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It depends on how men and women there are in the group of 9.

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Salary = Base + Commission
.
Company A:
Base = $200
Commission = 10% of sales = .1*S
A = .1S + 200
.
Company B:
Base = 400
Commission = .05*S
B = .05S + 400
.
We're looking for the point where they're equal.
.
200 + .1S = 400 + .05S
.
multiply by 100 to get rid of decimals
.
20000 + 10S = 40000 + 5S
Subtract 20000 from both sides
10S = 20000 + 5S
Subtract 5S from both sides
5S = 20000
Divide both sides by 5
S = 4000
.
So, at $4000 sales, the two jobs pay the same.
.
You have to always check your work. This can be done by plugging the values into the basic equations and determining if they check. Or we can graph the two lines and see where they cross.
.
Company A:
200 + .1(4000) = 200 + 400 = 600
Company B:
400 + .05(4000) = 400 + 200 = 600
.
Graphs

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When your put too much into a question, you limit the number of tutors willing to try to help. One question per email allows multiple people to provide help.
.
Slope intercept form is: y = mx + b
.
11.) 4x + y = 1
subtract 4x from both sides
y = -4x + 1
.
12.) x - y = 6
Add y to both sides
x = y + 6
Subtract 6 from both sides
x - 6 = y
y = x - 6
.
13.) 6 x - 3 y = -9
subtract 6x from both sies
-3y = -6x - 9
divide both sides by -3
y = 2x + 3
.
14.) -12 x - 4 y = 2
Add 12x to both sides
-4y = 12x + 2
Divide both sides -4
y = -3x - 1/2
.
15.) 2 x + 5 y = -10
Subtract 2x from both sides
5y = -2x -10
Divide both sides by 5
y = -2/5x -2
.
16.) -x - 10 y = 20
Add x to both sides
-10y = x + 20
Divide both sides by -10
y = -1/10x -2

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In solving word problems, always use as few unknowns as possible. Typically, this means you strive to define each unknown in terms of another.
.
We also consider what we know about the figure. We are told it is a rectangle. So we know:
Perimeter = 2L + 2W
Area = L * W
.
We also are told the dimensions of the base rectangle:
L = 24
W = 16
A = 24 * 16 = 384
.
We also are told that if you add the same amount, say 'c', to both dimensions, then the area is doubled.
The problem does not ask us to find 'c' so perhaps that doesn't matter. We could just use ratios.
.
24 * 16 = 384
which can be described as ratios
24/16 = 1.5
,so
1.5x*x = 384
.
Doubling the area = 2*384 = 768
.
1.5x^2 = 768
x^2 = 768/1.5 = 512
x = sqrt(512) = 22.6274 = W
.
L= 1.5x = 33.9411
.
Checking the area:
33.9411*22.6274 = 767.9988
.
So this estimate is off by only 12 in ten thousand.
Very close indeed...

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Substitute x=6 in the function
So

This defines the point (6,9).
.
Substituting x=0 in the function, we get y=0, which defines the point (0,0).
.
You can plot these two points and draw the graph.
.
The graph is:

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The two equations are:


.
For lines to parallel thus must have the same slope.
Perpendicular lines have slopes that are the negative inverse of each other.
.
The negative inverse of -2 is 1/2, so it appears these lines are perpendicular.
.
We can graph them as:

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Ratio can be express as, for example, 2 to 1 or 2:1 or 2/1.
In this case, it asks for a ratio of L to W, which can be shown as L/W.
20/15 is the ratio.
or
1.33 to 1
or
1 & 1/3 to 1
.
Is 20 really one-third more than 15? Or is is 1/4?
.
To check a change, the formula is

.

