unit6: algebraic expressions and equations

Indice Monomials Adding and subtracting Identities and Equations Solving Exercises

Algebraic expressions and equations

Matematicas 1o E.S.O.Alberto Pardo Milanes

1 Monomials

2 Adding and subtracting monomials

3 Identities and Equations

4 Solving

5 Exercises

Alberto Pardo Milanes Algebraic expressions and equations

Monomials

What´s a monomial?

A variable is a symbol.

An algebraic expression in variables x, y, z, a, r, t . . . k is anexpression constructed with the variables and numbers usingaddition, multiplication, and powers.

A number multiplied with a variable in an algebraic expression isnamed coefficient.

A product of positive integer powers of a fixed set of variablesmultiplied by some coefficient is called a monomial.

Examples: 3x,2

3xy2, x2y3z.

Monomials

Like monomials and unlike monomials

In a monomial with only one variable, the power is called its order,or sometimes its degree.Example: Deg(5x4)=4.

In a monomial with several variables, the order/degree is the sumof the powers.Example: Deg(x2z4)=6.

Monomials are called similar or like ones, if they are identical ordiffered only by coefficients.

Example: 2x3y2 and2

5x3y2 are like monomials. 4xy2 and 4y2x4

are unlike monomials.

Adding and subtractingmonomials

Adding and subtracting monomials

Adding and Subtracting

You can ONLY add or subtract like monomials.

To add or subtract like monomials use the same rules as withintegers.

Example: 3x+ 4x = (3 + 4)x = 7x.

Example: 20a− 24a = (20− 24)a = −4a.

Example: 7x+ 5y ⇐= you can´t add unlike monomials.

Identities and Equations

What´s an equation? Identities vs equations.

An equation is a mathematical expression stating that a pair ofalgebraic expression are the same.

If the equation is true for every value of the variables then it´scalled Identity.

An identity is a mathematical relationship equating one quantity toanother which may initially appear to be different.

Example: x2 − x3 + x+ 1 = 3x4 is an equation,3x2 − x+ 1 = x2 − x+ 2 + 2x2 − 1 is an identity.

Parts of an equation.

In an equation:the variables are named unknowns (or indeterminate quantities),the number multiplied with a variable is named coefficient, aterm is a summand of the equation, the highest power of theunknowns is called the order/degree of the equation.

Example: In the equation 2x3 + 4y + 1 = 4:the unknowns are x and y,the coefficient of x3 is 2andthe coefficient of y is 4,the order of the equation is 3.

Solving

Solution of an equation.

You are solving a equation when you replace a variable with avalue and the mathematical expressions are still the same. Thevalue for the variables is the solution of the equation.

Example: In the equation 2x = 10 the solution is 5, because2 · 5 = 10.

Example: Sam is 9 years old. This is seven years younger than hersister Rose’s age. We can solve an equation to find Rose’s age:x− 7 = 9, the solution of the equation is 16, so Rose is 16 yearsold.

Solving

The balance method.

To solve equations you can use the balance method, you mustcarry out the same operations in both sides and in the same order.You must use these properties:

• Addition Property of Equalities: If you add the same number toeach side of an equation, the two sides remain equal (note you canalso add negative numbers).Example: x+ 3 = 5 =⇒ x+ 3−3 = 5−3 =⇒ x = 2

• Multiplication Property of Equalities: If you multiply by the samenumber each side of an equation, the two sides remain equal (noteyou can also multiply by fractions).

Example:x

5= 6 =⇒ 5 · x

5= 5 · 6 =⇒ x = 30

Solving

The balance method.

Example:x

5= 6 =⇒ 5 · x

5= 5 · 6 =⇒ x = 30

Solving

The balance method.

Example:x

5= 6 =⇒ 5 · x

5= 5 · 6 =⇒ x = 30

Solving

The balance method.

Example:x

5= 6 =⇒ 5 · x

5= 5 · 6 =⇒ x = 30

Solving

The balance method.

