 3) is null, the equation is checked. It is thus the first solution.

x = - 3

If (x + 3) is not null, one can simplify the two members by (x + 3) and one finds :  

2x - 2 = x + 4

2x - x = 2 + 4

x = 6 which is the second solution

In short the two solutions are :

x = - 3

x = 6

We thus finish the elementary study of the algebra.

 

     

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