### Straight Line and Linear Equations

The two lines represent the equations 4x-6y=-4 and 2x+2y=6. Because the graphs of 4x-6y=12 and 2x+2y=6 are straight lines, its linear.

y=3x+25 y= f(x) = mx+c

f(x) represents a functional relationship to units sold, not specifying fn explicitly. mx+c represents a generalized version, m is the coefficient, though m and c are parameters. Parameters are determined outside the model called exogenous variables. y and x are external and determined inside the model, see my post on macroeconomics – models and introduction which explains exogenous and endogenous variables. y is the dependent variable and x is the independent variable.

**Draw T-chart:**

Choose values for x and plug them in the equation to satisfy the equation values for y. You may pick negative numbers though to keep things simple we won’t go into negatives. Choose at least three x values, and you may add a coordinates column.

A domain restriction, x can take any value within the number system, rule out negative values x0. Values of x are domain. Values of y are range. Domain independent, and range dependent. For a value of x we derive a value of y. All functions are relations but not all relations are functions.

**Graph:**

y intercept where function cuts x axis. x=0 to achieve y intercept, x=0 and y=25, (y=3m+25x). Coefficient to x is slope, rise over run = slope.

If one variable increases, the other follows, in Y as a result of 1 unit in X.

Base salary remains the same though salary of 3 changes to 4 slope increases y-intercept remains same. y intercept stays the same, slope constant:

A parallel shift as rise over run is the same:

Both increase not parallel shift, steeper:

Horizontal line, a parallel line m=0, y=2, equation x=0, can take any value:

Vertical line, a parallel line to y, x=2, equation y=0 can violate a function as you can have infinitely many:

### Quadratic Equations

For each value of x you get one value of y, its a function:

Two values of y:

y=ax² +bx+c

**Example 1- **The bread stall has estimated its income in the following equation: income=65t-2t²+200. Where, costs are $7 for materials and a fixed cost to cover labour and the stall of $300 per week.

(i) Plot the graph of his projected income.

(ii) What is his estimated income if 10 loaves are sold and 20 loaves sold?

(iii) Will I have profit or loss if I sell 10 loaves? Why?

(iii) Will I have profit or a loss if 20 loaves are sold? Why? How much of a profit or a loss?

(iii) What is the minimum number of loaves I must sell to break even?

Solution

Note if variable cost is not per unit divide it by number of units.

__1 ^{st} factorise using quadratic format ax²+bx+c__

Y=65t-2t²+200 C=300+7t

A=-2, B=65, C=200 As =-2t²+65t+200

A=-2, B=58, C=-100 As cancelling to left =-2t²+65t+200=7t+300

=-2t²+65t-7t+200-300=0

=-2t²+58t-100=0

__4 ^{th} Find other Coordinates or Middle points Between Break Even Points__

X (t number # of units) | Y (Income y=-2t²+65t+200) | C (Cost C=7t+300) | Description | |

Y Intercept | 0 | 200 | 300 | |

Minimum Break Even | 1.84 | 312.83 | 312.88 | Minimum break even at 2 loaves |

Income and Cost at 10 Pairs | 10 | 650 | 370 | Profit |

Maximum Turning Point | 16.25 | 728.13 | 413.75 | |

Income and Cost at 20 Pairs | 20 | 700 | 440 | Profit |

Maximum Break Even | 27.16 | 490.07 | 490.12 | 28 loaves break even |

Trend | 35 | 25 | 545 | End of graph lines |

Note when filling in table always calculate (brackets) first.

__5 ^{th} Plot Points on Graph and Estimate Axis cost and quantity__

Note from the above table X and Y ordered pair coordinates. C is where cost crosses the line.

**Example 2-** (i) y=x*²+x-6*

Set y=0 -> x²+x-6=0

Multiply 1st and 3rd terms: x²+3x-2x=6=0

x(x+3)-2(x+3)=0 -> (x+3)(x-2)=0

-6x² has two factors 3x + -2x

Either, x+3=0 -> x=-3 or, x-2=0 -> x=2

x intercepts (-3,0), and (2,0)

y intercepts -> x=0 -> y=-6 (0,-6)

**Example 3- **y=-2x²+3x+5

-2x²+3x+5=0

1st x 3rd -> -2x²+5x-2x+5=0 -> -x(2x-5)-1(2x-5)=0 -> (2x-5)(-x-1)=0

Either x=5/2 or, x=-1

**Example 4- **x²+2x+2=0

General form: ax²+bx+c

2x² has not got two factors, therefore it has no solution. -> x²=-22-2 -> x²+2x+2=0

**Example 5-** x²+x-6=0

a=1 b=1 c=-6

**Example 6-** (i) 2x²+3x+2 (ii) x²+2x+8

(i) and (ii) 2x²+3x+2=x²+2x+8

-> 2x²-x²+3x-2x+2-8=0 -> x²4x-6=0 -> x²+3x-2x-6=0 -> x(x+3)(x-2)=0

-> (x+3)(x-2)=0 -> x=-3 -> (i) y=2(-3)²+3(-3)+2 or, x=2

### Implicit Linear Functions

y=mx+c y=ax+b Parameters and variables a, b, m …

The slope is the coefficient of x and the y intercept is the constant.

