Algebraic expressions and equations, variable, identity, monomial, solving equations, the balance method,...

1. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Algebraic expressions and equations
Matem´ticas 1o E.S.O.
a
Alberto Pardo Milan´s
e
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2. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
1 Monomials
2 Adding and subtracting monomials
3 Identities and Equations
4 Solving
5 Exercises
Alberto Pardo Milan´s
e Algebraic expressions and equations

3. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Monomials
Alberto Pardo Milan´s
e Algebraic expressions and equations

4. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Monomials
What´s a monomial?
A variable is a symbol.
An algebraic expression in variables x, y, z, a, r, t . . . k is an
expression constructed with the variables and numbers using
addition, multiplication, and powers.
A number multiplied with a variable in an algebraic expression is
named coefficient.
A product of positive integer powers of a fixed set of variables
multiplied by some coefficient is called a monomial.
2 2 2 3
Examples: 3x, xy , x y z.
3
Alberto Pardo Milan´s
e Algebraic expressions and equations

5. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
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 sum
of the powers.
Example: Deg(x2 z 4 )=6.
Monomials are called similar or like ones, if they are identical or
differed only by coefficients.
2
Example: 2x3 y 2 and x3 y 2 are like monomials. 4xy 2 and 4y 2 x4
5
are unlike monomials.
Alberto Pardo Milan´s
e Algebraic expressions and equations

6. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Adding and subtracting
monomials
Alberto Pardo Milan´s
e Algebraic expressions and equations

7. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
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 with
integers.
Example: 3x + 4x = (3 + 4)x = 7x.
Example: 20a − 24a = (20 − 24)a = −4a.
Example: 7x + 5y ⇐= you can´t add unlike monomials.
Alberto Pardo Milan´s
e Algebraic expressions and equations

8. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Alberto Pardo Milan´s
e Algebraic expressions and equations

9. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
What´s an equation? Identities vs equations.
An equation is a mathematical expression stating that a pair of
algebraic expression are the same.
If the equation is true for every value of the variables then it´s
called Identity.
An identity is a mathematical relationship equating one quantity to
another 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.
Alberto Pardo Milan´s
e Algebraic expressions and equations

10. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Parts of an equation.
In an equation:
the variables are named unknowns (or indeterminate quantities),
the number multiplied with a variable is named coefficient, a
term is a summand of the equation, the highest power of the
unknowns 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 2
and
the coefficient of y is 4,
the order of the equation is 3.
Alberto Pardo Milan´s
e Algebraic expressions and equations

11. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Parts of an equation.
In an equation:
the variables are named unknowns (or indeterminate quantities),
the number multiplied with a variable is named coefficient, a
term is a summand of the equation, the highest power of the
unknowns 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 2
and
the coefficient of y is 4,
the order of the equation is 3.
Alberto Pardo Milan´s
e Algebraic expressions and equations

12. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Parts of an equation.
In an equation:
the variables are named unknowns (or indeterminate quantities),
the number multiplied with a variable is named coefficient, a
term is a summand of the equation, the highest power of the
unknowns 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 2
and
the coefficient of y is 4,
the order of the equation is 3.
Alberto Pardo Milan´s
e Algebraic expressions and equations

13. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Parts of an equation.
In an equation:
the variables are named unknowns (or indeterminate quantities),
the number multiplied with a variable is named coefficient, a
term is a summand of the equation, the highest power of the
unknowns 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 2
and
the coefficient of y is 4,
the order of the equation is 3.
Alberto Pardo Milan´s
e Algebraic expressions and equations

14. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Identities and Equations
Parts of an equation.
In an equation:
the variables are named unknowns (or indeterminate quantities),
the number multiplied with a variable is named coefficient, a
term is a summand of the equation, the highest power of the
unknowns 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 2
and
the coefficient of y is 4,
the order of the equation is 3.
Alberto Pardo Milan´s
e Algebraic expressions and equations

15. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
Alberto Pardo Milan´s
e Algebraic expressions and equations

16. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
Solution of an equation.
You are solving a equation when you replace a variable with a
value and the mathematical expressions are still the same. The
value for the variables is the solution of the equation.
Example: In the equation 2x = 10 the solution is 5, because
2 · 5 = 10.
Example: Sam is 9 years old. This is seven years younger than her
sister 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 years
old.
Alberto Pardo Milan´s
e Algebraic expressions and equations

17. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
Solution of an equation.
You are solving a equation when you replace a variable with a
value and the mathematical expressions are still the same. The
value for the variables is the solution of the equation.
Example: In the equation 2x = 10 the solution is 5, because
2 · 5 = 10.
Example: Sam is 9 years old. This is seven years younger than her
sister 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 years
old.
Alberto Pardo Milan´s
e Algebraic expressions and equations

18. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
Solution of an equation.
You are solving a equation when you replace a variable with a
value and the mathematical expressions are still the same. The
value for the variables is the solution of the equation.
Example: In the equation 2x = 10 the solution is 5, because
2 · 5 = 10.
Example: Sam is 9 years old. This is seven years younger than her
sister 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 years
old.
Alberto Pardo Milan´s
e Algebraic expressions and equations

19. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
To solve equations you can use the balance method, you must
carry 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 to
each side of an equation, the two sides remain equal (note you can
also add negative numbers).
Example: x + 3 = 5 =⇒ x + 3−3 = 5−3 =⇒ x = 2
• Multiplication Property of Equalities: If you multiply by the same
number each side of an equation, the two sides remain equal (note
you can also multiply by fractions).
x x
Example: = 6 =⇒ 5 · = 5 · 6 =⇒ x = 30
5 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

20. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
To solve equations you can use the balance method, you must
carry 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 to
each side of an equation, the two sides remain equal (note you can
also add negative numbers).
Example: x + 3 = 5 =⇒ x + 3−3 = 5−3 =⇒ x = 2
• Multiplication Property of Equalities: If you multiply by the same
number each side of an equation, the two sides remain equal (note
you can also multiply by fractions).
x x
Example: = 6 =⇒ 5 · = 5 · 6 =⇒ x = 30
5 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

21. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
To solve equations you can use the balance method, you must
carry 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 to
each side of an equation, the two sides remain equal (note you can
also add negative numbers).
Example: x + 3 = 5 =⇒ x + 3−3 = 5−3 =⇒ x = 2
• Multiplication Property of Equalities: If you multiply by the same
number each side of an equation, the two sides remain equal (note
you can also multiply by fractions).
x x
Example: = 6 =⇒ 5 · = 5 · 6 =⇒ x = 30
5 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

22. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
To solve equations you can use the balance method, you must
carry 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 to
each side of an equation, the two sides remain equal (note you can
also add negative numbers).
Example: x + 3 = 5 =⇒ x + 3−3 = 5−3 =⇒ x = 2
• Multiplication Property of Equalities: If you multiply by the same
number each side of an equation, the two sides remain equal (note
you can also multiply by fractions).
x x
Example: = 6 =⇒ 5 · = 5 · 6 =⇒ x = 30
5 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

23. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

24. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

25. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

26. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

27. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

28. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

29. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Solving
The balance method.
• Brackets: Sometimes you will need to solve equations involving
brackets. If brackets appear, first remove the brackets by expanding
each bracketed expression.
Example: 2(x − 3) = 2 =⇒ 2x − 6 = 2 =⇒ 2x − 6+6 = 2+6 =⇒
2x 8
2x = 8 =⇒ = =⇒ x = 4
2 2
Use all three properties to solve equations:
Example: Solve 4x + 3 · (x − 25) = 240:
First we remove brackets: 3 · (x − 25) = 3x − 75 so
4x+3x − 75 = 240.
Them we use addition property:
4x+3x−75+75 = 240+75 =⇒ 4x+3x = 240+75 =⇒ 7x = 315.
7x 315
Now we can use multiplication property: =
7 7
So the solution is x = 45.
Alberto Pardo Milan´s
e Algebraic expressions and equations

30. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Alberto Pardo Milan´s
e Algebraic expressions and equations

31. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
a 4x + 2 = 26
b 5(2x − 1) = 7(9 − x)
x 2x
c + =7
2 3
d 19 + 4x = 9 − x
e 3(2x + 1) = x − 2
x 3x 1
f − =
5 10 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

32. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
x 2x
a 4x + 2 = 26 c + =7
2 3
4x = 26 − 2 3x 4x 42
+ =
4x = 24 6 6 6
x = 24 : 4 3x + 4x = 42
x=6 7x = 42
x = 42 : 7
b 5(2x − 1) = 7(9 − x)
x=6
10x − 5 = 63 − 7x
10x + 7x = 63 + 5 d 19 + 4x = 9 − x
17x = 68 4x + x = 9 − 19
x = 68 : 17 5x = −10
x=4 x = −10 : 5
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

33. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
x 2x
a 4x + 2 = 26 c + =7
2 3
4x = 26 − 2 3x 4x 42
+ =
4x = 24 6 6 6
x = 24 : 4 3x + 4x = 42
x=6 7x = 42
x = 42 : 7
b 5(2x − 1) = 7(9 − x)
x=6
10x − 5 = 63 − 7x
10x + 7x = 63 + 5 d 19 + 4x = 9 − x
17x = 68 4x + x = 9 − 19
x = 68 : 17 5x = −10
x=4 x = −10 : 5
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

34. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
x 2x
a 4x + 2 = 26 c + =7
2 3
4x = 26 − 2 3x 4x 42
+ =
4x = 24 6 6 6
x = 24 : 4 3x + 4x = 42
x=6 7x = 42
x = 42 : 7
b 5(2x − 1) = 7(9 − x)
x=6
10x − 5 = 63 − 7x
10x + 7x = 63 + 5 d 19 + 4x = 9 − x
17x = 68 4x + x = 9 − 19
x = 68 : 17 5x = −10
x=4 x = −10 : 5
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

35. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
x 2x
a 4x + 2 = 26 c + =7
2 3
4x = 26 − 2 3x 4x 42
+ =
4x = 24 6 6 6
x = 24 : 4 3x + 4x = 42
x=6 7x = 42
x = 42 : 7
b 5(2x − 1) = 7(9 − x)
x=6
10x − 5 = 63 − 7x
10x + 7x = 63 + 5 d 19 + 4x = 9 − x
17x = 68 4x + x = 9 − 19
x = 68 : 17 5x = −10
x=4 x = −10 : 5
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

36. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
e 3(2x + 1) = x − 2
6x + 3 = x − 2
6x − x = −2 − 3
5x = −5
x = −5 : 5
x = −1
x 3x 1
f − =
5 10 5
2x 3x 2
− =
10 10 10
2x − 3x = 2
−1x = 2
x = 2 : (−1)
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

37. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 1
Solve the equations:
e 3(2x + 1) = x − 2
6x + 3 = x − 2
6x − x = −2 − 3
5x = −5
x = −5 : 5
x = −1
x 3x 1
f − =
5 10 5
2x 3x 2
− =
10 10 10
2x − 3x = 2
−1x = 2
x = 2 : (−1)
x = −2
Alberto Pardo Milan´s
e Algebraic expressions and equations

38. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 2
Find a number such that 2 less than three times the number is 10.
Alberto Pardo Milan´s
e Algebraic expressions and equations

39. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 2
Find a number such that 2 less than three times the number is 10.
Data: Let x be the number.
Three times the number is 3x.
2 less than three times the number is (3x − 2) and this is
10.
3x − 2 = 10
3x = 12
x = 12 : 3
x=4 Answer: The number is
x = 4.
Alberto Pardo Milan´s
e Algebraic expressions and equations

40. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 2
Find a number such that 2 less than three times the number is 10.
Data: Let x be the number.
Three times the number is 3x.
2 less than three times the number is (3x − 2) and this is
10.
3x − 2 = 10
3x = 12
x = 12 : 3
x=4 Answer: The number is
x = 4.
Alberto Pardo Milan´s
e Algebraic expressions and equations

41. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 2
Find a number such that 2 less than three times the number is 10.
Data: Let x be the number.
Three times the number is 3x.
2 less than three times the number is (3x − 2) and this is
10.
3x − 2 = 10
3x = 12
x = 12 : 3
x=4 Answer: The number is
x = 4.
Alberto Pardo Milan´s
e Algebraic expressions and equations

42. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 3
Mr. Roberts and his wife have 370 pounds. Mrs. Roberts has 155
pounds less than twice her husband’s money. How many pounds
does Mr. Roberts have? How many pounds does Mrs. Roberts
have?
Alberto Pardo Milan´s
e Algebraic expressions and equations

43. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 3
Mr. Roberts and his wife have 370 pounds. Mrs. Roberts has 155
pounds less than twice her husband’s money. How many pounds
does Mr. Roberts have? How many pounds does Mrs. Roberts
have?
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) = 370
x + 2x − 155 = 370
3x = 525
x = 525 : 3
x = 175 Answer: Mr. Roberts has 175
370 − 175 = 195 pounds. Mrs. Roberts has 195
pounds.
Alberto Pardo Milan´s
e Algebraic expressions and equations

44. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 3
Mr. Roberts and his wife have 370 pounds. Mrs. Roberts has 155
pounds less than twice her husband’s money. How many pounds
does Mr. Roberts have? How many pounds does Mrs. Roberts
have?
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) = 370
x + 2x − 155 = 370
3x = 525
x = 525 : 3
x = 175 Answer: Mr. Roberts has 175
370 − 175 = 195 pounds. Mrs. Roberts has 195
pounds.
Alberto Pardo Milan´s
e Algebraic expressions and equations

45. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 3
Mr. Roberts and his wife have 370 pounds. Mrs. Roberts has 155
pounds less than twice her husband’s money. How many pounds
does Mr. Roberts have? How many pounds does Mrs. Roberts
have?
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) = 370
x + 2x − 155 = 370
3x = 525
x = 525 : 3
x = 175 Answer: Mr. Roberts has 175
370 − 175 = 195 pounds. Mrs. Roberts has 195
pounds.
Alberto Pardo Milan´s
e Algebraic expressions and equations

46. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 4
The length of a room exceeds the width by 5 feet. The length of
the four walls is 30 feet. Find the dimensions of the room.
Alberto Pardo Milan´s
e Algebraic expressions and equations

47. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 4
The length of a room exceeds the width by 5 feet. The length of
the four walls is 30 feet. Find the dimensions of the room.
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) feet
and this is 30 feet.
x + (x + 5) + x + (x + 5) = 30
x + x + 5 + x + x + 5 = 30
x + x + x + x = 30 − 5 − 5
4x = 20
x = 20 : 4
x=5
5 + 5 = 10 Answer: The room is 5 feet
long and 10 feet wide.
Alberto Pardo Milan´s
e Algebraic expressions and equations

48. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 4
The length of a room exceeds the width by 5 feet. The length of
the four walls is 30 feet. Find the dimensions of the room.
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) feet
and this is 30 feet.
x + (x + 5) + x + (x + 5) = 30
x + x + 5 + x + x + 5 = 30
x + x + x + x = 30 − 5 − 5
4x = 20
x = 20 : 4
x=5
5 + 5 = 10 Answer: The room is 5 feet
long and 10 feet wide.
Alberto Pardo Milan´s
e Algebraic expressions and equations

49. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 4
The length of a room exceeds the width by 5 feet. The length of
the four walls is 30 feet. Find the dimensions of the room.
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) feet
and this is 30 feet.
x + (x + 5) + x + (x + 5) = 30
x + x + 5 + x + x + 5 = 30
x + x + x + x = 30 − 5 − 5
4x = 20
x = 20 : 4
x=5
5 + 5 = 10 Answer: The room is 5 feet
long and 10 feet wide.
Alberto Pardo Milan´s
e Algebraic expressions and equations

50. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 5
Maria spent a third of her money on food. Then, she spent e21 on
a present. At the end, she had the fifth of her money. How much
money did she have at the beginning?
Alberto Pardo Milan´s
e Algebraic expressions and equations

51. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 5
Maria spent a third of her money on food. Then, she spent e21 on
a present. At the end, she had the fifth of her money. How much
money did she have at the beginning?
x
Data: At the beginig She had x euros. She spent on food
3
x
and e21 on a present. At the end She had euros.
5
x x
x − − 21 = 7x = 315
3 5
x x 21 x x = 315 : 7
− − = x = 45
1 3 1 5
15x 5x 315 3x
− − =
15 15 15 15 Answer: At the beginig
15x − 5x − 315 = 3x She had e45.
15x − 5x − 3x = 315
Alberto Pardo Milan´s
e Algebraic expressions and equations

52. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 5
Maria spent a third of her money on food. Then, she spent e21 on
a present. At the end, she had the fifth of her money. How much
money did she have at the beginning?
x
Data: At the beginig She had x euros. She spent on food
3
x
and e21 on a present. At the end She had euros.
5
x x
x − − 21 = 7x = 315
3 5
x x 21 x x = 315 : 7
− − = x = 45
1 3 1 5
15x 5x 315 3x
− − =
15 15 15 15 Answer: At the beginig
15x − 5x − 315 = 3x She had e45.
15x − 5x − 3x = 315
Alberto Pardo Milan´s
e Algebraic expressions and equations

53. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 5
Maria spent a third of her money on food. Then, she spent e21 on
a present. At the end, she had the fifth of her money. How much
money did she have at the beginning?
x
Data: At the beginig She had x euros. She spent on food
3
x
and e21 on a present. At the end She had euros.
5
x x
x − − 21 = 7x = 315
3 5
x x 21 x x = 315 : 7
− − = x = 45
1 3 1 5
15x 5x 315 3x
− − =
15 15 15 15 Answer: At the beginig
15x − 5x − 315 = 3x She had e45.
15x − 5x − 3x = 315
Alberto Pardo Milan´s
e Algebraic expressions and equations

54. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 6
John bought a book, a pencil and a notebook. The book cost the
double of the notebook, and the pencil cost the fifth of the book
and the notebook together. If he paid e18, what is the price of
each article?
Alberto Pardo Milan´s
e Algebraic expressions and equations

55. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 6
John bought a book, a pencil and a notebook. The book cost the
double of the notebook, and the pencil cost the fifth of the book
and the notebook together. If he paid e18, what is the price of
each article?
Data: The notebook cost x euros. The book cost 2x euros.
The notebook and the book together cost (x + 2x) euros.
x + 2x
The pencil cost . He paid e18.
5
x + 2x 5x+10x+x+2x = 90
x + 2x + = 18
5 18x = 90
x 2x x + 2x 18 x = 90 : 18
+ + =
1 1 5 1 x=5
5x 10x x + 2x 90 2x = 10
+ + = x + 2x 15
5 5 5 5 = = 3.
5x + 10x + (x + 2x) = 90 5 5
Alberto Pardo Milan´s
e Algebraic expressions and equations

56. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 6
Data: The notebook cost x euros. The book cost 2x euros.
The notebook and the book together cost (x + 2x) euros.
x + 2x
The pencil cost . He paid e18.
5
x + 2x 5x+10x+x+2x = 90
x + 2x + = 18
5 18x = 90
x 2x x + 2x 18 x = 90 : 18
+ + =
1 1 5 1 x=5
5x 10x x + 2x 90 2x = 10
+ + = x + 2x 15
5 5 5 5 = = 3.
5x + 10x + (x + 2x) = 90 5 5
Answer: The notebook cost e5, the book e10 and the
pencil e3 .
Alberto Pardo Milan´s
e Algebraic expressions and equations

57. Indice Monomials Adding and subtracting Identities and Equations Solving Exercises
Exercises
Exercise 6
Data: The notebook cost x euros. The book cost 2x euros.
The notebook and the book together cost (x + 2x) euros.
x + 2x
The pencil cost . He paid e18.
5
x + 2x 5x+10x+x+2x = 90
x + 2x + = 18
5 18x = 90
x 2x x + 2x 18 x = 90 : 18
+ + =
1 1 5 1 x=5
5x 10x x + 2x 90 2x = 10
+ + = x + 2x 15
5 5 5 5 = = 3.
5x + 10x + (x + 2x) = 90 5 5
Answer: The notebook cost e5, the book e10 and the
pencil e3 .
Alberto Pardo Milan´s
e Algebraic expressions and equations