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First factor your denominators.

Your LCD is (x+1)(x-1) and x cannot = -1 or +1 or your denominator will equal 0. If you get -1 or +1 for your answer it is extraneous and can't be used.
Multiply every thing by (x+1)(x-1) and your denominators will all be eliminated.






The solution is all real numbers because this is an identity. No matter what you put in for x (within the domain) you will get both sides of the equation are equal. Pick some random numbers besides 1 and -1 and see for yourself.
:
There's easier ways to do this, but you would have to notice some things like if you subtract 1/(x^2-1) from both sides you get:
(x-1)/(x^2-1)=1/(x+1)
(x-1)/((x+1)(x-1))=1/(x+1) cancel and you have:
1/(x+1)=1/(x+1) They're identical.
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Rules we'll use:



:







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Expand (4x - 2)^2 Remember this is the same thing as:
(4x-2)(4x-2)
4x(4x-2)-2(4x-2) First distribute 4x times the second parenthesese, then distribute the -2
4x(4x)+4x(-2)-2(4x)-2(-2)
16x^2-8x-8x+4 Finally Combine your like terms.

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Find values for m and b in the following system so that the solution to the system
is (-3, 4).
L1) 5x + 7y = b
L2) mx + y = 22
Let x=-3 and y=4 in L1 and solve for b:
5(-3)+7(4)=b
-15+28=b
13=b

:
Now let x=-3 and y=4 in L2 and solve for m:
m(-3)+4=22
-3m+4=22
-3m+4-4=22-4
-3m=18
-3m/-3=18/-3

:
Check it out. Your system for which (-3,4) is a solution is:
5x+7y=13--->>5(-3)+7(4)=13---->>-15+28=13--->>13=13
-6x+y=22---->>>-6(-3)+(4)=22--->>18+4=22---->>22=22
We're Right!!!
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Solve each of the following systems by addition. If a unique solution does not exist, state
whether the system is inconsistent or dependent.
L1) 2x+3y=1
L2) 5x+3y=16
Multiply either L1 or L2 by -1 and add it to L1 and the y's will be eliminated and you can solve for x. I'm going with L2:
L2) -1(5x+3y)=-1(16) ---->> -5x-3y=-16
:
2x+3y=1
-5x-3y=-16
__________
-3x=-15
-3x/-3=-15/-3
x=5
Substitute that into either L1 or L2 and solve for y. I'm going with L1:
2(5)+3y=1
10+3y=1
-10+10+3y=-10+1
3y=-9
3y/3=-9/3
y=-3
The unique solution is: (5,-3)
:::::::::::::::::::::::::::::::::
and
L1)x+5y=10
L2)-2x-10y= -20
Multiply L1 by 2 and add to L2 in an attempt to eliminate the x's.
L1) 2(x+5y)=2(10)---->> 2x+10y=20
2x+10y=20
-2x-10y=-20
___________
0+0=0
0=0
There is no unique solution for this one. because 0 does = 0 this one is dependent. It has an infinite number of solutions because graphically they are the same line and have all points in common.
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Solve each of the following systems by substitution.
L1) 5x - 2y=-5
L2) y - 5x = 3
Solve L2 for y (because y doesn't have a coefficient to worry about dividing) and substitute that in for y in L1 and solve L1 for x:
L2) y-5x+5x=5x+3 ----->>> y=5x+3
L1) 5x-2(5x+3)=-5
5x-10x-6=-5
-5x-6=-5
-5x-6+6=-5+6
-5x=1
-5x/-5=1/-5
x=-1/5
Plug this value into L2 and solve for y:
L2) y-5(-1/5)=3
y+1=3
y+1-1=3-1
y=2
The solution is (x,y)=(-1/5,2)
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the distance between (2,3) and (6,0) is?
The distance formula is:
(x1,y1)=(2,3) and (x2,y2)=(6,0)





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-4y - (5y +6) = -7y + 3
-4y-5y-6=-7y+3
-9y-6=-7y+3
7y-9y-6=7y-7y+3
-2y-6=3
-2y-6+6=3+6
-2y=9
-2y/-2=9/-2
y=-9/2
:
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What is a five-digit number divisible by both 2 and 5?
Any five digit number that ends in a 0 would be divisible by 10 and thus divisible by both 2 and 5. I'm going with 9,2340. You pick another.
:
What is a three digit number divisible by 2 and not 4?
If the number ends in an even number, but the last two numbers aren't 0 or divisible by 4, then the number will be divisible by 2 and not 4. I'm going with 902. You find another.
:
What is a three-digit number not divisible by 2, 3, or 5?
The number can't end in an even number, or it's divisible by 2, the sum of the digits can't be divisible by 3 or the number will be divisible by 3, the number can't end in 0 or 5 or it's divisible by 5. I'm going with 931. What number can you think of?
:
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10y+15x=20
Let x=0 and solve for y
10y+15(0)=20
10y+0=20
10y=20
10y/10=20/10
y=2 Plot (0,2)
:
Let x=2 and solve for y
10y+15(2)=20
10y+30=20
10y+30-30=20-30
10y=-10
10y/10=-10/10
y=-1 Plot (2,-1)
For lines you only need to plot two points unless otherwise specified. If you need to plot more, let x be even numbers to avoid fractions.
Connect the two points and you have a line that looks like:

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Factor, (there are other methods.)



Set each parenthesese equal to 0 and solve for x.
and
and
and
and and and
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Formula used:
divided by Flip and multiply the second fraction.






