Place to Learn.....eStudypk: Calculus

Straight-Line Equations

Posted by Muhammad Atif Saeed | Tuesday, 20 December 2011 | Posted in Calculus, feature, Mathandstat

Straight-line equations, or "linear" equations, graph as straight lines, and have simple variable expressions with no exponents on them. If you see an equation with only x and y — as opposed to, say x² or sqrt(y) — then you're dealing with a straight-line equation.

There are different types of "standard" formats for straight lines; the particular "standard" format your book refers to may differ from that used in some other books. (There is, ironically, no standard definition of "standard form".) The various "standard" forms are often holdovers from a few centuries ago, when mathematicians couldn't handle very complicated equations, so they tended to obsess about the simple cases. Nowadays, you likely needn't worry too much about the "standard" forms; this lesson will only cover the more-helpful forms.

I think the most useful form of straight-line equations is the "slope-intercept" form:

y = mx + b

This is called the slope-intercept form because "m" is the slope and "b" gives the y-intercept. (For a review of how this equation is used for graphing, look at slope and graphing.)

I like slope-intercept form the best. It is in the form "y=", which makes it easiest to plug into, either for graphing or doing word problems. Just plug in your x-value; the equation is already solved for y. Also, this is the only format you can plug into your (nowadays obligatory) graphing calculator; you have to have a "y=" format to use a graphing utility. But the best part about the slope-intercept form is that you can read off the slope and the intercept right from the equation. This is great for graphing, and can be quite useful for word problems. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

Common exercises will give you some pieces of information about a line, and you will have to come up with the equation of the line. How do you do that? You plug in whatever they give you, and solve for whatever you need, like this:

Find the equation of the straight line that has slope m = 4
and passes through the point (–1, –6).

Okay, they've given me the value of the slope; in this case,

m = 4

Also, in giving me a point on the line, they have given me an

-value and a

-value for this line:

x = –1

and

y = –6

In the slope-intercept form of a straight line, I have

, and

. So the only thing I don't have so far is a value for is

(which gives me the

-intercept). Then all I need to do is plug in what they gave me for the slope and the

and

from this particular point, and then solve for

y = mx + b(–6) = (4)(–1) + b–6 = –4 + b–2 = b

Then the line equation must be "

y = 4x – 2

What if they don't give you the slope?

Find the equation of the line that passes through the points (–2, 4) and (1, 2).

Well, if I have two points on a straight line, I can always find the slope; that's what the

slope formula

is for.

Now I have the slope and two points. I know I can find the equation (by solving first for "

") if I have a point and the slope. So I need to pick one of the points (it doesn't matter which one), and use it to solve for

Using the point

(–2, 4)

, I get:

y = mx + b4 = (–²/₃)(–2) + b4 = ⁴/₃ + b4 – ⁴/₃ = b12/3 – ⁴/₃ = b b = 8/3

...so

y = ( –²/₃) x + ⁸/₃.

On the other hand, if I use the point

(1, 2)

, I get:

y = mx + b2 = (–²/₃)(1) + b2 = –²/₃ + b2 + ²/₃ = b⁶/₃ + ²/₃ = b b = ⁸/₃

So it doesn't matter which point I choose. Either way, the answer is the same:

y = (–²/₃)x + ⁸/₃

As you can see, once you have the slope, it doesn't matter which point you use in order to find the line equation. The answer will work out the same either way.

Point-Slope Form

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x₁, y₁) and a slope m, and have you plug it into this formula:

y – y₁ = m(x – x₁)

Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula:

Find the equation of the straight line that has slope m = 4 and passes through
the point (–1, –6).

This is the same line that I found on the

, so I already know what the answer is (namely,

y = 4x – 2

)

. But let's see how the process works with the point-slope formula.

They've given me

m = 4, x₁ = –1,

and

y₁ = –6

. I'll plug these values into the point-slope form, and solve for "

y – y₁ = m(x – x₁) y – (–6) = (4)(x – (–1)) y + 6 = 4(x + 1) y + 6 = 4x + 4 y = 4x + 4 – 6

y = 4x – 2

This matches the result I got when I plugged into the slope-intercept form. This shows that it really doesn't matter which method you use (unless the text or teacher specifies). You can get the same answer either way, so use whichever method works more comfortably for you.

