2º ESO EQUATIONS 1

Susana López

UNIT 6. EQUATIONS 1. INTRODUCTION Algebraic expressions are expressions involving numbers and letters, like polynomials. In this unit we will mostly deal with expressions with only one variable.

To deal with: tratar con, manejar. To involve: involucrar, conllevar. Definition. An equality is an algebraic expression with an equals sign in the middle. As an equality is an algebraic expression, an equality is also made up of terms. What is in both sides are called first and second members, respectively. Then, an equality is something like:

Equality: First member = Second member (Members are also called left-hand side of the equality and right-hand side of the equality) Example: 2435 −=+ xx The first member is 5x + 3. The first member has two terms: 5x and 3. The second member is 4x− 2. The second member has two terms: 4x, −2. The terms 3 and −2 are called constant terms. We can call 5x and 4x subject terms (subject to x), 3 and −2 are called non-subject terms.

Types of equalities 1. An equality which is always true is called an identity. Examples:

yyy ++=+ 112 is true for every value of y. !y( )

a! b = !b+ a is true for whatever a and b.

x2! y

2= x + y( ) " x ! y( ) is always true, as we have just studied in the previous unit.

2. An equation is an equality which is only true for one value of the variable involved. Examples: x ! 5= 0 is only true for x = 5.

512 =+y is only true for y = 2: 5122 =+⋅

Definitions. A solution or root of an equation is a number which makes the equation true. To solve an equation means to find its roots. That is why the variable of an equation is called unknown. Unknown: incognita, desconocido. Root: raíz.

Example.

93 =x Name the elements of this equation. x is the unknown. The first member is 3x, the second member is 9.

The equation is only true for 3=x . 3 is the solution or root.

2º ESO EQUATIONS 2

Susana López

2. EQUIVALENT EQUATIONS Definition. Equivalent equations are equations having the same solutions. Example. The equations x + 3=10, and x + 4 = 11 are equivalent. The solution is 7. Solving equations. We must work out the unknown, by finding an equivalent expresion such as

x = number. To work out: despejar.

Such as: como (lo que se muestra a continuación) Example. 7x + 8 = 57 is equivalent to x = 7. This very simple equation gives us the solution: 7. But, how??? The aim is to work out the unknown using this rule. Aim: objetivo.

Rule to get equivalent equations Whatever you do to one side of the equation, you must do to the other.

Whatever: lo que quiera que…, lo que sea. This means: - You can add or subtract a number on both sides of an equality. - You can multiply or divide both sides of an equality by the same number. Example.

2x + 4 =16 We must “unwrap” the x. To wrap: envolver What shall we do in order to get rid of 4? We subtract 4 from both terms: To get rid of: deshacerse de.

2x + 4 − 4 =16 − 4 We will write only:

2x = 16 − 4 This is why we say: +4 crosses the equals as −4. This is to transpose

2x = 12

What shall we do in order to get rid of 2? We divide by 2 both members: 2x / 2 = 12 / 2

We will write only: x = 12 / 2

This is why we say: 2 that multiplies crosses the “=” dividing. And finally we have:

x = 6 ←SOLUTION

2º ESO EQUATIONS 3

Susana López

3. STEPS TO SOLVE EASY LINEAR EQUATIONS

Step 1 Multiply out any brackets. Pay attention to the signs! Step 2 Rearrange the equation: all subject terms on one side of the “=” and all

non-subject terms on the other side.

Remember to reverse the +/− sign of any term that crosses the “=”. This is called to transpose.

Step 3 Collect like terms: combine them together on each side of the equation and reduce it to the form Ax = B, where A and B are just numbers.

Step 4 Work out the unknown: find its value. Exercise. Work out the following equations. Mind the brackets!!

a) 2 x – 7 = 3 x – 8 b) x – 7 = 2 ( x – 3) c) 3( 6 + x) = 2( x – 5) d) 9 ( x – 1) = 6 ( x + 3) e) 3 ( 3x +1 ) − ( x − 1 ) = 6 ( x + 10)

2º ESO EQUATIONS 4

Susana López

4. LINEAR EQUATIONS There is a general method for solving linear equations. We are going to study this method by solving an equation:

223

52

37

−=−

−−+ xxxx

Step 1 Put the same denominator in all the terms, by calculating the l.c.m of the denominators and getting the right numerators, also on the second member.

