42 linear equations

Linear Equations I

Example A.a. We order pizzas from Pizza Grande. Each pizza is $3. There is $10 delivery charge. How much would it cost if we want x pizzas delivered?

Recall example A from the section on expressions.(–Link this)Linear Equations I

Example A.a. We order pizzas from Pizza Grande. Each pizza is $3. There is $10 delivery charge. How much would it cost if we want x pizzas delivered? For x pizzas it would cost 3 * x = $3x.


Example A.a. We order pizzas from Pizza Grande. Each pizza is $3. There is $10 delivery charge. How much would it cost if we want x pizzas delivered? For x pizzas it would cost 3 * x = $3x. To have them delivered, it would cost 3x + 10 ($) in total.




b. Suppose the total is $34, how many pizzas did we order?



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24,



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24, each pizza is $3 so we must have ordered 8 pizzas.



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24, each pizza is $3 so we must have ordered 8 pizzas. In symbols, we've the equation 3x + 10 = 34,



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24, each pizza is $3 so we must have ordered 8 pizzas. In symbols, we've the equation 3x + 10 = 34, backtrack-calculation: 3x + 10 = 34 subtract 10 –10 –10



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24, each pizza is $3 so we must have ordered 8 pizzas. In symbols, we've the equation 3x + 10 = 34, backtrack-calculation: 3x + 10 = 34 subtract 10 –10 –10 so 3x = 24



b. Suppose the total is $34, how many pizzas did we order? We backtrack the calculation by subtracting the $10 for delivery so the cost for the pizzas is $24, each pizza is $3 so we must have ordered 8 pizzas. In symbols, we've the equation 3x + 10 = 34, backtrack-calculation: 3x + 10 = 34 subtract 10 –10 –10 so 3x = 24 divide by 3 so x = 8 (pizzas)

In the above examples, the symbolic method to find solution may seem unnecessarily cumbersome but for complicated problems, the symbolic versions are indispensable.

Linear Equations I

In the above examples, the symbolic method to find solution may seem unnecessarily cumbersome but for complicated problems, the symbolic versions are indispensable.An equation is two expressions set equal to each other. Equations look like: left expression = right expression or LHS = RHS

Linear Equations I


Linear Equations I

We want to solve equations, i.e. we want to find the value (or values) for the variable x such that it makes both sides equal.


Linear Equations I

We want to solve equations, i.e. we want to find the value (or values) for the variable x such that it makes both sides equal. Such a value is called a solution of the equation.


In the example above 3x + 10 = 34 is an equations and x = 8 is the solution for this equations because 3(8) + 10 is 34.

Linear Equations I



Linear Equations I


Where as we use an expression to calculate future outcomes,we use an equation to backtrack from known outcomes to the original input x, the solution for the equation.

In the example above 3x + 10 = 34 is an equations and x = 8 is the solution for this equations because 3(8) + 10 is 34.

Linear Equations IA linear equation is an equation where both the expressions on both sides are linear expressions such as 3x + 10 = 34, or8 = 4x – 6.

Linear Equations I

A linear equation does not contain any higher powers of x such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because of the x2.

A linear equation is an equation where both the expressions on both sides are linear expressions such as 3x + 10 = 34, or8 = 4x – 6.

Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is.

Linear Equations I



Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is. The easiest linear equations to solve are the single–step equations such as the following ones,

Linear Equations I



Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is. The easiest linear equations to solve are the single–step equations such as the following ones,x – 3 = 12,

Linear Equations I



Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is. The easiest linear equations to solve are the single–step equations such as the following ones,x – 3 = 12,12 = x + 3,

Linear Equations I



Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is. The easiest linear equations to solve are the single–step equations such as the following ones,x – 3 = 12,12 = x + 3,3*x = 12,12 = all four equation are one-step equations.

x3

Linear Equations I



Linear equations are the easy to solve, i.e. it’s easy to manipulate a linear equation, to backtrack the calculations, to reveal what x is. The easiest linear equations to solve are the single–step equations such as the following ones,x – 3 = 12,12 = x + 3,3*x = 12,12 = all four equation are one-step equations.

x3

Linear Equations I



12 = x – 3,x + 3 = 12, 12 = 3*x,x/3 = 12

These equations are the same, i.e. it doesn’t matter it’s A = B or B = A. Both versions will lead to the answer for x.

