D) Decouple like terms (letters left, use Switch Sides Switch Signs)
Our goal is to get x on the left equal to a number on the right. Right now, we have x’s and numbers all over the place. So use the Sync shortcut, Switch Sides Switch Signs, to get letters left and numbers right. Switch the signs of ONLY those terms that jump the
= sign. Just moving a term around on the same side does not switch its sign. Same Side Same Sign goes along with Switch Sides Switch Signs.
-11-9x-6=-4x-2x+11 4x-9x+2x=6+11-8
-5+7x=12x-1 7x-12x=-1+5
1-x=-5+6x+7-2x -x-6x+2x=-5-1+7
The SSSS shortcuts are based on Sync which is based on equality is just a special case of OOOS (Opposite Operation Opposite Side). Both the D and the B steps are similar because they end up moving something from one side to the other. The B step used multiplying and dividing, while the D step uses combining.
These shortcuts are well worth teaching students. Some books teach them to write out all the details of the combining on both sides, but the end result is a simple move. So I teach them to make the move based on the Sync Action. This results in simpler thinking and less writing, which results in less mistakes. Sometimes writing things down clarifies a student’s thinking and helps reduce mistakes, but not here. Writing down the simple and obvious actually increases mistakes. Of course, some students insist, so fine, I let them.
100) eliminate decimals
E) Eliminate decimals by multiplying both sides by biggest number of places
The rule is abbreviated for easier memorization and needs explaining. You can’t multiply by a decimal place, but you can multiply by 1 with a 0 for each place. So if your equation has a number with 1 decimal place, then multiply by 10, which is a 1 with one zero. If your equation has a number with 2 decimal places, then multiply by 100, which is a 1 with two zeroes. If you find a number with 3 places, multiply by 1000.
Make sure you use enough zeroes to match the number with the most places. It is better to multiply by a number with too many zeroes than too few so that you make sure to eliminate all decimal places. Reducing or rounding later will automatically take care of too many places. Also make sure not to forget to multiply the numbers with invisible decimal places. Everything on both sides must be multiplied for a proper Sync.
Also, make sure you have done step F first. Filling the parentheses will change the decimal places! So step E should never take place if ( ) are still there.
10(.2-3x)=(1.4+6x)10 2-30x=14+60x
places 1 0 1 0
Notice that I wrap each side in ( ) to remind myself that I must change the whole side when I Sync. Of course, Sync is required because I am inflating the sides by 10, 100, or whatever. Shifting, by the distributive property, then takes place when I multiply the sides.
Notice also that the E step on top, when done completely, gives me a clean D step on the next line. This is the purpose of E and F--to reduce the clutter and eliminate the complications. The E and F steps fill a similar role to In and Fun with expressions. They clean up the complicated, leaving us with just the basic operations for the D-A steps.
In both equations I multiply by a 1 with as many 0’s that match the most decimal places I found
100(.37+x)=(2.1+8x)100 37+100x=210+800x
places 2 0 1 0
101) eliminate fractions
E) Eliminate fractions by multiplying both sides by common multiple of the bottoms The technical names of what we are looking for is least common denominator or least common multiple. Remember that the denominators are the bottom numbers and we want to eliminate them so that the fractions become whole numbers, which means the bottoms must all become 1. To achieve that we must multiply all the fractions by a number that all the bottom numbers multiply into. If that seemed like a lot of words, basically we are going to look for common denominators, but stop short of combining the fractions.
If you have bottom numbers of 2 and 3, then times by 6. If you have bottoms of 3 and 4 and 12, then times by 12. Any number will work if all the bottoms go into it evenly.
Just like with decimals, it is better to multiply by a number to big, than one too small. Any common multiple will work, not just the lowest. Reducing in step A will auto-adjust.
Don’t forget the invisible 1’s on the bottom of whole numbers. The integers need to get multiplied by your number, just like the visible fractions.
Just like with decimals, make sure you have done step F before step E, because any parentheses will change your fractions.
6(4x+ 1 3 )=(5- 5 2 x)6
24x+2=30-15x
In both equations I multiplied by a number that all bottom numbers go into evenly. This Shifts every
fraction and turns them into whole numbers.
12(- 1 2 + 3 4 x)=( 5 6 x+ 1 3 )12 -6+9x=10x+4
102) Fill Parentheses
F) Fill parentheses by multiplying outside ^ inside (distribute)
There is nothing new here. This is the familiar Shift Action applied to parentheses. The outside number is the sprinkling can that waters every term inside and makes them grow.
The only thing you need to carefully notice is the change to fractions and decimals during this step. Because they change is why you must wait to eliminate them until the next step.
