YOU are better than YOU think. Show yourself how:
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Read logic
chapters 1 to 5 in online volume Three
Skills for Algebra for greater skills & confidence
in work |
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Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite.
Logic
Mastery
Amazing, Amusing, Amorous, Delicious, Delightful, Edifying,
Strengthening Elixir.
It eases work & learning difficulties Makes the hard easier. Opens eyes.
Leads to greater precision.
in reading and
writing
Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck.
After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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Caution: Site advice is approximately correct, for some circumstances, not all. That leaves room for thought |
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What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts.
Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice.
For online automated help in senior high school maths & calculus,
visit quickmath.com For Automatic
Calculus and Algebra Help with derivatives, integrals, graphs, linear equations,
matrix algebra, visit calc101.com
With overlap, each site quickmath
& calc101offers a different range of
services, some free, some not, all based on webmathematica. Good luck.
Intersection of Lines
Link with Systems of Equations
The point-slope, slope-intercept, two-point and vertical line forms of the equation of a line can be written in the form
a x + by = c
where both coordinates x and y have been written on the left hand side of the equation and the ordered pair (a,b) is non zero, that is not (0,0).
For example
y = 5 x -10
holds when and only when
y + (-5)x = -10
It has the above form with a = 1 (not written as 1 y = y), b = -5 = (-5) and c = -10. Here the ordered pair (a, b) = (1, -5) is not (0, 0).
The equation ax + by = c when b is nonzero is equivalent to y = -(a/b)x +(c/b). The latter is the slope-intercept form of the line with slope m = -a/b and y-intercept c/b. In the case b = 0, the equation becomes ax=c or x = c/a and it provides the equation of the vertical line with x intercept c/a. So in all cases where the ordered pair (a,b) of coefficients is nonzero, the solution set of the equation ax+by= c appears as a straight line in the plane.
In general, the y intercept of the equation ax+by =c is given by y = c/b when b is nonzero; and the x intercept of the same equation is given by x = c/a when a is nonzero. And if a = 0, the equation becomes by = c, the equation for the horizontal straight line y = c/b, provided b is nonzero.
If (a,b)=(0,0) then (x,y) is a solution of the equation a x + by = c when and only when 0x+0y = c. So if c has the value 0, all point (x,y) in the plane satisfy the equation, and the equation imposes no restriction on (x,y). To avoid that equation, we assume (a,b) is nonzero.
Pairs of Lines - Geometric Expectations
Suppose L1 and L2 denote lines in the plane. These lines could be the same (concident) if unwittingly we have denoted the same line twice. These lines could be parallel or intersecting. That is what we envision geometrically.
Geometrically, if two lines L1 and L2 meet, they can only meet in one point. If the L1 and L2 meet in two points they coincide.
Pairs of Lines - The Algebraic Model
Let lines L1 and L2 be described by two equations
L1 eq'n: ax + b y = c (keep the old actors)
L2 eq'n: cx + d y = e (introduce new actors)
in which (a,b) and (c,d) are both not equal to (0,0). In brief with less clutter we write
L1: ax + b y = c
L2: cx + d y = e
A point (x,y) which satisfies both equations will belong to both lines L1 and L2 and so provide an intersection point of the lines or the solution sets for both equations.
If solving simultaneous equations is easy for you, then a large part of high school equation solving becomes very simple -- too simple, as then solutions to problems become the task, hard or not, of finding a set of linear equations to solve for the missing information in a problem.
Algebraic Example of Intersecting Lines: Find the intersection point if any of the equations
L1: 2x + 3 y = 1
L2: 5x + 9 y = 4
Solution First Part (x-elimination): Multiply the first equation by c=5 and the second equation by a=2 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations
5*L1: 10x + 15 y = 5
2*L2: 10x + 18 y = 8
2*L2-5*L1: 3 y = 3
The third equation 3y =3 follows from taking the 5*L1 equation away from the 2*L2 equation. The third equation implies y =3/3 =1. So x-elimination yields the value of y.
Solution Second Part: (Get x): Now the first equation L1 (we could have used the second) implies 2x = 1 -3y which in turn implies the run-on set of equalities x = ½ (1-3y) = ½ (1-3*1) = ½ (-2) = -1. Mathematics notation in allowing run-on equalities allows a run-on sentence or sequence of assertions.
So the point (x,y) = (-1, 1) belongs or should satisfy both equations and thus belong to both lines L1 and L2.
Solution Third Part (Check Answer): That is, Check that (x,y) = (-1, 1) satisfies the two equations
L1: 2x + 3 y = 1
L2: 5x + 9 y = 4Whenever we obtain a solution of a set of equations, the possibility of an arithmetic or logical error in the steps that yield the solution suggests the solution should be checked. If the check fails, we correct the steps (or redo them) after checking the check to avoid looking for an error that is not in the solution.
Solution Postscripts
Uniqueness: Note the above sequence of steps that lead to the solution imply if (x,y) satisfies the two equations for L1 and L2 then (x,y) = (-1, 1). So the solution (-1,1) of the above two equations is unique. There is not a second intersection point.