Basic principle for solving one- step-equations:Linear Equations I

Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.

Linear Equations I

Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.Example B. Solve for x a. x – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.Example B. Solve for x a. x – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

This says “x take away 3 gives 12”, hence add 3 to get back to x.

Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.Example B. Solve for x a. x – 3 = 12 Add 3 to both sides + 3 + 3

b. x + 3 = –12

c. 3x = 15

Linear Equations I


Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.Example B. Solve for x a. x – 3 = 12 Add 3 to both sides + 3 + 3 x = 15

b. x + 3 = –12

c. 3x = 15

Linear Equations I


Basic principle for solving one- step-equations:To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation.Example B. Solve for x a. x – 3 = 12 Add 3 to both sides + 3 + 3 x = 15 check: 15 – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

?



b. x + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?



b. x + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?


This says “3 added to x gives –12”, hence subtract 3 to get back to x.


b. x + 3 = –12 Subtract 3 from both sides –3 –3

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?




b. x + 3 = –12 Subtract 3 from both sides –3 –3 x = –15

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?




b. x + 3 = –12 Subtract 3 from both sides –3 –3 x = –15 check: –15 + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

?





c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?





c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?



This says “triple the x gives 15”, hence divide by 3 to get back to x.



3x3

153=

c. 3x = 15 Both sides divided by 3

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?






3x3

153=

x = 5


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?






3x3

153=

x = 5 check: 3(5) = 15


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

?






3x3

153=

x = 5 check: 3(5) = 15


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

15 = 15 (yes)

?




x3

–12=d. Linear Equations I

This says “x divided by 3 gives –12”, hence multiply by 3 to get back to x.

x3

–12=d. Multiply both sides by 3

x3 –12=( (3)

)

Linear Equations IThis says “x divided by 3 gives –12”, hence multiply by 3 to get back to x.

x3


x3 –12=( (3)

)

x = –36


x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36


x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36


Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution.

x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36


Next we solve equations that require two steps. These are the ones that we have to collect the x-terms (or the number–terms) first with addition or subtraction, then multiply or divide to get x.


x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36


Next we solve equations that require two steps. These are the ones that we have to collect the x-terms (or the number–terms) first with addition or subtraction, then multiply or divide to get x.Example C. Solve for x a. 4x – 6 = 30


x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36


Next we solve equations that require two steps. These are the ones that we have to collect the x-terms (or the number–terms) first with addition or subtraction, then multiply or divide to get x.Example C. Solve for x a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides


+6

+6

x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36




+6

+6 4x = 36

x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36




+6

+6 4x = 36 4 4

Divide both sides by 4

x3


x3 –12=( (3)

)

x = –36 Check: 3–12=– 36




+6

+6

x = 9

4x = 36

(Check this is the right answer.)4 4

Divide both sides by 4

Linear Equations IExample C. Solve for x b. x – 6 = 3x

Linear Equations IExample C. Solve for x b. x – 6 = 3x Collect the x's by subtracting x from both sides

–x

–x


–x

–x –6 2x=


–x

–x

–3 = x

–6 2x=2 2 Divide by 2


–x

–x

–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often.

Divide by 2


–x

–x

–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often. Example D. To make a cheese sandwich, we use two slices of bread each having 70 calories and slices of cheeses with cheese where each slices of cheese is 90 caloriesa. How many calories are there in the sandwich with 2 slices of cheese?