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factor completely with respect to the intergers

Replace middle term with two terms that multiply to give you the product of the outer terms: x^4*36=36x^4, but add to give you the middle term:-13x^2. Then factor by grouping.


Both parenthesese are the differences of perfect squares.

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HINT: When there are four terms, they usually want you to factor by grouping.
factor comletely with respect to the intergers
Group the first 2 terms and the last four terms
Factor the GCF out of each group
Factor the group out

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Sum the polynomials 2x^2 - 3x - 4, 2x - 5 and 3x^3 + 2x^2 - 6

Combine like terms.

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Two rules to remember on this one:
Factorization of difference of perfect squares
Sums of perfect squares are prime.
:
factor completely with respect to the intergers
Factor out GCF of 10
Factor the difference of perfect squares a=x^2 and b=4.
Factor the difference of perfect squares a=x and b=2.

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The distributive property says:
a(b+c)=ab+bc the a gets multiplied by everything in the parenthesese
For instance:
2(4+3)=2(4)+2(3)=8+6=14
:
Proof that it's true:
2(4+3)=2(7)=14
:
Ex:
3(x+4)=3x+3(4)=3x+12
:
There are too many ways to factor. You'll have to give me an example of what you're working on. I don't want to discourage you with something you're not up to yet.
:
Since you asked about the distributive property, I will tell you that factoring out the GCF (greatest common factor) is the opposite of factoring:
3x+12=3(x)+3(4)=3(x+4)
:
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I'm guessing this is probably mis-typed, but if not:
4n-6n=2n
-2n=2n
-2n+2n=2n+2n
0=4n
0/4=4n/4

Check:
4(0)-6(0)=2(0)
0-0=0
0=0 We're right!!!
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Am i correct in that y=x^2-2x+4 has no x-intercept?
Yes. An easy way to check that is to take the inside part of the square root in the quadratic equation b^2-4ac, and if it's negative, there is no x intercept.
a=1, b=-2, and c=4
(-2)^2-4(1)(4)
4-16
-12 <---no x-intercept.
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simplify =?
A lot of people use the acronym FOIL to remember to distribute everything (First, Outer, Inner, Last). I like to just distribute the first term then distribute the second term in the first parenthesis throught the second parenthesis.


Combine your like terms:

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joe purchased a jacket at a 10% discount sale for 121.50 what was the original price of the jacket
Let the original price of the jacket be:x
The discount then is: (10/100)x=.10x
The amount he paid: 121.50
:
Equation to solve:
x-.10x=121.50
.90x=121.5
.90x/.90=121.50/.90
x=1215/9
x=$135
:
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The formula for finding a slope (m) is:
Slope formula:
when given two points: (x1,y1) and (x2,y2)
:
If you're ginven an equation and asked for a slope, or given an angle of elevation and asked for a slope we get it differently.
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Solve the following systems by substitution:
L1) 4x - 12y = 5
L2) -x + 3y = -1
Solve L2) for x and substitute that in for x in L1):
-x+3y-3y=-1-3y
-x=-1-3y
-(-x)=-(-1-3y)
x=1+3y
:
4(1+3y)-12y=5
4+12y-12y=5
4+0=5
4=5
Your variables were eliminated and 4 does not = 5, there is no solution. These are parallel lines that will never intersect.
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Given f(x)=-2(x+4)^2+5, find the vertex, domain,range and axis of symmetry.
:
This is in vertex form:, vertex=(h,k), domain is all real numbers, axis of symmetry is x=h, if a is positive, the range is (h,infinity) because it opens upward, if a is negative, the range is (-infinity,h).
:
a=-2, h=-4, k=5
The vertex is: (h,k)=(-4,5)
Domain: all real numbers, (-infinity,infinity)
Range, because a=-2 is: (-infinity,5)
Axis of symmetry is: x=-4
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Hi Ashley,
Solve the following systems by substitution:
L1) 8x - 4y = 16
L2) y = 2x - 4
Substitute 2x-4 for y in L2) and solve for x:
8x-4(2x-4)=16
8x-8x-16=16
0+16=16
16=16
Your variable was eliminated and 16 = 16, therefore there is an infinite number of solutions. The lines are lying on top of each other, they are graphically the same line and have all points in common.
Some books list the solution to something with an infinite number of solutions as (x,2x-4). Most are happy with "all real numbers" or "infinite solutions."
:
Just to prove a point, watch what happens when you try to solve L1 for y:
8x-4y=16
-8x+8x-4y=-8x+16
-4y=-8x+16
-4y/-4=-8x/-4+16/-4
y=2x-4 It turns right into L2!
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graph: y=(3/5)x-4
Let x=0 and solve for y.
y=(3/5)(0)-4
y=0-4
y=-4 Plot (0,-4)
:
Let x=5 and solve for y
y=(3/5)(5)-4
y=3-4
y=-1 Plot (5,-1)
Connect the two points and you have a line that looks like:

You can plot more points if your teacher requires it, but you only need two points when graphing a line. If you pick more points, make x numbers that are divisible by 5 and you'll avoid plotting fractions.
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Rules I will use:



:
Write the expression as one logarithm.
2ln(x)-4ln(1/y)-3ln(xy)






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The x comes first.
They're in alphabetical order: (x,y).
Later on you will have a z and they will still be in alphabetical order: (x,y,z)
That observation has helped me with a few formulas too.
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Find the value of x
The LCD is 2*5 so multiply both sides by 10








: check by letting x=5 and seeing if both sides equal.


We're right!!!
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3(w+2)-(12-w) distribute the - through 12-w
3w+6-12+w
3w+w+6-12 combine like terms
4w-6
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