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When you work with coins, you have to keep in mind that there will be counts of the coins and values of the coins to balance.
n = number of nickels
5n = the value of the nickels in cents
d = number of dimes
10d = value of the dimes in cents
q = quarters
25q = value of the quarters in cents
.
We are told the total value:
5n + 10d + 25q = 335 cents
.
Note that everything needs to be in terms of cents.
.
We are told:
n + d + q = 30.
.
We also are told:
n = 2q
, which means the number of nickels is twice the number of quarters.
.
So we have defined 3 equations and we have 3 unknowns...hmmm...
.
5n + 10d + 25q = 335
n + d + q = 30
n = 2q
.
Substituting 2q for n, we have
.
5(2q) + 10d + 25q = 335
2q + d + q = 30
.
Simplifying, we have
10q + 10d + 25q = 335
10d + 35q = 335
.
2q + d + q = 30
3q + d = 30
.
d = 30 - 3q
, which seems obvious if you think about it.
.
10(30-3q) + 35q = 335
300 - 30q + 35q = 335
5q = 35
q = 7
, so we think we have 7 quarters
.
n = 2q
n = 14 nickels
.
Since there are 30 coins, then there must be 9 dimes.
.
We can check the values to see if they total 335.
5n + 10d + 25q = 335 cents
5n = 5(14) = 70 cents
7q = 7(25) = 175 cents
9d = 9(10) = 90 cents
70 + 175 + 90 = 335 cents
So it checks.
Done.

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You multiply the weight by the distance from the fulcrum.
70 lb * 5 ft = 350 ft-lb
.
On the other end, the person is 85 lb.
.
We want them to balance, so...
.
85 lb * x ft = 350 ft-lb
.
85x = 350
.
Divide both sides by 85
.
x = 4.1176471 ft

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Zelda is 18 years older than her son.
Z = S + 18
.
A year ago, she was 3 times her son's age.
We have to say Z-1 and S-1 to represent their ages 1 year ago.
Z-1 = 3*(S-1)
Z-1 = 3S -3
Z = 3S -2
.
We now have Z defined by two equations.
.
S+18 = 3S-2
.
Subtract S from both S
18 = 2S-2
.
Add 2 to both sides
20 = 2S
2S = 20
.
Divide both sides by 2
S = 10, so the son is 10 years old.
.
Looking back at the setup equations.
Z = S + 18
Z = 10 + 18 = 28, so Zelda is 28.
.
How old were they last year?
S = 10-1 = 9
Z = 28-1 = 27
.
To check, was Zelda 3 times as old as her son a year ago?
3*9 = 27
Yes, she was.
.
Done.

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multiply everything by 3
2x + 12 = 6x
subtract 2x from both sides
12 = 6x
6x = 12
divide by 6
x = 2
.
But there is no y

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c = number of cones
110c = revenue generated by the sale of cones in cents
s = number of sundaes
235s = revenue generated by the sale of sundaes in cents
.
110c + 235s = 29440 (given)
.
And we have another fact:
c + s = 170
.
Two equations and two unknowns is solvable.
.
We can easily determine c.
c = 170 - s
.
Now substitute into the other equation.
110c + 235s = 29440
110(170-s) + 235s = 29440
18700 - 110s + 235s = 29440
125s = 10740
s = 85.92 sundaes sold
2.35s = 201.912
.
c = 170 - s = 170 - 85.92
c = 84.08
1.10c = 92.488
.
Total = 201.912 + 92.488
Total = 294.4
.
These are hard numbers to work with, but the answer checks.

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If there are no exponents, then it is linear.
Standard form is Ax + By = C
.
3x - 5y = 0

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We start with what we know about distance problems.
d = r * t, where d=distance, r=rate, and t=time
.
Upstream info:
60=r*5
r = 12 mph
.
Downstream
60 = r*3
r = 20
.
We need to assume the boat runs as fast as possible in both directions, otherwise we cannot solve the problem.
.
Let's call the upstream rate: U
Let's call the downstream rate: D
Let's call the unknown maximum rate: X
And finally, the current is flowing at C mph.
.
Upstream:
U = X-C
In English, the upstream rate = max speed of the boat MINUS the current flowing against it.;
X - C = 12, as shown above.
.
Downstream:
D = X + C
In English, the downstream rate = max speed of the boat PLUS the current pushing it along.
X+C=20
.
So we have two equations and two unknowns. Piece of cake. Right?
X - C = 12
X + C = 20
.
Let's add them.
2X = 32
X = 16
.
So the unknown maximum speed X is 16.
.
Now we check our work, as we always do...
.
Looking back at what we know,
X-C = 12
16-C=12
C=4
OK
.
And we know
X+C = 20
16 + C = 20
C=4
.
Checking the upstream rate...
60/(16-4)= 60/12 = 5, which is how long we were told it took.
Checking the downstream rate...
60/(16+4) = 60/20 = 3, which is how long we were told it took.
Done.