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Straight-Line Equations:
Point-Slope Form (page 2 of 3)
Sections: Slope-intercept form, Point-slope form, Parallel and perpendicular lines

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x₁, y₁) and a slope m, and have you plug it into this formula:

y – y₁ = m(x – x₁)

Find the equation of the straight line that has slope m = 4 and passes through
the point (–1, –6).

This is the same line that I found on the

, so I already know what the answer is (namely,

y = 4x – 2

)

. But let's see how the process works with the point-slope formula.

They've given me

m = 4, x₁ = –1,

and

y₁ = –6

. I'll plug these values into the point-slope form, and solve for "

y – y₁ = m(x – x₁) y – (–6) = (4)(x – (–1)) y + 6 = 4(x + 1) y + 6 = 4x + 4 y = 4x + 4 – 6

y = 4x – 2

Find the equation of the line that passes through the points (–2, 4) and (1, 2).

I've already

answered

this one, but let's look at the process. I

should get the same result (namely,

y = ( –²/₃) x + ⁸/₃

Given two points, I can always find the slope:

Then I can use either point as my

(x₁, y₁)

, along with this slope Ive just calculated, and plug in to the point-slope form. Using

(–2, 4)

as the

(x₁, y₁)

, I get:

y – y₁ = m(x – x₁) y – (4) = ( –²/₃)(x – (–2)) y – 4 = ( –²/₃)(x + 2) y – 4 = ( –²/₃) x – ⁴/₃y = ( –²/₃) x – ⁴/₃ + 4 y = ( –²/₃) x – ⁴/₃ + ¹²/₃

y = ( –²/₃) x + ⁸/₃

This is the same answer I got when I plugged into the slope-intercept form. So, unless your text or teacher specifies the method or format to use, you should use whichever format suits your taste, because you'll get the same answer either way.
Parallel and Perpendicular Lines

There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is the usual format for the question:

Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that
are (a) parallel to the given line and (b) perpendicular to it.

In other words, they've given me a reference line —

2x – 3y = 9

— that I'll be comparing to, and some point somewhere else on the plane — namely,

(4, –1)

. Then they want me to find the line through

(4, –1)

that is parallel to (that has the same slope as)

2x – 3y = 9

. On top of that, they then want me to find the line through

(4, –1)

that is perpendicular to (that has a slope that is the negative reciprocal of the slope of)

2x – 3y = 9

Clearly, the first thing I need to do is

solve

2x – 3y = 9

" for "

=", so that I can find my reference slope:

2x – 3y = 9 –3y = –2x + 9 y = (²/₃)x – 3

So the reference slope from the reference line is

m = ²/₃

Since a parallel line has an identical slope, then the parallel line through

(4, –1)

will have slope

m = ²/₃

. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:

y – (–1) = (²/₃)(x – 4) y + 1 = (²/₃) x – ⁸/₃y = (²/₃) x – ⁸/₃ – ³/₃

y = (²/₃) x – ¹¹/₃

This is the parallel line that they asked for.

For the perpendicular line, I have to find the perpendicular slope. The reference slope is

m = ²/₃

, and, for the perpendicular slope, I'll flip this slope and change the sign. Then the perpendicular slope is

m = –³/₂

. So now I can do the point-slope form. Note that the only change from the calculations I just did is that the slope is different now.

y – (–1) = ( –³/₂)(x – 4) y + 1 = ( –³/₂) x + 6

y = ( –³/₂) x + 5

Then the full solution to this exercise is:

parallel:

y = (²/₃) x – ¹¹/₃

perpendicular:

y = ( –³/₂) x + 5

http://www.guangxiedu.net/wp-content/uploads/2011/07/2204126.jpg

Solving Literal Equations

Posted by Muhammad Atif Saeed | | Posted in Calculus, Mathandstat

Sometimes you have a formula, such as something from geometry, and you need to solve for some variable other than the "standard" one. For instance, the formula for the perimeter P of a square with sides of length s is P = 4s. You might need to solve this equation for s, so you can plug in a perimeter and figure out the side length.