4(9x-6)=-2(13-6x) 36x-24=-26+12x
These examples show why it is so important to do step F before step E. Also, if students try to do step E before step F, they usually multiply the outside factor and the inside factors, which is often wrong, because it is a double multiplication to that superterm. Other terms outside any ( ) they will only multiply once. That puts their equation out of Sync.
In the fraction example below, they will ^20 inside and ^20 outside, which is multiplying by 400. If they do that on both sides, they are fine. However, let’s say they have a problem where there is a +4 next to the (x-7). They will multiply the 4 by 20 only once, while all the ( ) terms are ^ 400. That is out of Sync. This is why F is before E, to help them avoid that minefield of complexity altogether.
3 2 ( 3 5 + 3 4 x)= 1 2 (x-7)
10 9 + 9 8 = 1 2 x- 7 2
It looks like I should multiply by 20 on this line, but after the ( ) I see I really need to multiply by 40.
.3(.4x-1.6)=.25(2x+.5) .12x-.48=.5x+.125
Should I multiply by 2 places or 3 places?
These questions, and risk of mistakes, can be avoided by doing step F before step E.
103) Flip complex fractions
F) Flip complex fractions, this turns them into regular fractions
Once in a great while I have seen complex fractions in equations. These are very easily turned into regular fractions and then you can continue with the regular FA steps.
A complex fraction is fractions within a fraction. Instead of number over number, it is fraction over fraction. Just as the top number is divided by the bottom number, so the top fraction is divided by the lower fraction. So really, this is just a fancy way of writing fraction _ fraction. Therefore, flip the bottom fraction and turn it into two regular fractions multiplying each other.
7 2 3 4
x= 3 5 x+ 1 4
1 7
7 2 * 4 3 x= 3 5 x+ 1 4 * 7 1
Fraction over fraction turns into fraction ^ flipped fraction
multiply right away
104) Figure functions
F) Figure functions
The reason for step F is to take care of the complicated first. Step E cleans up what step F can’t handle. That is the basic principle that applies even to things not covered by step F. Simplify whatever looks complicated so you can turn the equation into something familiar and easy. Use the calculator!
The one point not to miss is that this applies only to numbers. If you have a power or root of x, or a trig or log function of x, you can do nothing. The equation is not even a V1D1 equation. The degree is more or less than 1, but it is not 1. These FA steps apply only to degree 1.
3x*cos45=9 2 x+log 2 8
3x*.7071=81x+3
(sin90)x=4 3 +log100*x 1x=64+2x
You may not at first see how to solve an equation, but do what you can to get it to a point where it looks familiar and solvable
105) Proportions
Proportions are a special type of equation that is simply fraction = fraction. These occur often enough in real life and on tests that it is worth learning a shortcut for them. Of course, you can use the FA method starting on step F, but the shortcut is faster and is the natural result of step F.
The basic idea is shown here.
a b = d c becomes ad=bc
If you were to follow the normal F step and multiply both sides by the common denominator, bd, then cancel, you end up with the same result. This works with polynomials of any degree in the fractions. The only limitation is that it must be fraction = fraction.
There can be no other terms combined with either fraction.
The shortcut is commonly called cross-multiplying, similar to cross-canceling with fractions. Once the cross-multiplying is done you end up with a regular equation that is usually on step D or B. You might even have a quadratic (x2) equation on your hands. We will cover those in the next chapter.
x+1 4 = 7 x Ž 7(x+1)=4x x+1 2 = x-4 8 Ž (x+1)(x-4)=16
AlGeBrA: QuAdrATIC eQuATIons
Quadratic equations are V1D2 equations. They have one variable with degree two. By Syncing and Shifting you can make it look like ax2+bx+c=0. Many of these equations will factor and all of them can be solved with the quadratic formula.
106) Fill, Flip, or Figure
This step is identical to step F for linear equations. You can ignore the x2 for now because you are dealing only with the numbers. Just as with method 1, fill parentheses, flip complex fractions, or figure functions. Everything should look and feel the same, except there will be an x2 along with the x.
4x 2 *tan45=9x+log 2 16 4x 2 *1=9x+4
.3x(.4x-1.6)=.25(2x 2 +.5) .12x 2 -.48x=.5x 2 +.125
4(9x-6)=-2x(13-6x)
36x-24=-26x+12x 2
107) eliminate fractions or decimals
Again, this step is identical to step E for linear equations. Continue to treat the x2 as a regular x, because you are dealing only with the fractions and decimals.
12(- 1 2 + 3 4 x 2 )=( 5 6 x+ 1 3 )12 -6+9x 2 =10x+4
100(.37+x)=(2.1+8x 2 )100 37+100x=210+800x 2
places 2 0 1 0
108) descending order = 0
Here is where the FA method for quadratic equations takes off in a different direction from linear equations. Because you have an x2 in the equation, instead of decoupling the letters to the left and the plain numbers to the right, bring everything to the left. Combine like terms if you need to. Then put the terms in descending order of exponents in the familiar pattern of ax2+bx+c=0. This is just like what you did in the polynomial chapter. You are now ready for factoring or for the quadratic formula.
