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Okay from
1 1/3x-3 1/2=41/18, let us rewrite it in improper fractions
(4/3)x - 7/2 = 41/18
If I multiply everything by the lowest common denominator, 18, I can cleasr all fractions...thus
18[(4/3)x - 7/2 = 41/18]
24x - 63 = 41
now add 63 and divide by 24
24x = 104
x = 104/24 = 13/3

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From
8x - 3y = -11
2x - 5y = 27
let us multiply the second equation by 4 and subtract it from the first, thus eliminating the x's...
8x - 3y = -11
-(8x - 20y = 108)
and we get
17y = -119
y = -7
and then by substituting we get
2x - 5(-7) = 27
2x + 35 = 27
2x = -8
x = -4

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There is a special form of a line not taught very often called the intercept form and it goes like this
x/a + y/b = 1
where a and b are the x- and y-intercepts...
Here we have such a case and so
x/3 + y/2 = 1
now multiply by 6 and get
2x + 3y = 6

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Let us rearrange and coomlete the square twice, so from
x^2 + y^2 - 16x + 4y + 59 = 0 we get
x^2 - 16x + 64 + y^2 + 4y + 4 = -59 + 64 + 4
(x - 8)^2 + (y + 2)^2 = 9
This is a circle with center at (8, -2) and radius 3.

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The area is given as 9 -- is that what they really want? I doubt it...
The facts they give us are
h = 2b
and
A = 9
but we know that
A = (1/2)bh
so now substitute in everything and get
9 = (1/2)b(2b)
b^2 = 9
b = 3
Then h = 6.
This is probably what they wanted.

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From x+2/x + 1/3 = 2, we multiply all by 3x to clear fractions
(x+2)/x + 1/3 = 2
3(x+2) + x = 6x
3(x+2) = 5x
x + 2 = 5x/3
now subtract off x (which is 3x/3)
2 = 2x/3
x = 3

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From 7a/15b-2b/5 we will multiply the second fraction by 3b/3b to have common denominators...thus
7a/15b - (2b/5)(3b/3b) =
7a/15b - 6b^2/15b =
(7a - 6b^2) / 15b

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If you mean
(3rad(a) - 2rad(b)) * (3rad(a) + 2rad(b))
then you have a sum times a difference which equals the difference of their squares, so you get
[3rad(a)]^2 - [2rad(b)]^2 =
9a - 4b

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From
P = (2pt) / (t + 4)
and assuming that P and p are different variables, multiply both sides by t + 4
P(t + 4) = 2pt
Pt + 4P = 2pt
now collect terms with t in them
Pt - 2pt = -4P
t(P - 2p) = -4P
t = -4p / (P - 2p)
If P and p are meant to be the same, then
t = 4

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From
y^2 - 4y - 4x = 0
we complete the square, also add 4x to both sides
y^2 - 4y + 4 = 4x + 4
(y - 2)^2 = 4(x + 1)
now divide by four and subtract one and get
(1/4)(y - 2)^2 = x + 1
x = (1/4)(y - 2)^2 - 1
This is a wider-than-normal parabola, concave right, with vertex at (-1, 2).

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From
4x^2 + 25y^2 - 100 = 0
add 100 and then divide by 100
4x^2 + 25y^2 = 100
x^2 / 25 + y^2 / 4 = 1
This is an ellipse with semi-major axis of 5 and semi-minor axis of 2.
Intercepts at (5, 0), (-5, 0), (0, 2) and (0, -2).

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From
3x + 2y = 16
2x + y = 9
solve the second equation for y and substitute it into the first equation...
y = -2x + 9
then
3x + 2(-2x + 9) = 16
3x - 4x + 18 = 16
-x = -2
x = 2
then
y = -2(2) + 9
y = 5

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Form (x/x-9) - 4 = (9/x-9) we multiply all by (x-9) to clear fractions and get
x - 4(x - 9) = 9
x - 4x + 36 = 9
-3x = -27
x = 9
but notice x is not allowed to be nine...the denominators would go to zero....therefore there is No Solution.

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Well 1/50 of 120 bottles is 2.4 bottles...therefore he can allow two defective ones per hour before rejecting the whole line...the 3rd bad one and it's over...

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Okay from
a(b^-1) / c
we know that negative exponents on the top become positive exponents on the bottom, so this becomes
a / bc.
If the original problem was (ab)^-1 / c (it is not clear), then your answer is
1/abc

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Just posted this solution...scroll thru my solutions...

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Call the shortest leg x.
The hypotenuse is x + 18.
The other leg is x + 14 since it is only 4 less than the hypotenuse.
Now apply the Pythagorean Theorem: a^2 + b^2 = c^2 and we get
x^2 + (x + 14)^2 = (x + 18)^2
x^2 + x^2 + 28x + 196 = x^2 + 36x + 324
now collecting and solving we get
x^2 - 8x - 128 = 0
(x - 16)(x + 8) = 0
x = 16 or x = -8
but distance is never negative so the short leg is 16 and the longer leg is 30. The hypotenuse is 34.