This process of solving a formula for a given variable is called "solving literal equations". One of the dictionary definitions of "literal" is "related to or being comprised of letters", and variables are sometimes referred to as literals. So "solving literal equations" seems to be another way of saying "taking an equation with lots of letters, and solving for one letter in particular."

At first glance, these exercises appear to be much worse than your usual solving exercises, but they really aren't that bad. You pretty much do what you've done all along for solving linear equations and other sorts of equation; the only substantial difference is that, due to all the variables, you won't be able to simplify your answers as much as you're used to. Here's how "solving literal equations" works:

Solve A = bh for b

If they'd asked me to solve

3 = 2

for

, I'd have divided both sides by

. Following the same reasoning, I get:

Solve d = rt for r Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Solve P = 2l + 2w for w

If they'd asked me to solve

3 = 2 + 2

for

, I'd have moved the "free"

to the other side, and then divided through by the

on the variable. Following the same reasoning, I get:

Solve Q = ^{(c + d)}/₂ for d

–

Solve V = ^3k/_t for t

If they'd asked me to solve

5 = 3 /

for

, I'd have multiplied through by

, and then divided both sides by

. Following the same reasoning, I get:

This next exercise requires a little "trick" to solve it:

Solve Q = 3a + 5ac for a

The "trick" came in the second line, where I factored out the a. This technique doesn't come up often, but it's just about guaranteed to come up in your homework once or twice, and almost-certainly on your next test, precisely because so many students don't see the "trick". So keep in mind: When you can't isolate the desired variable because it is a factor in two or more terms, collect those terms together on one side of the "equals" sign, factor out the desired variable, and then divide off whatever is left.

Solve A = ( ¹/₂ )ah – ( ¹/₂ )bh for h

This example used the same "trick" as the previous one. In the fourth line, I factored out the h. You should expect to need to know how to do this!

The area A of a sector (a pie-wedge-shaped section) of a circle is given by:

...where r is the radius of the circle and S is the angle measure (in degrees) of the sector. Solve this equation for S.

Probably one of the more important classes of literal equations you will need to solve will be linear equations. For instance, it is common that you are given problems of this type:

What is the slope of the line with equation 3x + 2y = 8?

In order to find the

slope

, it is simplest to put this line equation into

slope-intercept form

. If I rearrange this line to be in the form "

y = mx + b

", it will be easy to read off the slope

. So I'll solve:

3x + 2y = 8

2y = –3x + 8

y = ( ^–3/₂ ) x + 4

Then

the slope is m = ^–3/₂ .

Warning: There are many contexts, such as graphing and systems of equations, in which you will need to be able to solve a linear equation for "y =", so make sure you are comfortable with these techniques.

Find the slope and y-intercept of the line with equation 2x – y = 5.

I'll solve for "

y =

2x – y = 5

2x = y + 5

2x – 5 = y

Then

y = 2x – 5

, and, from the slope-intercept form of

y = mx + b,

I can see that:

the slope is m = 2 and the y-intercept is b = –5.

Find the slope and y-intercept of the line with equation x – 2y = 5.

I'll solve for "

y =

x – 2y = 5

x = 2y + 5

x – 5 = 2y

( ¹/₂ ) x – ( ⁵/₂ ) = y

Then

y = ( ¹/₂ ) x – ( ⁵/₂ )

, so:

the slope is m = ¹/₂ and the y-intercept is b =

^–5/₂

Find the slope and y-intercept of the line with equation 4x + 5y = 12.

I'll solve for "

y =

4x + 5y = 12

5y = – 4x + 12

y = ( ^–4/₅ ) x + ( ¹²/₅ )

Then the slope is m = ^–4/₅ and the y-intercept is b = ¹²/₅ .

Don't let literal equations "throw" you. Solving literal equations is just like solving linear (and other sorts of) equations, except that the answers don't simplify as much. The techniques involved are otherwise exactly the same. Just take your time and be sure to write out all your steps clearly

Systems of Simultaneous Equations

Posted by Muhammad Atif Saeed | | Posted in Calculus, Mathandstat

In a previous chapter, solving for a single unknown in one equation was already covered. However, there are situations when more than one unknown variable is present in more than one equation. When in a given problem, more than one algebraic equation is true at a time, it is said there is a system of simultaneous equations which are all true together at once. Such sets of multiple equations may help solve for more than one unknown variable in a problem, since having more than one unknown in one equation is typically not enough information to "solve" any of the unknowns.