5x-8x 2 =7+2x -8x 2 +5x-2x-7=0
-8x 2 +3x-7=0 1-2x=9+2x 2 -4x -2x 2 -2x+4x+1-9=0
-2x 2 +2x-8=0
109) Common factor
This is nothing but the old familiar common factoring from the polynomials chapter. You may not be able to find a common factor, but if you can find one, you should use it. If the number in front of the x2 is negative, then you should factor by a negative, because that will make the next steps easier for you.
-2x
2+2x-8=0 -2(x
2-x+4)=0
x
2-x+4=0 4x
2+6x+18=0
2(2x
2+3x+9)=0 2x
2+3x+9=0
Now notice something. You can divide both sides by the common factor. Since the right side is 0, it entirely disappears. Look at your equation as if it is at the B step in FA method 1. The common factor in front of the ( ) is the coefficient, and the ( ) is the tag. Dividing both sides by the coefficient cancels it out. You are now prepared to bifactor.
15x
2-5x-20=0 5(3x
2-x-4)=0
3x
2-x-4=0
Divide both sides by 5 and it disappears because of
the 0 on the right (More clutter gone!)
110) Bifactor
Again, this is a repeat of the bifactoring from the polynomials chapter. The new part is that if you can bifactor, then go ahead and set each factor equal to 0, then solve those mini-equations. Notice that the answers will always be the opposite sign of the factors from which they come.
x
2-7x+10=0 (x-5)(x-2)=0 x-5=0 x-2=0
x=5 x=2
2x
2-5x-33=0 (2x-11)(x+3)=0 2x-11=0 x+3=0
x=
11/
2x=-3
x
2+8x+12=0 (x+6)(x+2)=0 x+6=0 x+2=0
x=-6 x=-2
3x
2-11x-20=0 (3x+4)(x-5)=0 3x+4=0 x-5=0
x=
-4/
3x=5 Factor ^ factor = 0, therefore when either factor is 0 the equation is true.
So set each factor = 0.
111) Answer formula
Of course, this is the quadratic formula. If none of the factoring got you to the answers then the quadratic formula will. It always will give you an answer, even if it has to use imaginary numbers. Just make sure your equation looks like ax2+bx+c=0
-b± @ b 2 -4ac 2a
-3± @ 6
2-4*3*-8 2*3
7± @ (-7)
2-4*5*2 2*5
3x ax 2 2 +6x-8=0 5x +bx+c=0 ax 2 2 -7x+2=0 +bx+c=0
a=3 b=6 c=-8 a=5 b=-7 c=2
2 answers in one formula First from the +
Second from the -x=1.41 x=2.41 x=1 x=.4
AlGeBrA: oTher eQuATIons
112) linear
Technically, any equation of degree 1 is a linear equation because its graph is a straight line. However, this lesson is focusing on the equations that are based on y=mx+b.
Most algebra textbooks will have at least one whole chapter dedicated to these types of equations. They come in different forms and you usually have to change them from one form to another. This is where the FA method comes in handy. If you can think of the y as being the variable you want and the x as a constant, like &, then you can use the FA method for any degree 1 equation.
The D step will look a little different, but the concept is the same. The variable you want, y, goes on the left. The unwanted variable, x, and all the numbers, go on the right.
The C step is often skipped, but if there are like terms, go ahead and combine them.
You just won’t be able to combine everything on the right side to a single term. Combine them as far as you can and put the x term first followed by the plain number term.
3x=
1/
4(2y-5)
y goes left, all else right
^ both sides by -2
113) rational
These equations look almost like proportions. They look enough like proportions that many students get fooled and start doing very wild forms of cross multiplying. What throws them off is the fact that there are variables on the bottom, but a proportion is only fraction = fraction and nothing else. These other equations are really just step E equations that could be linear or quadratic. They reveal themselves after step E is done.
x+2 3 - 1 x = 4x 2
4x(x+2) ( x+2 3 - 1 x = 4x 2 )
3(4x)-4(x+2)=2(x+2) 12x-4x-8=2x+4 12x-4x-2x=8+4
6x=12 x=2
Step E: Eliminate fractions Common denominator is 4x(x+2).
The x in the second fraction is factor of 4x already. x2 is extra.
All tops must be ^ by factors they are missing on the bottom, just like regular and polynomial fractions.
Follow steps E through A as normal.
The above equation proved to be a linear equation, but not this next one. At first, it is not obvious that this equation contains an x2 but after the E step it is clear. Therefore, switch to the FA method for quadratic equations.
