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Published on February 6, 2014

Author: Tzenma

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Basic Equations http://www.lahc.edu/math/frankma.htm

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“,

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, x–4=6

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, x–4=6 This is an equation.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, Note that an x–4=6 This is an equation. equation must have two sides because we are comparing two things.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, Note that an x–4=6 This is an equation. equation must +4 +4 Add 4 to both sides have two sides to recover x. because we are comparing two things.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, Note that an x–4=6 This is an equation. equation must +4 +4 Add 4 to both sides have two sides to recover x. x = 10 because we are comparing two things.

Basic Equations We introduce “equations” in this section by starting with the basic rules of backtracking-a procedure that we utilize daily. Example A. We left the house in the morning with unknown amount of money, say \$x. We bought a \$4-burger for lunch so we have \$(x – 4) left. Upon returning home we realized that we had \$6 left. Then by adding the \$4 back to \$6, we may conclude that we must have x = \$10 in the morning. To record example A with symbols, we equate x – 4 to be the same as \$6 using the equal sign “=“, Note that an x–4=6 This is an equation. equation must +4 +4 Add 4 to both sides have two sides to recover x. x = 10 because we are comparing two things. Hence if we know the difference “x – c”, then by adding the number c, we could backtrack and recover the number x, i.e. x–c +c x

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled.

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled.

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16,

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. x + 12 = 16

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. x + 12 = 16 This is an equation.

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. x + 12 = 16 This is an equation. In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x,

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. x + 12 = 16 This is an equation. In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. x + 12 = 16 –12 –12 This is an equation. Subtract 12 from both sides to find x. In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. This is an equation. x + 12 = 16 –12 –12 Subtract 12 from both sides to find x. x = 4 In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. This is an equation. x + 12 = 16 –12 –12 Subtract 12 from both sides to find x. x = 4 In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x The answer for x is called a solution for the equation.

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. This is an equation. x + 12 = 16 –12 –12 Subtract 12 from both sides to find x. x = 4 In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x The answer for x is called a solution for the equation. In particular if we replace the x in the equation by the solution, then the two sides should come out equal,

Basic Equations Example B. We have x gallons of gas left in the gas tank. After pumping 12 gallons of gas, we have x + 12 gallons and our 16-gallon tank is filled. By subtracting the 12 gallons from 16 gallons, we see that we had x = 4 gallons before it’s filled. To record example B, we equate x + 12 to be the same as 16, i.e. This is an equation. x + 12 = 16 –12 –12 Subtract 12 from both sides to find x. x = 4 In general, if we know the sum “x + c”, then by subtracting c, we could recover the number x, i.e. x+c –c x The answer for x is called a solution for the equation. In particular if we replace the x in the equation by the solution, then the two sides should come out equal, e.g. by setting x = 4 in x + 12 = 16, we obtain 4 + 12 = 16 or 16 = 16.

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution.

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations.

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x Conversely if we divide x by a, then multiply their quotient by a, we get back the number x. (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x Conversely if we divide x by a, then multiply their quotient by a, we get back the number x. For example, 10 ÷ 5 * 5 = 2 * 5 = 10. (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x Conversely if we divide x by a, then multiply their quotient by a, we get back the number x. For example, 10 ÷ 5 * 5 = 2 * 5 = 10. x÷a (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x Conversely if we divide x by a, then multiply their quotient by a, we get back the number x. For example, 10 ÷ 5 * 5 = 2 * 5 = 10. x ÷ a* a x (The */÷ Principle for Solving equation)

Basic Equations We say that we solve an equation when we manipulate it step by step to obtain the solution. There are two principles that we use to solve equations. (The +/– Principle for Solving equation) From examples A and B, we see that to solve an equation, we apply addition to undo a subtraction and we apply subtraction to undo an addition. Likewise, if we multiply x by a, then divide their product by a, we get back the number x. For example, 2 * 5 ÷ 5 = 10 ÷ 5 = 2. x*a ÷a x Conversely if we divide x by a, then multiply their quotient by a, we get back the number x. For example, 10 ÷ 5 * 5 = 2 * 5 = 10. x ÷ a* a x (The */÷ Principle for Solving equation) To solve an equation, we apply division to undo a multiplication and we apply multiplication to undo an division.