Be careful with the signs!- use brackets.

l.c.m.(3, 5, 2) = 30 ⇒ 30

)2(3030

)3(1526)7(10 −⋅=

−⋅−⋅−+⋅ xxxx

Step 2 Get rid of the denominators: get everything off the bottom.

( )230)3(1526)7(10 −⋅=−⋅−⋅−+⋅ xxxx

Step 3 Multiply out any brackets. Pay attention to the signs!

45127010 −−+ xx + 603015 −= xx

Step 4 Rearrange the equation: all subject terms on one side of the “=” and all non-subject terms on the other side. Remember to reverse the +/− sign of any term that crosses the “=”. Subject terms: terms with x.

45706030151210 +−−=−+− xxxx

Step 5 Collect like terms: combine them together on each side of the equation and

reduce it to the form Ax = B, where A and B are just numbers.

8517 −=− x

you could also write 8517 =x , multiplying by − 1. ←IMPORTANT

Step 6 Work out the unknown: find its value. (Sliding the coefficient of x underneath the constant term)

51785

=−

−=x

Step 7 Check the answer by substituting it back into the equation to see if it works.

YES 2531242

535

523

75

??? 252

535

523

75 22

35

23

7

−==+−=−

−⋅

−+

−=−

−⋅

−+

→−=−

−−+ xxxx

Exercise 1. Work out the following equations.

a) 146

3132

=−

−+ xx b)

824

4)2(2 xxx

=−

−− c)

52

5212

41 −

=−

−− xxx

3º ESO UNIT 4 5

Susana López

5. QUADRATIC EQUATIONS Definition. A quadratic equation is an equation of the form:

02 =++ cbxax , where 0≠a .

a, b and c are called the coefficients of the quadratic equation. a is the coefficient of x2. b is the coefficient of x. c is the constant term. In the definition of quadratic equation 0≠a means that x2 appears. Example. 16x2 ! 24x + 9 = 0 has as coefficients: 16=a , 24−=b and 9=c . Example. 042 2 =+− xx , has coefficients a = 2, b = −1 and c = 4. The standard form of a quadratic equation is 02 =++ cbxax .

It means that a polynomial of degree two appears equated to zero P(x) = 0.

If not, we have to rearrange so that it becomes like that. That is called “to standardize”.

Before we start solving, we always have to put the equation in the form P(x) = 0,

that is,

there must be a zero on the right-hand side.

Examples.

The equation 32 −= xx is not in the standard form. We have to rearrange it:

032 =+− xx . Now it is in the standard form, we have standardized the equation.

Before solving the equation ( ) 152 =−⋅ xx , we have to multiply out the brackets and to rearrange the quadratic to get the standard form:

⇒=− 1522 xx 01522 =−− xx .

Now we have to learn how to solve this equation.

3º ESO UNIT 4 6

Susana López

5.1. CLASSIFICATION and METHODS. A. Incomplete quadratic equations.

b or/and c are equal to zero.

We will study these equations through examples:

Type 1. 5 x2 = 0

There is only one term, the term with x2 5 x2 = 0 5 times x2 give 0, then:

Either 5 is 0 (imposible) or x2 is zero ⇒ x2 = 0, hence x = 0 THERE IS ONE SOLUTION x = 0.

Exercise 1. Work out the following equations: a)

!

3x2

= 0 b)

!

"4x2

= 0 c) x(x +1)+3x = 4x d)

!

125x2

= 0 e) 4x2 + 5= 5

Type 2. 2x2 − 18 = 0,

There are two terms: x2 and constant term, but no x in the equation

Step 1. x2 must be worked out:

2 x2 = 18 ⇒ x2 = 182= 9 ,

Step 2. Worked out x by taking the square root of the second member. A “plus or minus sign” ±( ) must be put in front of the root. Hence

x = ± 9 =x1= 9 = 3

x2= ! 9 = !3

"#$

%$

There are two solutions because both 3 and − 3 squared are 9.

Example: 3x2 + 75 = 0⇒ 3x2 = −75 ⇒ x2 = ! 753

= −25

This equation has no solution because x2 can never be negative.

THERE CAN BE TWO DIFFERENT SOLUTIONS OR NO SOLUTION

Exercise 2. Work out the following equations: a)

!

3x2" 27 = 0 b)

!

3x2

+ 27 = 0 c)

!

2x2" 32 = 0 d)

!

4x2

= 36 e)

!

4x2

+ 36 = 0 f)

!