An unknown quantity is something that needs algebraic information in order to solve it. An equation involving the unknown is typically a piece of information which may provide the information to "solve" the unknown, i. e. to determine a specific number value (or limited number of discrete values) that the unknown is (or can be) equal to. Some equations provide little or no information and so do little or nothing to narrow down the possibilities for solutions of the unknowns. Other equations make it impossible to satisfy an unknown with any real number, so the solution set for the unknown is an empty set. Many other useful equations make it possible to solve an unknown with one or just a few discrete solutions. Similar statements can be made for systems of simultaneous equations, especially regarding the relationships between them.

Linear Simultaneous Equations with Two Variables

In the previous module, linear equations with two variables were discussed. A single linear equation having two unknown variables is practically insufficient to solve or even narrow down the solutions for the two variables, although it does establish a relationship between them. The relationship is shown graphically as a line. Another linear equation with the same two variables may be enough to narrow down the solution to the two equations to one value for the first variable and one value for the second variable, i. e. to solve the system of two simultaneous linear equations. Let's see how two linear equations with the same two unknowns might be related to each other. Since we said it was given that both equations were linear, the graphs of both equations would be lines in the same two-dimensional coordinate plane (for a system with two variables). The lines could be related to each other in the following three ways:

1. The graphs of both equations could coincide giving the same line. This means that the two equations are providing the same information about how the variables are related to each other. The two equations are basically the same, perhaps just different versions or forms of each other. Either one could be mathematically manipulated to produce the other one. Both lines would have the same slope and the same y-intercept. Such equations are considered dependent on each other. Since no new information is provided, the addition of the second equation does not solve the problem by narrowing the solution set down to one solution.

Example: Dependent linear equations

$6x - 3y = 12 \$ $y = 2x - 4 \$

The above two equations provide the same information and result is the same graph, i. e. lines which coincide as shown in the following image.

Let's see how these equations can be mathematically manipulated to show they are basically the same.

Divide both sides of the first equation $6x - 3y = 12 \$ by 3 to give

$2x -y = 4 \$

Now add y to both sides

$2x = 4 + y \$

Now subtract 4 from both sides

$y = 2x - 4 \$

This is the same as the second equation in the example. This is the slope-intercept form of the equation, from which a slope and a y-intercept unique to the line can be compared with any other equations in the slope-intercept form.

2. The graphs of two lines could be parallel although not the same. The two lines do not intersect each other at any point. This means there is no solution which satisfies both equations simultaneously, i. e. at the same time. The solution set for this system of simultaneous linear equations is the empty set. Such equations are considered inconsistent with each other and actually give contradictory information if it is claimed they are both true at the same time in the same problem. The parallel lines have equal slopes but different y-intercepts.

Sets of equations which have at least one common point which might provide a solution set are consistent with each other. For example, the dependent equations mentioned previously are consistent with each other.

Example: Inconsistent linear equations

$3x - 2y = -2 \$ $3x -2y = 2 \$

To compare slopes and y-intercepts for these two linear equations, we place them in the slope-intercept forms. Subtract 3x from both sides of both equations.

$3x - 2y = -2 \qquad \qquad 3x -2y = 2$ $-2y = -3x - 2 \qquad \qquad -2y = -3x + 2$

Divide both sides of both equations by -2 and simplify to get slope-intercept forms for comparison.

$(-2y)/(-2) = (-3x - 2)/(-2) \qquad (-2y)/(-2) = (-3x + 2)/(-2)$

$y = \frac{3}{2} x + 1 \qquad \qquad \qquad \qquad \qquad y = \frac{3}{2} x - 1$

Now, both slopes are equal at 3/2, but the y-intercepts at 1 and -1 are different.
The lines are parallel. The graphs are shown here:

3. If the two lines are not the same and are not parallel, then they would intersect at one point because they are graphed in the same two-dimensional coordinate plane. The one point of intersection is the ordered pair of numbers which is the solution to the system of two linear equations and two unknowns. The two equations provide enough information to solve the problem and further equations are not needed. Such equations intersecting at a point providing a solution to the problem are considered independent of each other. The lines have different slopes but may or may not have the same y-intercept. Because such equations provide at least one solution point, they are consistent with each other.

