42 linear equations

Linear Equations I

Back to Algebra–Ready Review Content.

http://www.lahc.edu/math/algebra/alg-ready-placement-review.html

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.




For x pizzas it would cost 3 * x = $3x.


Linear Equations I

Example A.





To have them delivered, it would cost 3x + 10 ($) in total.


Linear Equations I

Example A.







Linear Equations I

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

Example A.







Linear Equations I


We backtrack the calculation by subtracting the $10 for delivery

so the cost for the pizzas is $24,

Example A.







Linear Equations I



so the cost for the pizzas is $24, each pizza is $3 so we must

have ordered 8 pizzas.

Example A.







Linear Equations I





In symbols, we've the equation 3x + 10 = 34,

Example A.







Linear Equations I






backtrack-calculation: 3x + 10 = 34 subtract 10

–10 –10

Example A.







Linear Equations I







–10 –10

so 3x = 24

Example A.







Linear Equations I







–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




An equation is two expressions set equal to each other.

Equations look like:

left expression = right expression

or

LHS = RHS

Linear Equations I







or

LHS = RHS

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.







or

LHS = RHS

Linear Equations I



equal. Such a value is called a solution of the equation.







or

LHS = RHS

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










or

LHS = RHS

Linear Equations I




Where as we use an expression to calculate future outcomes,

we use an equation to help us 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, or

8 = 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


3x + 10 = 34, or

8 = 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



of the x2.



3x + 10 = 34, or

8 = 4x – 6.



to reveal what x is. The easiest linear equations to solve are

the single–step equations such as the following ones,

Linear Equations I



of the x2.



3x + 10 = 34, or

8 = 4x – 6.





x – 3 = 12,

Linear Equations I



of the x2.



3x + 10 = 34, or

8 = 4x – 6.





x – 3 = 12,

12 = x + 3,

Linear Equations I



of the x2.



3x + 10 = 34, or

8 = 4x – 6.





x – 3 = 12,

12 = x + 3,

3*x = 12,

12 =

all four equation are one-step equations.

x3

Linear Equations I



of the x2.



3x + 10 = 34, or

8 = 4x – 6.





x – 3 = 12,

12 = x + 3,

3*x = 12,

12 =

all four equation are one-step equations.

x3

Linear Equations I



of the x2.



3x + 10 = 34, or

8 = 4x – 6.

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


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




Example B. Solve for x

a. x – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I





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.





a. x – 3 = 12 Add 3 to both sides

+ 3 + 3

b. x + 3 = –12

c. 3x = 15

Linear Equations I

This says








+ 3 + 3

x = 15

b. x + 3 = –12

c. 3x = 15

Linear Equations I

This says








+ 3 + 3

x = 15 check: 15 – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

?

This says








+ 3 + 3

x = 15 check: 15 – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

This says








+ 3 + 3

x = 15 check: 15 – 3 = 12

b. x + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

This says



This says

“3 added to x gives –12”,

hence subtract 3 to get

back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12

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

–3 –3

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

This says



This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

This says



This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

?

This says



This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

This says



This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

c. 3x = 15

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

This says



This says



back to x.

This says

“triple the x gives 15”,

hence divide by 3 to get

back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

3x3

153

=

c. 3x = 15 Both sides divided by 3

Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

This says



This says



back to x.

This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

3x3

153

=

x = 5


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

This says



This says



back to x.

This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

3x3

153

=

x = 5 check: 3(5) = 15


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

?

This says



This says



back to x.

This says



back to x.






+ 3 + 3

x = 15 check: 15 – 3 = 12


–3 –3

x = –15 check: –15 + 3 = –12

3x3

153

=

x = 5 check: 3(5) = 15


Linear Equations I

12 = 12 (yes)

?

–12 = –12 (yes)

?

15 = 15 (yes)

?

This says



This says



back to x.

This says



back to x.

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 I

This says



back to x.

x3


x3

–12=( (3) )

x = –36

Linear Equations I

This says



back to x.

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.

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

Linear Equations I

This says



back to x.

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.



solution.

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.




Example C. Solve for x

a. 4x – 6 = 30



solution.

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.





a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides



solution.

+6 +6

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.








solution.

+6 +6

4x = 36

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.








solution.

+6 +6

4x = 36 4 4

Divide both sides by 4

x3


x3

–12=( (3) )

x = –36 Check:3

–12=– 36

Linear Equations I

This says



back to x.








solution.

+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


b. x – 6 = 3x Collect the x's by subtracting x from both sides–x –x



–6 2x=



–3 = x

–6 2x=2 2

Divide by 2



–3 = x

–6 2x=2 2

In real–life, we encounter linear equations often.

Divide by 2



–3 = x

–6 2x=2 2


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 calories

a. How many calories are there in the sandwich with 2 slices of

cheese?

Divide by 2



–3 = x

–6 2x=2 2






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



–3 = x

–6 2x=2 2






cheese?



b. What is the expression that calculate the number of calories

of a sandwich with x slices of cheese?

Divide by 2



–3 = x

–6 2x=2 2






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?


The total calories 14 + 90x is 500,


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

140 + 90x = 500


Subtract 140 from

both sides–140 –140


140 + 90x = 500


Subtract 140 from


90x = 360


140 + 90x = 500


Subtract 140 from


90x = 360 Divide both sides by 9090 90

x = 4


140 + 90x = 500


Subtract 140 from



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


140 + 90x = 500


Subtract 140 from



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.


140 + 90x = 500


Subtract 140 from



x = 4


#x ± # = #x ± #,

where # can be any number. We solve it by following steps:



140 + 90x = 500


Subtract 140 from



x = 4


#x ± # = #x ± #,


1. Add or subtract to move the x-term to one side of the

equation and get: #x ± # = # or # = #x ± #



140 + 90x = 500


Subtract 140 from



x = 4


#x ± # = #x ± #,




2. Add or subtract the # to separate the number-term from the

x-term to get: #x = # or # = #x.



140 + 90x = 500


Subtract 140 from



x = 4


#x ± # = #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



140 + 90x = 500

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



Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2

–3x –3x

– 4 = 2x + 2

–2 –2



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




the other side.

Linear Equations I

Example E.

Solve 3x – 4 = 5x + 2

–3x –3x

– 4 = 2x + 2

–2 –2

– 6 = 2x

2–6

22x

=




the other side.

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




the other side.

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




the other side.

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

–x14. 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 – 8

19. –x = 3 – 2x 20. –5x = 6 – 3x 21. –x + 2 = 3 + 2x

22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 24. –2x + 2 = 9 + x

Education

42 linear equations