• Brackets: Sometimes you will need to solve equations involvingbrackets. If brackets appear, first remove the brackets by expandingeach bracketed expression.Example: 2(x− 3) = 2 =⇒ 2x− 6 = 2 =⇒ 2x− 6+6 = 2+6 =⇒

2x = 8 =⇒ 2x

2=⇒ x = 4

Use all three properties to solve equations:Example: Solve 4x+ 3 · (x− 25) = 240:First we remove brackets: 3 · (x− 25) = 3x− 75 so4x+3x− 75 = 240.Them we use addition property:4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.

Now we can use multiplication property:7x

7So the solution is x = 45.

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Solving

The balance method.

2x = 8 =⇒ 2x

2=⇒ x = 4

Exercises

Exercise 1

Solve the equations:

a 4x+ 2 = 26

b 5(2x− 1) = 7(9− x)

d 19 + 4x = 9− x

e 3(2x+ 1) = x− 2

5− 3x

Exercises

Exercise 1

a 4x+ 2 = 264x = 26 − 24x = 24x = 24 : 4x = 6

b 5(2x− 1) = 7(9− x)10x − 5 = 63 − 7x10x + 7x = 63 + 517x = 68x = 68 : 17x = 4

63x + 4x = 427x = 42x = 42 : 7x = 6

d 19 + 4x = 9− x4x + x = 9 − 195x = −10x = −10 : 5x = −2

Exercises

Exercise 1

a 4x+ 2 = 264x = 26 − 24x = 24x = 24 : 4x = 6

b 5(2x− 1) = 7(9− x)10x − 5 = 63 − 7x10x + 7x = 63 + 517x = 68x = 68 : 17x = 4

63x + 4x = 427x = 42x = 42 : 7x = 6

d 19 + 4x = 9− x4x + x = 9 − 195x = −10x = −10 : 5x = −2

Exercises

Exercise 1

a 4x+ 2 = 264x = 26 − 24x = 24x = 24 : 4x = 6

b 5(2x− 1) = 7(9− x)10x − 5 = 63 − 7x10x + 7x = 63 + 517x = 68x = 68 : 17x = 4

63x + 4x = 427x = 42x = 42 : 7x = 6

d 19 + 4x = 9− x4x + x = 9 − 195x = −10x = −10 : 5x = −2

Exercises

Exercise 1

a 4x+ 2 = 264x = 26 − 24x = 24x = 24 : 4x = 6

b 5(2x− 1) = 7(9− x)10x − 5 = 63 − 7x10x + 7x = 63 + 517x = 68x = 68 : 17x = 4

63x + 4x = 427x = 42x = 42 : 7x = 6

d 19 + 4x = 9− x4x + x = 9 − 195x = −10x = −10 : 5x = −2

Exercises

Exercise 1

e 3(2x+ 1) = x− 26x + 3 = x − 26x − x = −2 − 35x = −5x = −5 : 5x = −1

5− 3x

102x − 3x = 2−1x = 2x = 2 : (−1)x = −2

Exercises

Exercise 1

e 3(2x+ 1) = x− 26x + 3 = x − 26x − x = −2 − 35x = −5x = −5 : 5x = −1

5− 3x

102x − 3x = 2−1x = 2x = 2 : (−1)x = −2

Exercises

Exercise 2

Find a number such that 2 less than three times the number is 10.

Exercises

Exercise 2

Data: Let x be the number.Three times the number is 3x.2 less than three times the number is (3x − 2) and this is10.

3x − 2 = 103x = 12x = 12 : 3

x = 4 Answer: The number isx = 4.

Exercises

Exercise 2

3x − 2 = 103x = 12x = 12 : 3

Exercises

Exercise 2

3x − 2 = 103x = 12x = 12 : 3

Exercises

Exercise 3

Mr. Roberts and his wife have 370 pounds. Mrs. Roberts has 155pounds less than twice her husband’s money. How many poundsdoes Mr. Roberts have? How many pounds does Mrs. Robertshave?