Basic Equations Example C. Let y be the price of an apple.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 to solve for the answer.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer. Example D. Three people shared z pieces of candies so each person has z ÷ 3 pieces of candies.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer. Example D. Three people shared z pieces of candies so each person has z ÷ 3 pieces of candies. Suppose each person received 8 pieces, then by multiplying 8 by 3 we may conclude there were 8 x 3 = 24 pieces originally.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer. Example D. Three people shared z pieces of candies so each person has z ÷ 3 pieces of candies. Suppose each person received 8 pieces, then by multiplying 8 by 3 we may conclude there were 8 x 3 = 24 pieces originally. In symbols, we equate z ÷ 3 with 8, z÷3 = 8

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer. Example D. Three people shared z pieces of candies so each person has z ÷ 3 pieces of candies. Suppose each person received 8 pieces, then by multiplying 8 by 3 we may conclude there were 8 x 3 = 24 pieces originally. In symbols, we equate z ÷ 3 with 8, z÷3 = 8 Multiply both sides by 3 z÷3*3 =8*3 to solve for the answer.

Basic Equations Example C. Let y be the price of an apple. We bought 5 apples, so it cost \$5y. Suppose we paid \$20 for them, then dividing 20 by 5 we may conclude that each apple was 20 ÷ 5 = \$4. In the language of equations, we equate 5y with \$20, 5y = 20 5y ÷ 5 = 20 ÷ 5 Divide both sides by 5 y=4 to solve for the answer. Example D. Three people shared z pieces of candies so each person has z ÷ 3 pieces of candies. Suppose each person received 8 pieces, then by multiplying 8 by 3 we may conclude there were 8 x 3 = 24 pieces originally. In symbols, we equate z ÷ 3 with 8, z÷3 = 8 Multiply both sides by 3 z÷3*3 =8*3 to solve for the answer. z = 24

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group.

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people.

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people.

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people.

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group?

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, subtract \$45 from \$133,

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, –45 –45 subtract \$45 from \$133,

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, –45 –45 subtract \$45 from \$133, 8x = 88

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, –45 –45 subtract \$45 from \$133, 8x = 88 then divide the result by 8. 8x ÷ 8 = 88 ÷ 8

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, –45 –45 subtract \$45 from \$133, 8x = 88 then divide the result by 8. 8x ÷ 8 = 88 ÷ 8 x = 11

Linear Equations Example E. A tour bus charges \$45 flat fee plus \$8/ticket for each per person in the group. a. Write the expression for the total cost to rent a bus for a group of x people. At \$8/ticket, it cost 8x for the tickets for x people. Adding the \$45 flat fee, we have 8x + 45 as the total cost to rent the bus for x people. b. We paid \$133 to rent a bus, how many people do we have in the group? By equating the cost formula 8x + 45 to the actual cost of \$133, we have the equation. 8x + 45 = 133 To solve for x, –45 –45 subtract \$45 from \$133, 8x = 88 then divide the result by 8. 8x ÷ 8 = 88 ÷ 8 Hence there are 11 people x = 11 in the group.

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life.

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2 5x = 20

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2 5x = 20

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2 5x = 20 (Divide by 5) 5x ÷ 5 = 20 ÷ 5

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2 5x = 20 (Divide by 5) 5x ÷ 5 = 20 ÷ 5 (The solution) x= 4

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Add 2) +2 +2 5x = 20 (Divide by 5) 5x ÷ 5 = 20 ÷ 5 (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Subtract 4) (Add 2) –4 –4 +2 +2 5x = 20 (Divide by 5) 5x ÷ 5 = 20 ÷ 5 (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Subtract 4) (Add 2) –4 –4 +2 +2 12 = 6x 5x = 20 (Divide by 5) 5x ÷ 5 = 20 ÷ 5 (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Subtract 4) (Add 2) –4 –4 +2 +2 (Divide by 6) 12 = 6x 5x = 20 (Divide by 5) 12 ÷ 6 = 6x ÷ 6 5x ÷ 5 = 20 ÷ 5 (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Subtract 4) (Add 2) –4 –4 +2 +2 (Divide by 6) 12 = 6x 5x = 20 (Divide by 5) 12 ÷ 6 = 6x ÷ 6 5x ÷ 5 = 20 ÷ 5 2= x (The solution) (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18