2x-6 x - 2x 4 = x-3 4
2x(x-3) ( 2x-6 x - 2x 4 = x-3 4 )
x(x)+4(x-3)=4(2x) x
2+4x-12=8x x
2+4x-8x-12=0
x
2-4x-12=0 (x-6)(x+2)=0 x=6 and x=-2
Step E: Eliminate fractions Common bottom is 2x(x-3).
2x-6 factors to 2(x-3).
All tops must be ^ by factors they are missing on the bottom, just like regular and polynomial fractions.
Follow steps E through A as normal.
114) Multi-variable
A linear equation is a multi-variable equation with two variables, but you will also encounter equations with three or more variables. The directions will tell you to solve for x or solve for n or some other variable. If the degree of the variable you need to find is 1, then you can use the same basic steps of the FA method for linear equations. Treat all the unwanted variables as constants.
Solve for x 5b-3x+6=8x-b+7 -3x-8x=-5b-6-b+7
-11x=-6b+1 x=
6b-111
Solve for n
4n-6m+s=7s-2+2n 4n-2n=6m-s+7s-2
2n=6m+6s-2 n=3m+3s-1 D C
B A
D C B A
all but x goes right COLT _ both sides by -11 fraction is reduced
all but n goes right
COLT _ both sides by 2
no more to do
115) exponential
A special set of equations that appear in many textbooks are exponential equations.
These equations have variables in their exponents. This makes it impossible to use a calculator because how do you enter something like 4x ? Neither you nor the calculator knows what x is. But the equations are setup to have similar bases, which means their exponents must equal.
For example, 6x=62 We now use the fact that this equation is already in Sync.
Therefore, x=2 because the exponents must be equal to keep the sides equal and in Sync.
8
x+1=8
4Because the equation is in Sync and the bases are the same, the exponents must equal. Therefore, x+1=4, so x=3. Check it on your calculator!
If the bases are not equal, then Shift and use the exponent rules to make them equal.
Then you can set the left exponent = to the right exponent.
2
x-4=8 becomes 2
x-4=2
3Now you can set the exponents equal to each other. x-4=3 therefore x=7.
6
x-.3=6
7if n
expression1=n
expression2then expression1=expression2 Now solve the new equation using the FA method
D
116) Inequalities
Inequalities are just regular equations with one small twist. Whenever you Sync both sides by multiplying or dividing by a negative number, you need to reverse the direction of the arrow in the inequality. That’s it! Remember this one thing and solve inequalities just like equations.
-4x Þ 36
x Ý -9 -5x Û -20 x Ü 4
Whenever you sync by ^ or _ with a negative number reverse the direction of the inequality arrow.
x+6 Ü 9+4x x-4x Ü -6+9
-3x Ü 3 x Û -1
2x+8 Ý 9x-6 2x-9x Ý -6-8
-7x Ý -14
x Þ 2
117) radical
Algebra is just the beginning of a whole new world of math. There are many new types of problems beyond what you have seen that will require new knowledge and new strategies. However, their foundation will be the ten Algebra Actions.
One example is a type of equation known as radical equations. This is when a variable, not a number, is inside the root symbol. In this type of equation, you will “solve it twice.”
In the first stage, pretend that the root is a plain x and follow the FA steps down to the B step. Now Sync by squaring both sides. Do not individually square the terms on each side, but put ( ) around each entire side and square the whole side at once. In really complicated cases you will need to use distribution. After squaring, you can then start the second stage at step F.
One note of caution. If you ever get a square root equal to a negative number, then you need to stop or use imaginary numbers if you know them. In the world of real numbers there is no such thing as a square root of a negative, because - ^ - = + and + ^ + = +.
Not even your calculator can give you the square root of a negative number!
5@x=15
system of equations
Visually, a system of equations is two or three lines on a grid that intersect at a point.
We solve the equations to find the coordinates of that point. We are not going to graph in this course. Graphing and other topics, like trigonometry, will be covered in another book.
What we want to look at in this chapter are systems of two or three linear equations with or without quadratic equations. Some textbooks include other combinations, but these are the major systems all books cover as well as both methods of solving.
The overall strategy to solve a system is not always one of the Actions, but along the way the equations are Subbed then solved using the FA method.
118) systems by substitution
This method will solve all the standard problems in textbooks and on tests, but can take longer than the elimination method that sometimes works. It’s a matter of choice.
We are looking for the unique combination of x and y that will make two equations work. In other words, we want that special (x,y) coordinate that is on both lines at the same time. It stands to reason that if we find the x that works, then we can substitute that value back into the equations to find the y that works.
By solving one of the equations for x I will find all the y’s that go with it. Then if I Sub that value of x into the other equation I will find the y that works in that equation as well.
By solving one of the equations for x I will find all the y’s that go with it. Then if I Sub that value of x into the other equation I will find the y that works in that equation as well.