**Example 1- **(i) 6x-2y+8=0 Either x or y could be independent or dependent making the implicit function linear.

-2y=-6x-8 -> y=3x+4

Example 2- 3×4.5y=150 -> 4.5y=3x+150 -> y=-3/4.5x+150/4.5 -> =-3/45/10x+150/45/10

3/45/10 = 3x 10/45 -> x=-30/45x+1500/45 -> x=-2/3x+33.33 which is maximum produced.

Slope -> if production of x is ^ increased by 1 unit, production of y decreased by 2/3 unit.

y intercept -> a maximum of 33.33 units of y can be produced if production of x is zero.

x intercept -> set y=0 -> 0=-2/3x + 33.33 -> -2/3x+33.33 -> x=33.33×3/2 -> x=50

50 units maximum of x if allocate all resources to the production of 0 to y.

### Generalisation

Explicit linear function is ax+by+c=0 Explicit form is y=-1x-

B cannot equal zero, as leads to infinity .

Slope = – y int. = – Whether – or + it depends on a and b.

**Example 1- **The # of customers need to be predicted for the bread stall, and can be predicted with the following equation: p=275000 + 7500t

Where p = # of customers annually, and t = years (measured from current year).

t=0 identifies current session. Slope y=mx+c, where y intercept =27000 + 7500t slope.

(i) Graph the equation

(ii) Identify the slope and y intercept

(iii) Interpret the meaning of slope and y intercept in this application

(i)

(ii) Additional customers add 7500 = slope # of customers annually. y intercept 275000 = # of customers in current year is 275000.

ax+by+c=0 -> ax+by=-c -> by=-ax-c

y=x – -> by+c=0 -> by=-c -> y=-=12

**Example 2- **The total cost c of renting and marketing the market stall is estimated by the equation: c=0.40x+1800

Where, c=total cost in $ and, x= # of years operated.

(i) Identify the slope

(ii) Identify the y intercept

(iii) Interpret a and b

(i)

(ii) y intercept cost is 1800 at idle, x=0

(iii) a is the cost of $1800 if the stall is rented for 0 years. b is the additional 0.40 per year increase.

**Example 3- **The # of ingredients supplied each month can be estimated by the equation: c=1200-12.5p, where c= the # of ingredients supplied on time and p= # of suppliers assigned to deliver goods on time. If c is graphed on the vertical axis.

(i) Identify the slope and explain its meaning

(ii) Identify the c intercept and explain its meaning

(iii) Identify the p intercept and explain its meaning

(i) c=1200-12.5p, where c= # of ingredients monthly, and p= # of suppliers. The slope is the -12.5p note it is the coefficient of p. Time of delivery will drop by 12.5 (approx) for one additional supplier.

(ii) y at 0 suppliers has 1200 hours for ingredients to be supplied, likely from your own labour.

(iii) The p intercept is 12.5. To find how many suppliers are needed for delivery time to get to zero, we set c=0 =1200-12.5p=0, where p=1200/12.5 =96. Delivery time will drop to 0 at 96 suppliers.

**Example 4- **The value of the bread machine is expressed by the equation: v=60000-7500t, where v = the value in $ and t = the age of the machine expressed in years.

(i) Identify the t and v intercepts

(ii) Interpret the meaning of the intercepts

(iii) Interpret the meaning of the slope

(i) Identify the v intercepts, v=60000 and t=0. At zero years value is 60000 (y intercept). At 8 years value is 0 (x intercept).

(ii) y intercept is 60000 value, x intercept declines 7500 every year as ages.

(iii) v=60000-7500p, where v= the value in $, and p= the age of the machine expressed in years. The slope is the -7500p note it is the coefficient of p.

**Example 5- **The market stall produces two products, white and brown bread. Weekly labour availability equals 150 labour hours. Each unit of white bread requires 3 labour hours and each unit of brown bread requires 4.5 labour hours. If management wishes to use all labour hours the equation is: 3x+4.5y=150, where x equals the # of units produced of white bread and y equals the # of units produced of brown bread.