Example: Consistent independent linear equations

$y = 3x - 5 \$ $y = -x - 1 \$

Both of these equations are given in the slope-intercept, so it is easy to compare slopes and y-intercepts. For these two linear functions, both slopes are different and both y-intercepts are different. This means the lines are neither dependent nor inconsistent, so on a two-dimensional graph they must intersect at some point. In fact, the graph shows the lines intersecting at (1,-2), which is the ordered pair solution to this system of independent simultaneous equations. Visual inspection of a graph cannot be relied on to give perfectly accurate coordinates every time, so either the point is tested with both equations or one of the following two methods is used to determine accurate coordinates for the intersection point.

Independent equation intersecting lines.PNG

[edit] Solving Linear Simultaneous Equations

Two ways to solve a system of linear equations are presented here, the addition method and the substitution method. Examples will show how two independent linear simultaneous equations with two unknown variables could be solved for both unknown variables using these methods.

Elimination by Addition Method

The elimination by addition method is often simply called the addition method. Using the addition method, one of the equations is added (or subtracted) to the other equation(s), usually after multiplying the entire equation by a constant, in order to eliminate one of the unknowns. If the equations are independent, then the resulting equation(s) should be one(s) which will have one less unknown. For an original system of two equations and two unknowns, the resulting equation with one less unknown would have one unknown left which could easily be solved for. For systems with more than two equations and two unknowns, the process of elimination by addition continues until an equation with one unknown results. This unknown could then be solved for and the solved value then substituted into the other equations resulting in a system with one less unknown. The elimination by addition process is repeated until all of the unknowns are solved.

If a system has two equations which are dependent, then the addition of the equations could or would eliminate both unknowns at once. If the equations are parallel lines which are inconsistent, then a contradictory equation could result. The addition method is useful for solving systems of simultaneous linear equations, particularly if the equations are given in the form Ax + By = C, where x and y are the two unknown variables and A, B, and C are constants.

Example: Solve the following system of two equations for unknowns x and y using the addition method:

$x + 2y = 4 \$ $3x - y = 5 \$

Solution: We can either multiply the first equation by -3 and add the result to the second equation to eliminate x, or we can multiply the second equation by 2 and add the result to the first equation to eliminate y. Let's multiply [both sides of ] the second equation by 2.

$2 \cdot (3x - y) = 2 \cdot 5$

$2 \cdot 3x - 2 \cdot y = 10$

$6x \ - 2y = 10 \$

Now we add this resulting equation to the first equation; i. e. each of the two sides of the equations are added together to give a combined equation as shown here:

$x \ + \ 2y = 4 \$ $+ \ ( 6x - 2y = 10 ) \$ _____________________

$7x + 0\cdot y = 14 \$

This means that we add x + 2y and 6x - 2y to get 7x + 0·y and we add 4 and 10 to get 14.

This eliminates y from the combined equation to give an equation in x only:

$7x = 14 \$

Now we solve for x:

$x = 14/7 = 2 \$

Now that we have x, we can substitute the value for x into either of the original two equations and then solve for y. Let's pick the first equation for the substitution into x.

$2 + 2y = 4 \$

Solving for y:

$2y = 4 - 2 = 2\$ $y = 2/2 = 1 \$

So the solution set consists of the ordered pair ( 2,1) which is the point of intersection for the two linear functions as shown here:

[edit] Elimination by Substitution Method

The elimination by substitution method is often simply called the substitution method. With the substitution method, one of the equations is solved for one of the unknowns in terms of the other unknown(s). Then that expression for the first unknown is substituted into the other equation(s) to eliminate it such that the equation(s) then have only the other unknown(s) left. If the equations are independent, then the resulting equation(s) should be one(s) which will have one less unknown. For an original system of two equations and two unknowns, the resulting equation with one less unknown would have one unknown left which could easily be solved for. For systems with more than two equations and two unknowns, the process of elimination by substitution is repeated until an equation with one unknown results. This unknown could then be solved for and the solved value then substituted into the other equation(s), resulting in a system with one less unknown. The process of elimination by substitution continues until all of the unknowns are solved.