"x2

+ 4 = 0 g) x2 = !4

3º ESO UNIT 4 7

Susana López

Type 3. x2 − 5x = 0 ,

There are two terms: with x and x2, but no constant term. Step 1. We must factorize by taking out the common factor x from both terms.

⇒ x (x − 5) = 0 Step 2. Then we have two possibilities:

⇒

=−

=

05or 0either

xx

Step 3. Work out this last linear equation, we have the second solution.

hence:

=

=

50

xx

THESE EQUATIONS ALWAYS HAVE TWO DIFFERENT SOLUTIONS. Exercise. Work out the following equations: a)

!

x2" 2x = 0 b)

!

x2

+ 2x = 0 c)

!

2x2" x = 0

d)

!

x2" 3x = 0 e)

!

3x2

+ 2x = 0 f)

!

6x2" 5x = 0

g)

!

5x(x+ 2) = 3x(x "1) h)

!

4x " (6 # 2x) = #2x2

3º ESO UNIT 4 8

Susana López

B. Complete quadratic equations

02 =++ cbxax , 0,0 ≠≠ cb .

It is solved by using the “quadratic formula”:

aacbbx

242 −±−

=

Example: Consider the equation 352 2 =+ xx

First, we standardize the equation: 0352 2 =−+ xx

So 2=a , 5=b , 3−=c .

Now we apply the formula: a

acbbx2

42 −±−=

( )475

4495

424255

2232455 2 ±−

=±−

=+±−

=⋅

−⋅⋅−±−=x

We have two possibilities, one with plus sign and another one with minus sign:

−=−

=−−

=

==+−

==

3412

475

21

42

475

2

1

x

xx

Hence there are two solutions for x:

21

1 =x and 32 −=x .

Check the solutions: Remark: every equation can be solved as a complete equation, given that the missing coefficient can be considered to be zero. Examples: 0904094 22 =−+≈=− xxx , 00202 22 =+−≈=− xxxx Exercise. Solve the following quadratic equations:

a)

!

x2" 2x " 3= 0

b) x2 + 2x – 8 = 0 c) 3x2 + 9x + 6 = 0 d)

!

x2" 5x+ 6 = 0

e)

!

x2" 2x+1= 0

f) 6x2– 7x + 2 = 0

g) x2+ x +1= 0

h) x2! 6x + 9 = 0

i) x2! 5x = 6

j) x !3( ) " x + 6( ) = 0

3º ESO UNIT 4 9

Susana López

Equations Worksheet

1. I think of a number, double it and add 4. The answer is 16. What is my number? 2. I think of a number, multiply it by 5 and then add 2. The answer is 17. What is my number? 3. I think of a number, double it and add 3, then multiply the result by 4. The answer is 52. What is my number? 4. A triangle has sides of length x cm, x + 1 cm and 2x − 3 cm. The triangle has a perimeter of 18 cm. Find the length of its sides. 5. Organizing a party costs £20 plus £5 per person. a) Nick has a birthday party for 12 people. How much does it cost? b) Jean pays £100 for her birthday party. How many people went to the party? 6. The three angles of a triangle are: 2x, 7x and 3x. By forming an equation find the value of the angles of the triangle. 7. The width of a rectangle is 4 cm less than the length. The perimeter of the rectangle is 20 cm, find the dimensions of the rectangle. 8. A bottle of drink costs x pence. A cake costs 7 pence more than a drink. The total cost of two drinks and a cake is 97 pence. Find the cost of a cake. 9. Fifty-one more than 9 times a number is 114. What is the number? 10. 508 exceeds six times a number by 70. What is the number?

3º ESO UNIT 4 10

Susana López

11. Sixty-three more than four-fifths of a number equals 111. What is the number? Sol. 60. 12. What are two consecutive integers, such that seven times the larger minus three times the smaller is 95? 13. The sum of two consecutive integers is 65. What is the second number? 14. A rectangle has length x + 2 cm and width x + 1 cm. The rectangle has area of 6 cm2. Find the value of x. These two rectangles have the same area. Find the length of

BC.

16. A natural number equals its square minus 12. Find the number. 17. Find two successive natural numbers such that the sum of their squares is 613. (consecutive numbers are one unit apart) 18. The product of two consecutive natural numbers is 1122. What are the numbers?

A B

C D

x cm

x + 3 cm

x + 1 cm

(not to scale)

2 cm