(i) Identify the slopes and intercepts and interpret the meaning.

x= # of units of white bread y= # of units of brown bread labour= 150 hours (weekly)

y=mx+c 4.5=3x+150 4.5+150=3x 154.5=3x

### Simultaneous Equations

**Example 1-** (i) y=3x (ii) y=x+10 y is expressed in terms of x.

3x=x+10 -> 2x=10 -> x=5

Then plug in to equation: y=3(5)=15

Find coordinates: (x,y) = (5,15)

Check (i) y 15=3(5) 15=15

**Example 2-** (i) y=2x-10 (ii) 5-y/3=x

Method 1

5-y/3=x -> 5-y=3x (iii) y=-3x+5

(i) & (ii) -> 2x-10-3x+5 -> 5x=15 -> x=3

(i) y=-4 -> (x,y) = (3,-4)

Method 2 (Substitution)

(ii) 5-(2x-10)/3 =x

5-2x+10=3x -> 15=5x -> x=3

(x,y) = (3,-4)

**Example 3-** Eliminate 1 value: (i) 8x+4y=12 (ii)_(iii) -2x+y=9

Method 1

(ii) isolate y -> y=9+2x -> y=2x+9

Eliminate y, plug in (i) 8x+4(2x+9)=12

-> 8x+8x+36=12 -> 16x=-24 -> x=-24/16 = -6/4 = -3/2

y=6 (iii) y=2(-3/2)+9=6

(x,y) = (-3/2, 6)

Alternatively

(i) 8x+4y=12 (ii) -2+4y-9

(4×2=8) (ii) x4 -> -8x+4y=36 (iii)

(i) + (iii) 8x+4y=12 + -8x+4y=3 here, the + and – 8 cancel out and = 8y=48 -> y=6

Plug into equation 8x+4(6)=12 -> 8x+26=12 -> x=12/8 =-3/2

(x,y) = (-3/2, 6)

### Substitution

Simultaneous equations are two equations with two unknowns. Find the values of x and y:

**Example 1-** (i) 2y+x=8 (ii) 1+y=2x

From [i] y=2x-1 subtract 1 from each side substituting this value for y into [i] gives:

2(2x-1)+x=8

4x-2+x=8 expand the brackets

5x-2=8 tidy up

5x=10 add 2 to each side

x=2 by dividing both sides by 5 the value of x is found.

Substitute the value of x into y=2x-1 gives y=4-1=3, so x=2 and y=3.

**Example 2-** (i) y-2x=1 (ii) 2y-3x=5 Rearranging we get y=1+2x

We can replace the y in [ii] by substituting it with 1+2x. [ii] becomes 2(1+2x)-3x=5

2+4x-3x=5

2+x=5

x=3, Substituting x=3 into [i] gives us y-6=1, so y=7.

**Example 3-** (i) 2x+y=7 (ii) 3x-y=8

Add the two equations to eliminate the y values:

2x+y=7 3x-y=8

5x = 15

x = 3

Now you can put x=3 in either of the equations. Substitute x=3 into the equation 2x+y=7, 6+y=7, y=1

### Elimination

Multiplying (or dividing) the expression on each side by the same number does not alter the equation. Adding two equations produces another valid equation. 2x=x+10 (x=10) and x-3=7 (x also =10). Adding the equations gives 2x+x-3x+10+7 (x also=10). When combined either x or y is eliminated.

**Example 1-** In this example the x term will drop out giving a solution for y.

(i) 2y+x=8 (ii) 1+y=2x Rearrange the equation so it is similar to the other:

(ii) y-2x=-1 also, 2x (i) gives 4y+2x=16 which we call (iii).

(ii) y-2x=-1 (iii) 4y+2x=16

(ii) + (iii) gives 5y=15 so, y=3

Substituting y=3 into (i) gives 1+(iii)=2x so, 2x=4 giving x=2 and y=3.

### Two Cases (Unique No Solution)

**1-** Lines parallel, same slope

(i)y=2x+3 (ii) y=2x-2

(i) & (ii) 2x+3=2x-2 -> 3=-2 -> False statement, equations are inconsistent, no solution

**2-** Lines of top of each other, same slope and y intercept dependent variable, infinite solutions.

(i) y=2x+3 (ii) x=1/2y-3/2 -> y=2x+3 -> y=2(1/2-3/2)+3 -> =y-3+3 -> y=y

Alternatively

(ii) x=1/2y-3/2 -> 1/2y=x+3/2 -> (iii) y=2x+3

(i) & (iii) are the same and not independent.

### Unique Solution

### No Solution

### Infinite Solution

Not Independent variable as of 3 possible outcomes.

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*Copyright ©* 2016 Zoë-Marie Beesley

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