Exercises

Exercise 3

Data: They have 370 pounds. Mr. Roberts has x pounds.Twice Mr. Roberts’ money is 2x pounds.Mrs. Roberts has (2x − 155) pounds.

x+(2x−155) = 370x + 2x − 155 = 370

3x = 525x = 525 : 3x = 175

370 − 175 = 195

Answer: Mr. Roberts has 175pounds. Mrs. Roberts has 195pounds.

Exercises

Exercise 3

x+(2x−155) = 370x + 2x − 155 = 370

3x = 525x = 525 : 3x = 175

370 − 175 = 195

Exercises

Exercise 3

x+(2x−155) = 370x + 2x − 155 = 370

3x = 525x = 525 : 3x = 175

370 − 175 = 195

Exercises

Exercise 4

The length of a room exceeds the width by 5 feet. The length ofthe four walls is 30 feet. Find the dimensions of the room.

Exercises

Exercise 4

Data: The width of the room is x feet.The length of the room is (x + 5) feet.The length of the four walls is x+(x+5)+x+(x+5) feetand this is 30 feet.

x+(x+5)+x+(x+5) = 30x + x + 5 + x + x + 5 = 30x + x + x + x = 30 − 5 − 5

4x = 20x = 20 : 4

x = 55 + 5 = 10 Answer: The room is 5 feet

long and 10 feet wide.

Exercises

Exercise 4

x+(x+5)+x+(x+5) = 30x + x + 5 + x + x + 5 = 30x + x + x + x = 30 − 5 − 5

4x = 20x = 20 : 4

Exercises

Exercise 4

x+(x+5)+x+(x+5) = 30x + x + 5 + x + x + 5 = 30x + x + x + x = 30 − 5 − 5

4x = 20x = 20 : 4

Exercises

Exercise 5

Maria spent a third of her money on food. Then, she spent e21 ona present. At the end, she had the fifth of her money. How muchmoney did she have at the beginning?

Exercises

Exercise 5

Data: At the beginig She had x euros. She spentx

3on food

and e21 on a present. At the end She hadx

5euros.

x −x

3− 21 =

1515x − 5x − 315 = 3x15x − 5x − 3x = 315

7x = 315x = 315 : 7

x = 45

Answer: At the beginigShe had e45.

Answer:

Exercises

Exercise 5

3on food

5euros.

x −x

3− 21 =

1515x − 5x − 315 = 3x15x − 5x − 3x = 315

7x = 315x = 315 : 7

x = 45

Answer:

Exercises

Exercise 5

3on food

5euros.

x −x

3− 21 =

1515x − 5x − 315 = 3x15x − 5x − 3x = 315

7x = 315x = 315 : 7

x = 45

Answer:

Exercises

Exercise 6

John bought a book, a pencil and a notebook. The book cost thedouble of the notebook, and the pencil cost the fifth of the bookand the notebook together. If he paid e18, what is the price ofeach article?

Exercises

Exercise 6

John bought a book, a pencil and a notebook. The book cost thedouble of the notebook, and the pencil cost the fifth of the bookand the notebook together. If he paid e18, what is the price ofeach article?

Data: The notebook cost x euros. The book cost 2x euros.The notebook and the book together cost (x + 2x) euros.

The pencil costx + 2x

5. He paid e18.

x + 2x +x + 2x

x + 2x

55x + 10x + (x + 2x) = 90

5x+10x+x+2x = 9018x = 90x = 90 : 18

x = 52x = 10

x + 2x

Answer: The notebook cost e5, the book e10 and thepencil e3 .

Exercises

Exercise 6

5. He paid e18.

x + 2x +x + 2x

x + 2x

55x + 10x + (x + 2x) = 90

5x+10x+x+2x = 9018x = 90x = 90 : 18

x = 52x = 10

x + 2x

Exercises

Exercise 6

5. He paid e18.

x + 2x +x + 2x

x + 2x

55x + 10x + (x + 2x) = 90

5x+10x+x+2x = 9018x = 90x = 90 : 18

x = 52x = 10

x + 2x

unit6: algebraic expressions and equations

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