If a system has two equations which are dependent, then applying the substitution method would either eliminate two unknowns at once or result in an equation which do not yield single values for the remaining unknown(s). If the equations are parallel lines which are inconsistent, then a contradictory equation could result.

Example: Solve the following system of two equations for unknowns x and y using the substitution method:

$x - y = -1 \$ $x + 2y = -4 \$

Solution: We can start by solving for either x or y in terms of the other unknown in either one of the equations. Let's start by solving for x in terms of y in the first equation.

$x = y - 1 \$

Next, we substitute this expression for x into the other equation in order to eliminate x from the equation.

$(y - 1) + 2y = -4 \$ $3y - 1 = -4 \$

We have eliminated x and now we have an equation in terms of y only. We now solve for y in this equation.

$3y = -4 + 1 = -3 \$ $y = -3/3 = -1 \$

We have found the solution for y to be -1. We substitute this value for y into the expression for x in terms of y we determined from the first equation earlier.

$x = y - 1 = -1 - 1\$

Finally, we calculate the value of x.

$x = -2 \$

So the solution set consists of the ordered pair (-2,-1) which is the point of intersection for the two linear functions as shown here:

Slopes of Parallel and Perpendicular Lines

In a two-dimensional Cartesian coordinate plane, linear functions which are dependent or whose graphs are parallel lines will have the same slope. CONVERSELY, linear functions having equal slopes are either dependent or have graphs that are parallel lines in a two-dimensional Cartesian coordinate plane. Of course, vertical parallel lines of the general form x=c are not functions and have no defined slopes.

This paragraph restates itself. It should be reworded.

In a two-dimensional Cartesian coordinate plane, two lines that are perpendicular to each other will form right angles (90° angles) with each other at the point where they intersect. When the slopes of two linear functions whose graphs are lines that are perpendicular are multiplied together, the product of the two slopes equals -1. Conversely, if multiplying the slopes of two linear functions gives a product equal to -1, then their graphs are perpendicular lines on a two-dimensional Cartesian coordinate plane.

In other words, if two perpendicular lines have slopes m₁ and m₂, then

$m_1 m_2 = -1 \$ .

If a pair of perpendicular lines consists of a horizontal line (of the form y = c) and a vertical line (of the form x = c), then the preceding rule does not apply. A vertical line has no slope and the slope of a horizontal line = 0.

Example: Find the slope-intercept form of a [new] line which intersects y = (1/2)x - 3 at (4,-1) and is perpendicular to it.

Solution: First, find slope of the new line from slope of the given line. Let m = slope of the new line.

$\left ( \frac{1}{2} \right ) m = -1$

$2\cdot \left ( \frac{1}{2} \right ) m = 2\cdot (-1)$

$m = -2 \$

The slope-intercept form of the new line will be:

$y = -2x + b \$

where b is the y-intercept of the new line. Next, solve for y-intercept of new line using the intersecting point (4,-1) and the new slope of -2. Substitute x = 4 and y = -1 into the preceding equation and solve for b.

$-1 = -2 \cdot 4 + b \$

$-1 = -8 + b \$

$b = -1 + 8 = 7 \$

Finally, the slope-intercept form of the new perpendicular line is :

$y = -2x + 7 \$ .

Graph showing perpendicular lines in above example.

Solving Systems of Simultaneous Equations Involving Equations Of Degree 2

The substitution method should be used for efficiency when solving nonlinear simultaneous equations, unless other methods such as the graphing method provide clear and simple solutions quickly (when they would be faster than substitution).

Example: Solve the system of simultaneous equations.

$y^2 + (2x+3)^2 = 10 \$ $2x + y = 1 \$

With the second equation, make a given term (here, 2x should be used) the subject.

$2x = 1 - y \$

Substitute the third equation into the first, and through factorization of the resulting, simplified quadratic with one variable the solutions can be found.