Divide by 2


–x

–x

–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often. Example D. To make a cheese sandwich, we use two slices of bread each having 70 calories and slices of cheeses with cheese where each slices of cheese is 90 caloriesa. How many calories are there in the sandwich with 2 slices of cheese? There are 140 cal in the bread and 2 * 90 = 180 cal to make a total of 140 + 180 = 320 calories in the cheese.

Divide by 2


–x

–x

–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often. Example D. To make a cheese sandwich, we use two slices of bread each having 70 calories and slices of cheeses with cheese where each slices of cheese is 90 caloriesa. How many calories are there in the sandwich with 2 slices of cheese? There are 140 cal in the bread and 2 * 90 = 180 cal to make a total of 140 + 180 = 320 calories in the cheese.b. What is the expression that calculate the number of calories of a sandwich with x slices of cheese?

Divide by 2


–x

–x

–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often. Example D. To make a cheese sandwich, we use two slices of bread each having 70 calories and slices of cheeses with cheese where each slices of cheese is 90 caloriesa. How many calories are there in the sandwich with 2 slices of cheese? There are 140 cal in the bread and 2 * 90 = 180 cal to make a total of 140 + 180 = 320 calories in the cheese.b. What is the expression that calculate the number of calories of a sandwich with x slices of cheese?There are 140 + 90x calories in the sandwich.

Divide by 2

Linear Equations Ic. How many slices of cheese are there in a 500–cal sandwich?

Linear Equations Ic. How many slices of cheese are there in a 500–cal sandwich?The total calories 14 + 90x is 500,

Linear Equations Ic. How many slices of cheese are there in a 500–cal sandwich?The total calories 14 + 90x is 500, i.e. 140 + 90x = 500


Subtract 140 from both sides–140 –140

The total calories 14 + 90x is 500, i.e. 140 + 90x = 500



90x = 360




90x = 360 Divide both sides by 9090 90

x = 4





x = 4So there are 4 slices of cheese in a 500–cal sandwich.





x = 4

The more general linear equations have the form #x ± # = #x ± #, where # can be any number.

So there are 4 slices of cheese in a 500–cal sandwich.





x = 4

The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps:






x = 4

The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps:1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± #






x = 4

The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps:1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± #2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x.






x = 4

The more general linear equations have the form #x ± # = #x ± #, where # can be any number. We solve it by following steps:1. Add or subtract to move the x-term to one side of the equation and get: #x ± # = # or # = #x ± #2. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x. 3. Divide or multiply to get x: x = solution or solution = x



Example E.

Solve 3x – 4 = 5x + 2

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x

subtract 3x to remove the x from one side.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2

subtract 3x to remove the x from one side.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2

subtract 3x to remove the x from one side.subtract 2 to move the 2 to the other side.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x


Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x

2–6

22x=


divide by 2 get x.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x

2–6

22x=

–3 = x


divide by 2 get x.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2 –3x –3x – 4 = 2x + 2 –2 –2 – 6 = 2x

2–6

22x=

–3 = x


divide by 2 get x.

Linear Equations I

Exercise A. Solve in one step by addition or subtraction .

Linear Equations I

1. x + 2 = 3 2. x – 1 = –3 3. –3 = x –5

4. x + 8 = –15 5. x – 2 = –1/2 6. = x – 32

21

B. Solve in one step by multiplication or division. 7. 2x = 3 8. –3x = –1 9. –3 = –5x

10. 8 x = –15 11. –4 = 2x 12. 7 = 3

–x

13. = –43–x 14. 7 = –x 15. –x = –7

C. Solve by collecting the x’s to one side first. (Remember to keep the x’s positive.) 16. x + 2 = 5 – 2x 17. 2x – 1 = – x –7 18. –x = x – 819. –x = 3 – 2x 20. –5x = 6 – 3x 21. –x + 2 = 3 + 2x22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 24. –2x + 2 = 9 + x

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42 linear equations

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