Linear AlgebraAchievement Standard 1.4

- Terms containing the variable (x) should be placed on one side (often left)

e.g. Solvea) 5x = 3x + 6 b) -6x = -2x + 12

-3x -3x 2x = 6

÷2÷2 x = 3

You should always check your answer by substituting into original equation

+2x +2x -4x = 12

÷-4÷-4 x = -3

Always line up equals signs and each line should contain the variable and one equals sign

- Numbers should be placed on the side opposite to the variables (often right)

e.g. Solvea) 6x – 5 = 13 b) -3x + 10 = 31

+5 +5 6x = 18

÷6÷6 x = 3

-10 -10 -3x = 21

÷-3÷-3 x = -7

Always look at the sign in front of the term/number to decide operation

Don’t forget the integer rules!

SOLVING EQUATIONS- Remember that addition/subtraction undo each other as do multiplication/division

- Same rules apply for combined equations

e.g. Solvea) 5x + 8 = 2x + 20 b) 4x - 12 = -2x + 24

-2x -2x 3x + 8 = 20

-8-8 3x = 12

+2x +2x 6x - 12 = 24

÷6÷6 x = 6

÷3÷3 x = 4

+12

+12 6x = 36

- Answers can also be negatives and/or fractions

e.g. Solvea) 8x + 3 = -12x - 17 b) 5x + 2 = 3x + 1

+12x +12x 20x + 3 = -17

-3-3 20x = -20

-3x -3x 2x + 2 = 1

÷2÷2 x = -1 2

÷20÷20 x = -1

-2-2 2x = -1

Make sure you don’t forget to leave the

sign too!

Answer can be written as a decimal but easiest to leave as a fraction

- Expand any brackets first

e.g. Solve

a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8

-3-3 3x = 3

÷3÷3 x = 1

3x+ 3= 6 6x - 2 = x + 8-x-x

5x - 2 = 8+2+2

5x = 10÷5÷5

x = 2- For fractions, cross multiply, then solve

e.g. Solve

a) x = 9 4 2

2x = 36÷2÷2

x = 18

b) 3x - 1 = x + 3 5 2

2(3x - 1)= 5(x + 3)6x - 2 = 5x + 15

-5x-5x x - 2 = 15

+2+2 x = 17

- For two or more fractions, find a common denominator, multiply it by each term, then solve

e.g. Solve 4x - 2x = 10 5 3

5 × 3 = 15×15×15×15

60x

5

- 30x

3

= 150

Simplify terms by dividing numerator by denominator

12x

- 10x

= 150

2x = 150 ÷2÷2

x = 75e.g. Solve 5x - (x + 1) = 2x 6 4

×24×24×24

120x

6

- (24x + 24)

4

= 48x

20x

- 6x – 6

= 48x

14x – 6 = 48x -48x-48x

-34x – 6 = 0+ 6+ 6

-34x = 6÷ -34 ÷ -34

x = -6 34

e.g. Write an equation for the following information and solvea) A rectangular pool has a length 5m longer than its width. The perimeter of

the pool is 58m. Find its width

Draw a diagram

Let x = width-10-10

Therefore width is 12 m

x x

x + 5

x + 5

x + 5 + x + x + 5 + x = 58 4x + 10 = 58 4x = 48

÷4÷4x = 12

WRITING AND SOLVING

b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number?

Let n = a number n = n + 15-4n-4n

3n = 15

Therefore the number is 5

7 4

÷3÷3n = 5

- Inequations contain one of four inequality signs: < > ≤ ≥ - To solve follow the same rules as when solving equations- Except: Reverse the direction of the sign when dividing by a negative

e.g. Solvea) 3x + 8 > 24 b) -2x - 5 ≤ 13

-8-83x > 16

÷3÷3x > 16 3

+5+5-2x ≤ 18

÷-2÷-2 Sign reverses as dividing by a negative

x ≥ -9

As answer not a whole number, leave as a fraction

SOLVING INEQUATIONS

e.g. At an upcoming tournament, Jake has got $80 to spend. Jake wants to have at least $30 left by the time he returns home so he can buy a CD. At the tournament he is only allowed to spend his money on ‘V’ drinks which each cost $3.50. Form and solve an inequation to find out the maximum number of ‘V’ drinks he can buy at the tournament. 3.5x + 30 < 80 Let x = number

of V drinks-30-30 3.5x < 50

÷3.5÷3.5 x < 14.3

Therefore the maximum number of V drinks is: 14

- Involves replacing variables with numbers and calculating the answer- Remember the BEDMAS rules

e.g. If m = 5, calculate m2 – 4m - 3 = 52 – 4×5 - 3= 25 – 4×5 - 3= 25 – 20 - 3= 2