Linear Equations The equation 8x + 45 = 133 in the last example is of the form #x ± # = #, (or # = #x ± #) which occur often in real life. To solve for x in this type of equations, we apply the following two steps. i. Add or subtract the appropriate number to both sides to transform the equation to the form of #x = # (or # = #x). ii. Divide (or multiply) both sides by the appropriate number to both sides to obtain the solution x = # (or # = x). Example F. Solve the following equations. a. 5x – 2 = 18 b. 16 = 6x + 4 (Subtract 4) (Add 2) –4 –4 +2 +2 (Divide by 6) 12 = 6x 5x = 20 (Divide by 5) 12 ÷ 6 = 6x ÷ 6 5x ÷ 5 = 20 ÷ 5 2= x (The solution) (The solution) x= 4 Check: Set x = 4 5(4) – 2 = 20 – 2 = 18 Check: Set x = 2 6(2) + 4 = 12 + 4 = 16

Linear Equations What is the purpose of equations since we can just backtrack using our regular languages to obtain the answers as we did in the above examples?

Linear Equations What is the purpose of equations since we can just backtrack using our regular languages to obtain the answers as we did in the above examples? As we shall see later, formulating and solving equations allow us to find solutions to complicated problems that are impossible to be solved by backtracking using our everyday language and thoughts.

Linear Equations What is the purpose of equations since we can just backtrack using our regular languages to obtain the answers as we did in the above examples? As we shall see later, formulating and solving equations allow us to find solutions to complicated problems that are impossible to be solved by backtracking using our everyday language and thoughts. The language of equations reduces complicated problems to mechanical procedures and preserve our precious mental energy for other tasks.

Linear Equations Exercise A. Solve in one step by addition or subtraction . 3. –3 = x –5 1. x + 2 = 3 2. x – 1 = –3 4. x + 8 = –15 5. x – 2 = –1/2 6. 2 = x – 1 3 2 B. Solve in one step by multiplication or division. 7. 2x = 3 8. –3x = –1 11. –4 = x 2 14. 7 = –x 9. –3 = –5x 12. 7 = –x 3 15. –x = –7 10. 8 x = –15 13. –x = –4 3 C. Solve by collecting the x’s to one side first. (Remember to keep the x’s positive.) 18. –x = x – 8 16. x + 2 = 5 – 2x 17. 2x – 1 = – x –7 19. –x = 3 – 2x 20. –5x = 6 – 3x 22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 21. –x + 2 = 3 + 2x 24. –2x + 2 = 9 + x

Linear Equations D. Solve for x by first simplifying the equations to the form of #x ± # = #x ± #. 25. 2(x + 2) = 5 – (x – 1) 26. 3(x – 1) + 2 = – 2x – 9 27. –2(x – 3) = 2(–x – 1) + 3x 28. –(x + 4) – 2 = 4(x – 1) 29. x + 2(x – 3) = 2(x – 1) – 2 30. –2(x – 3) + 3 = 2(x – 1) + 3x + 13 31. –(x + 4) – 2(x+ 1) = 4(x – 1) – 2 32. x + 1 + 2(x – 3) = 2(x – 1) – (2 – 2x) 33. 4 – 3(2 – 2x) = 2(4x + 1) – 14 34. 5(x – 2) – 3(3 –x) = – 3(x +2) + 2(4x + 1) 35. –3(2 – 2x) + 3(3 – x) = 5(x – 1) + 2(2 – 3x) 36. 6(2x – 5) – 4 (3x +2) = 2x + 6(–3x – 4) – 8

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August 13, 2018

August 13, 2018

August 13, 2018

August 13, 2018

August 13, 2018

August 13, 2018

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