Transcript of "Unit6: Algebraic expressions and equations"
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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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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises 1 Monomials 2 Adding and subtracting monomials 3 Identities and Equations 4 Solving 5 ExercisesAlberto Pardo Milan´s e Algebraic expressions and equations
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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises MonomialsAlberto Pardo Milan´s e Algebraic expressions and equations
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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. 3Alberto Pardo Milan´s e Algebraic expressions and equations
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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
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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises Adding and subtracting monomialsAlberto Pardo Milan´s e Algebraic expressions and equations
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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
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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises Identities and EquationsAlberto Pardo Milan´s e Algebraic expressions and equations
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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
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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
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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
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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
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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
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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
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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises SolvingAlberto Pardo Milan´s e Algebraic expressions and equations
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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
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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
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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
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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
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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
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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
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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
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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
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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
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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
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Indice Monomials Adding and subtracting Identities and Equations Solving Exercises ExercisesAlberto Pardo Milan´s e Algebraic expressions and equations
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = −2Alberto Pardo Milan´s e Algebraic expressions and equations
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 = 315Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = 315Alberto Pardo Milan´s e Algebraic expressions and equations
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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 = 315Alberto Pardo Milan´s e Algebraic expressions and equations
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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
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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 5Alberto Pardo Milan´s e Algebraic expressions and equations
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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
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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
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