- Formulas can also have more than one variable

e.g. If x = 4 and y = 6, calculate 3x – 2y

= 3×4 - 2×6= 12 - 12= 0

e.g. If a = 2, and b = 5, calculate 2b – a 4 Because the top needs to

be calculated first, brackets are implied

= (2 × 5 – 2) 4= (10 – 2) 4= 8 4= 2

SUBSTITUTION

Plotting Points- To draw straight line graphs we can use a rule to find and plot co-ordinatese.g. Complete the tables below to find co-ordinates in order to plot the following straight lines:a) y = 2x b) y = ½x – 1 c) y = -3x + 2 x y = 2x y = ½x –

1 -2

-1

0

1

2x y = -3x +

2 -2

-1

0

1

2

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

2 x -1 2 x -2

02

-4

4

-2 ½ x -1 – 1 ½ x -2 – 1

-1-½

-2

0

-1 ½

-3 x -1 + 2 -3 x -2 +

2

2-1

8

-4

5

Gradients of Lines- The gradient is a number that tells us how steep a line is.- The formula for gradient is:Gradient = rise

run

1st point

2nd point

rise

run

 e.g. Write the gradients of lines A and B

A

B

A =

B =

4

6 8

4

4 = 18 2

6 = 34 2

 e.g. Draw lines with the following gradientsa) 1 b) 3 c) 2 2 5

To draw, write gradients as fractions

= 3 1

When calculating gradients it is best to write as simplest fraction

y = mx- This is a rule for a straight line, where the gradient (m) is

the number directly in front of the x- When drawing graphs of the form y = mx, the line always

goes through the origin i.e. (0,0)e.g. Draw the following lines:a) y = 2x b) y = 4x c) y = 3x 5 4

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

1. Step off the gradient from the origin (0,0) 2. Join the plotted point back to the origin

= 4x 1

To draw, always write gradients as fractions

gradient

Negative Gradients e.g. Write the gradients of lines A and B

A =

B =

-3

2

10

-5

-5 = -110 2

-3 2

A

B

When calculating gradients it is best to write as simplest fraction

e.g. Draw the following lines:a) y = -2x b) y = -4x c) y = -3x 5 4

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

1. Step off the gradient from the origin (0,0) 2. Join the plotted point back to the origin

= -4x 1

To draw, always write gradients as fractions

gradient

Intercepts- Is a number telling us where a line crosses the y-axis (vertical axis)i.e. The line y = mx + c has m as the gradient and c as the intercept e.g. Write the intercepts of the lines A, B and C

x-10 -8 -6 -4 -2 2 4 6 8 10

y

-10-8-6-4-2

2468

10

A

B

C

A =

B =

C =

8

4

-3

Drawing Lines: Gradient and Intercept Method- A straight line can be expressed using the rule y = mx + ce.g. Draw the following lines:a) y = 1x + 2 b) y = -3x – 2 c) y = -

4x + 8 2 7

x-10 -8 -6 -4 -2 2 4 6 8 10

y

-10-8-6-4-2

2468

10To draw:1. Mark in intercept2. Step off gradient3. Join up points

= -3x – 2 1

Note: Any rule with no number in front of x has a gradient of 1 1e.g. y = x – 1

Writing Equations of Lines- A straight line can be expressed using the rule y = mx + c

e.g. Write equations for the following lines

x-10 -8 -6 -4 -2 2 4 6 8 10

y

-10-8-6-4-2

2468

10

A

B

C

A: B: C:m = c = m = c = m = c = y = 3x – 6 4

y = -2x + 1 3

y = 4x + 4 1

34

-2 3

41 -6 +1 +4

Horizontal and Vertical Lines- Horizontal lines have a gradient of:

0Rule: y = c (c is the y-axis intercept)- Vertical lines have a

gradient that is:undefined

Rule: x = c (c is the x-axis intercept)e.g. Draw or write equations for the following lines:

a) y = 2 b) c) x = 4 d)

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

x = -1y = -3

b)

d)

Writing Equations Cont.When you are given two points and are expected to write an equation:- One method is set up a set of axes and plot the two points.

e.g. Write an equation for the line joining the points A=(1, 3) and B = (3, -1)

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345 m = -2

1c = 5

y = -2x + 5

Sometimes when plotting the points, you may need to extend the axes to find the intercept.

