Linear AlgebraAchievement Standard 1.4
- Terms containing the variable (x) should be placed on one side (often left)
e.g. Solve
a) 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. Solve
a) 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 EQUATIONSSOLVING EQUATIONS- Remember that addition/subtraction undo each other as do multiplication/division
- Same rules apply for combined equations
e.g. Solve
a) 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. Solve
a) 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 = 75
e.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÷4
x = 12
WRITING AND SOLVINGWRITING 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. Solve
a) 3x + 8 > 24 b) -2x - 5 ≤ 13-8-8
3x > 16÷3÷3
x > 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 INEQUATIONSSOLVING 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
SUBSTITUTIONSUBSTITUTION
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
0
2
-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 LinesGradients 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 = mxy = 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 GradientsNegative 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
InterceptsIntercepts- 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
2
4
6
8
10
A
B
C
A =
B =
C =
8
4
-3
Drawing Lines: Gradient and Intercept MethodDrawing Lines: Gradient and Intercept Method- A straight line can be expressed using the rule y = mx + c
e.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
2
4
6
8
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 LinesWriting 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
2
4
6
8
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 LinesHorizontal 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.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’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
1
2
3
4
51. 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
ApplicationsApplicationse.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 equation
2. 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?
$504b. How is this shown by the graph?
Where the line crosses the y-axis
SIMULTANEOUS EQUATIONSSIMULTANEOUS EQUATIONS- Are pairs of equations with two unknowns
To 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÷3
x = 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÷5
x = 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 = 2
Now 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
1As 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 INEQUALITIESGRAPHING LINEAR INEQUALITIES- For < and > we used a dotted line ------------------- - For ≤ and ≥ we used a solid line
- Shading in includes the region that contains the solutione.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)
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