Unequal Slopes: From the first of the two equations
L1: 2x + 3 y = 1
L2: 5x + 9 y = 4
we see that the slope of L1 is m1 = -2/3 while from the second of the two equations we see that the slope m2 = - 5/9 =\= -2/3 = - 6/9.
The case where the slopes are equal lead to parallel lines or coincident lines. Details follow.
The following explanations assume some knowledge of direct and indirect use of implication rules for arriving at conclusions. See the first logic chapters and first logic postscript in the online book Three Skills for Algebra
Algebraic Example of Parallel Lines: Find the intersection point if any of the equations
L1: -3x + 2 y = 1
L2: -9x + 6 y = 4
Solution First Part (try x-elimination): Multiply the first equation by c=3 and the second equation by a=1 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations
3*L1: -9x + 6 y
= 3
1*L2: -9x + 6 y = 4
Here we see that the left-sides of both equations are the same but the right
hand side are different. So we are looking for a point (x,y) such that the
expression -9x + 6 y gives two different values, namely 3 and 4, when
computed. That is impossible. So there is no solution. So we are done. Our
conclusion is no intersection.
Another way to see the impossibility of the two lines L1 and L2 intersecting is as follows. The first of the two equation
L1: -3x + 2 y = 1
L2: -9x + 6 y = 4
implies y = (3/2)x + 1 while the second implies y = (9/6)y + 4/6 = (3/2)x +(2/3). Now we have two new equations for L1 and L2, namely
L1: y = (3/2)x + 1
L2: y = (3/2)x + 2/3
The slope-intersection form of these equations tells us that both lines have the same slope 3/2, the y-intercept of the first is 1 while the while intercept of the second is 2/3. We see for each point x on the x-axis, the y coordinate of a point (x,y1) on the first line L1 is 1/3 more than the y-coordinate of a point (x,y1) on the second line L2. See the diagram
The intersection of a vertical line with L1 (the blue line) is 1/3 of a unit above its intersection with L2 (the red line). So there is no intersection. (An intersection point (p, q) would give a vertical line x = p in which the intersection with L1 and L2 have the same height q, but the intersections with L1 and L2 of x=p always have different heights.
Algebraic Example of Coincident Lines: Find the intersection point if any of the equations
L1: 3x + 9 y = 6
L2: 5x + 15 y = 10
Solution First Part (try x-elimination): Multiply the first equation by c=5 and the second equation by a=5 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations
5*L1: 15x + 45 y
= 30
3*L2: 15x + 45 y =
30
Here we see that the left-sides and rights of the resulting equations are
the same. So we are looking for a point (x,y) such that the expression
15x + 45 y gives the value 30. Here they are many solutions,
the set or line described by the equation
15x + 45 y = 30
Now let us retreat and rewrite each equation for L1 and L2 in slope-intercept form.
- The L1 equation 3x + 9 y = 6 gives y = (-3x + 6)/9 or y = -(1/3) x + 2/3
- The L2 equation : 5x + 15 y = 10 gives y = (-5x +10)/15 or equivalently y = -(1/3) x + 2/3
So L1 and L2 equations both represent the line y = -(1/3) x + 2/3. All points on this line satisfy the L1 and L2 equations. That implies L1 and L2 denote the same line.
www.whyslopes.com
Analytic Geometry
Area Entrance
Entrance + Pages Below this page
Pages at Current Level(L) Numericall Intro (L) Deriving Eq'ns (L) Perpendicular Lines (L) 3 Eqn Forms (L) Algebraic View (L) Finding Intersection Points (L) Coordinate Only Geometry. (L) Exercises (L) Lines Summary (L) Lessons Elsewhere
Area Entrance (C) Complex Numbers (FN) What are Functions? (FN) Functions - More SZM: Sign Analysis (L) Lines Summary (P) Polynomials (*,+,-) (Q) Quadratics (D) Simplify Square Roots (T) Unit Circle Trig Conic Sections Links More Links
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Extras: Not all perfect.
Equal Sign Use/Abuse Real Numbers Say More Positive Linear Inequalities Triangle Inequality Absolute Value |x| |x| Eq'ns & Inequalities Rectangular Coords Shortest Path Distance Formulas Add & Multiply Points Polar Coordinates Radians (A) Vectors (A) Coordinate Arithmetic (A) Navigation on Maps (A) Addition Geometrically (A) Rotation (PT) Translations (PT) Dilatations PT: Rotations
Links to Site Pages outside this site area follow - co- and pre- requisites.
Easy Consequences of this (newest) Complex Number. Starter Lesson follow below to provide an alternate development of HS or college maths.
Vec & Cmplx No Applet
B2 C. Conjugates
B3 Pythagoras
B4 Distance
B5 Rt Triangle Similarity
B6 Trig., Functions
B7 Dot & Cross Products
B8 Cosine Law
B9 Exponential & cis fns
B10 Easy Trig Identities
B11 Set Viewpoint Arithmetic Videos - Real Player Format
Decimal Addition Methods
Decimal Subtraction Methods
Decimal Multiplication Methods
Decimal Division Methods
Fractions
Primes
Greatest Common Divisor
Divisors
Least Common Multiples
Square Root
SimplificationUsing formulas forwards & Backwards - A unifying theme for algebra skill development - the 4th skill in Volume 2, Three Skills for Algebra!