- Or, substitute the gradient and a point into y = mx + c to find ‘c’, the intercept

m = -2 1

using point

(1, 3)

y = mx + c 3 = -2 x 1 + c 3 = -2 + c 5 = c

+2 +2

y = -2x + 5

Equations in the Form ‘ax + by = c’- Can use the cover up rule to find the two intercepts:

e.g. Draw the following lines:a) 2y – x = 4 b) 4x – 3y =12

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

1. Cover up ‘y’ term to find x intercept

- x = 4÷ -1 ÷ -1

x = -4

2. Cover up ‘x’ term to find y intercept

2y = 4÷ 2 ÷ 2

y = 2

3. Join up intercepts with a straight line

4x = 12÷ 4 ÷ 4

x = 3

-3y = 12÷ -3 ÷ -3

y = -4

It is also possible to rearrange equations into the form y = mx + ce.g. Rearrange 2x – y = 6

-2x -2x- y = 6 – 2x÷ -1 ÷ -1

y = -6 + 2xy = 2x – 6

x1 2 3 4 5 6 7 8 9 10

y

102030405060708090

100110120130140

Applicationse.g. A pizzeria specializes in selling large size pizzas. The relationship between x, number of pizzas sold daily, and y, the daily costs is given by the equation, y = 10x + 50

1. Draw a graph of the equation2. What are the costs if they sell 8 pizzas?$1303a. What is the cost per pizza?$103b. How is this shown by the graph?The gradient of the line4a. What are the costs

if they sell no pizzas?$50

4b. How is this shown by the graph?Where the line crosses the y-axis

SIMULTANEOUS EQUATIONS- Are pairs of equations with two unknownsTo solve we can use one of three methods:1. ELIMINATION METHOD- Line up equations and either add or subtract so one variable disappearse.g. Solvea) 2x + y =

20 x – y = 4

To remove the ‘y’ variable we add as the signs are opposite.

+ (

)3x = 24

÷3÷3x = 8 Now we substitute

x-value into either equation to find ‘y’2 × 8 + y =

2016 + y = 20-16

-16y = 4

Check values in either equation8 – 4 = 4

b) 2x + y = 7

x + y = 4

To remove the ‘y’ variable we subtract as the signs are the same.

- ( )x = 3

Now we substitute x-value into either equation to find ‘y’

2 × 3 + y = 7 6 + y = 7

-6-6y = 1

Check values in either equation3 + 1 = 4

- You may need to multiply an equation by a number to be able to eliminate a variablee.g. Solvea) 2x – y = 0 x + 2y =

5 + (

)5x = 5

÷5÷5x = 1

Now we substitute x-value into either equation to find ‘y’

2 × 1 – y = 0 2 – y = 0

-2-2– y = -2

Check values in either equation 1 + 2 × 2

= 5

Multiply the 1st equation by ‘2’ then add to eliminate the y

× 2× 1

4x – 2y = 0 x + 2y = 5

÷-1

÷-1 y = 2

Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting

b) 4x – 2y = 28

3x + 3y = 12

- ( )-18y = 36 ÷-

18÷-18y = -2

Now we substitute y-value into either equation to find ‘x’

4x – 2 × -2 = 28 4x + 4 =

28 -4-44x = 24

Check values in either equation 3 × 6 + 3 × -2 =

12

Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtractto eliminate the x

× 3× 4

12x – 6y = 84

12x + 12y = 48

÷4÷4x = 6

Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding

2. SUBSTITUTION METHOD- Make x or y the subject of one of the equations- Substitute this equation into the second e.g. Solvea) y = 3x +

1 9x – 2 y

= 4

b) x – y = 2 y = 2x +

3

Substitute what the subject equals in for that variable in the other equation

9x – 2(3x + 1) = 49x – 6x – 2 = 4

3x – 2 = 4+2+2

3x = 6÷3÷3

x = 2Now we substitute x-value into first equation to find ‘y’

y = 3×2 + 1y = 7

To check values you can substitute both values into second equation

x – (2x + 3) = 2x – 2x – 3 = 2

-x – 3 = 2+3+3

-x = 5÷-1

÷-1x = -5

1 As we are subtracting more than one term, place in brackets and put a one out in front.

Now we substitute x-value into second equation to find ‘y’

y = 2×-5 + 3y = -7

To check values you can substitute both values into second equation

GRAPHING LINEAR INEQUALITIES- For < and > we used a dotted line ------------------- - For ≤ and ≥ we used a solid line- Shading in includes the region that contains the solution

e.g. Solve by Shading in, and show the solution to the following system of inequalities

1. y < 22. x ≥ 3 3. y < 3x - 2

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5-4-3-2-1

12345

Test which region you will shade by substituting (0,0) into the equation.1. Is 0 < 2 Yes, so shade below the

line2. Is 0 ≥ 3 No, so shade to the right of the line (excluding the

point (0,0)2. Is 0 < 3x0 - 2No, so shade to the right of the line (excluding the

point (0,0)