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From x^2+y^2-6x-2y-6=0 we complete the squares...
x^2 - 6x + ___ + y^2 - 2y + ___ -6 = 0 + ___ + ___
x^2 - 6x + 9 + y^2 - 2y + 1 = 6 + 9 + 1
(x - 3)^2 + (y - 1)^2 = 16
center at (3, 1) and radius = 4

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This is an important fact in algebra...NO MATTER what the two quantities are (aside from a zero denominator),
(x - y) / (y - x) = -1
Negative One!
So (6-x) / (x-6) is, you guessed it!
-1

Okay, I added, but am having trouble simplifiying...
could you please help me?
This is what I have so far...

   x²       3x - 70
———————— + —————————
 x + 10      x + 10

I got   

   x² + 3x - 70
  ——————————————
      x + 10

how do I simplify that?
Thank you in advance for your help.

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Okay from
x^2 / x+10 + 3x-70 / x+10
we can add the numerator as they are since the denominators are the same, or what are called "like" denominators, so we simply get
(x^2 + 3x - 70) / (x + 10)
Now we factor the numerator and get
(x + 10)(x - 7) / (x + 10)
oh, what a surprise, the x + 10's cancel, leaving
the answer: x - 7

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I teach my students a method called the "criss-cross" method...
Whenever you have fractions like this you can apply a short-cut...
Goes like this...
A/B + C/D = (AD + BC) / BD
no matter what A, B, C, and D are and it works for adding or subtracting...
so from
5/(y-7) - 1/(y+1)
we criss cross and get
[5(y + 1) - (y - 7)] / [(y - 7)(y + 1)] =
5y + 5 - y + 7 / [(y - 7)(y + 1)] =
4y + 12 / [(y - 7)(y + 1)]
and we're done...

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We use the work problem formula
t/A + t/B = 1
t = time worked together
A = first guy's time alone
B = second guy's time alone
1 = one job or task
so plug in and get
3/4 + 3/B = 1
multiply all by 4B and get
3B + 12 = 4B
and
B = 12 hours

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I have solved this one already. Please scroll through my solutions.

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Any value that makes the denominator go to zero...that is if
4x - 5 = 0 or x + 1 = 0, we must exclude that value for x.
Thus x cannot be 5/4 or -1. We say these are excluded from the domain of x.

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From
4x/)x^2-18x+72) + 4/(x-6)
we factor the denominator of the first fraction and get
4x/(x-12)(x-6) + 4/(x-6)
Notice that if we multiply the second fractaion by (x-12)/(x-12), we will have common denominators...so
4x/(x-12)(x-6) + 4(x-12)/(x-6)(x-12) =
(4x + 4x - 48) / (x-6)(x-12) =
8x-48 / (x-6)(x-12) =
8(x-6) / (x-6)(x-12) =
8 / (x-12)

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From (x+2)/x / (x-2)/3x
we invert the denominator and then multiply and get
(x+2)/x * (3x/(x-2)
All that cancels are the x's, leaving
3(x+2) / x-2

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Let q = # of quarters. Thus their value is 25q.
Let d = # of dimes. Thus their value is 10d.
From the facts in the problem, we have
q + d = 38
25q + 10d = 680 ($6.80 expressed in cents)
We also have a third equation, not common in these problems (q = d + 7) which incidentally is impossible...
Let us solve the first for d and substitute into the second equation...
d = 38 - q
25q + 10(38 - q) = 680
25q + 380 - 10q = 680
15q = 300
q = 20 and then
d = 18

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Since the border all along the picture is x, the length of just the picture is
40 - 2x (since x on top and bottom).
The width is then 24 - 2x, by the same logic.
And of course, A = LW = 192.
So, by substituting, we get
192 = (40 - 2x)(24 - 2x)
192 = 960 - 128x + 4x^2 or
4x^2 - 128x + 768 = 0
x^2 - 32x + 192 = 0
(x - 24)(x - 8) = 0
x = 24 or x = 8
but the border cannot be 24, that's too big.
So the border is 8 inches, and the dimensions of just the picture are
40 - 16 = 24 inches
by
24 - 16 = 8 inches

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Let the numbers be x and y.
Then from the facts, we have
x - y = 11
sqrt(x) - sqrt(y) = 1
Let us solve the first one for x:
x = y + 11
Let us substitute that into the second equation.
sqrt(y + 11) - sqrt(y) = 1
We can rearrange this and then square both sides.
sqrt(y + 11) = sqrt(y) + 1
y + 11 = y + 2*sqrt(y) + 1
10 = 2*sqrt(y)
sqrt(y) = 5
y = 25
Thus x equals 36.

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Well the first line is y = 5x + 6 and the
second line is y = 6x - 18 (when simplified)
They are both lines with positive slopes.
They share the point (24, 126) is common, that is, they intersect there.

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For every right triangle, a^2 + b^2 = c^2...
Let x = the length of the shorter leg.
Thus x + 7 is the longer leg and
x + 9 is the length of the hypotenuse...
So substituting into the Pythagorean Theorem, we have
x^2 + (x + 7)^2 = (x + 9)^2
x^2 + (x^2 + 14x + 49) = x^2 + 18x + 81
now collect like terms and solve for x
2x^2 + 14x + 49 = x^2 + 18x + 81
x^2 - 4x - 32 = 0
(x - 8)(x + 4) = 0
x = 8 and x = -4
but distances cannot be negative, so
x = 8
The other sides must be 15 and 17.