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You have 3 angles: a, b & c.
a + b + c = 180
.
To define the equations, you can just start with a.
a = ?
.
b = 2a - 10 (twice a minus 10)
.
c = b + 25 (25 more than b)
.
Substitute for b
c = (2a-10) + 25
.
Now substitute back into a+b+c=180
a + (2a-10) + (2a-10+25) = 180
.
Eliminate the parentheses
a + 2a - 10 + 2a - 10 + 25 = 180
.
Combine like terms
5a + 5 = 180
.
Subtract 5 from both sides
5a = 175
.
Divide both sides by 5
a = 35
.
Substitute into the equation for b.
b = 2a -10 = 2*35 - 10 = 70 - 10 = 60
.
Add 25 to b to get c.
c = 60 + 25 = 85
.
Check by adding them up.
35 + 60 + 85 = 180
Done.

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Assuming you trying to simplify

.
You can see that = + 9 or - 9.
.
The square root of is
.
So the

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L = 2W + 3 (length = twice the width + 3)
A = L * W = 5 square inches
.
Substituting what we know.
A = (2W +3) * W = 5
2W^2 + 3W = 5
.
Subtract 5 from both sides
.
2W^2 + 3W - 5 = 0
.
We do not need to use the quadratic to solve this because it can be factored.
.
(2W + 5)(W - 1) = 0
.
So we have two candidate values for W: W = 1 and W = -5/2.
.
Since it is not possible to have a negative value for the width, W =1 is the candidate answer.
.
L = 2W + 3
L = 2(1) + 3
L = 5
.
Our candidate answer is:
L = 5
W = 1
.
Checking the work.
Is L*W = 5? yes.
So, we accept that answer.

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You have two integers, say, x and y.
Let y = 2x, which means it is twice the value of x.
We are told to solve x + y < 30.
So we start with the equivalent equation:
x + 2x = 30
x = 10
y = 2x = 20
.
So the two largest integers that solve the equivalent equation x+y = 30 are x=10 and y=20.
BUT
We are solving an inequality, not an equation. So, looking back, we see that x+y = 30 is NOT an acceptable answer. The requirement is that are x+y < 30.
.
Since we are still dealing only with integers, we can subtract 1 from x, which makes x = 9.
Then we can find y = 2x = 18
.
So, our proposed answer is:
x =9
y = 18
.
Checking our work to make sure x+y < 30...
x + y = 9 + 18 = 27 < 30. Which is fine.
.
Note that there is an intuitive value in the 27 in that we subtracted 1 from x, so we subtracted 2*1 from y, or 3 in total from 30.
.
A non-algebraic, brute force approach could be used with a problem like this. Simply set up table and plug in values to find the maximum value of x + 2x that is less than 30. BUT if you have to show your algebra teacher your work, the "brute force" approach will not earn much, if any, credit.
.
Done.

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The difference of two positive numbers is 3.
x - y = 3
.
x^2 + y^2 = 45
.
We can use the first equation to solve the second.
x - y = 3
.
Add y to both sides
x = y + 3
.
Substitute in the second equation and solve.
x^2 + y^2 = 45
(y+3)^2 + y^2 = 45
(y^ + 6y + 9) + y^2 = 45
.
Collect terms
2y^2 + 6y + 9 = 45
.
Subtract 45 from both sides.
2y^2 + 6y - 36 = 0
.
Divide by 2
.
y^2 + 3y - 18 = 0
.
Can we factor 18 into two terms that are 3 apart? Yes, 3 & 6.
.
(y + 6)(y - 3) = 0
.
So we have two candidate answers: y = -6 and y = 3.
.
The problem says the numbers are positive, so we assume y = 3 is one answer.
.
Given x^2 + y^2 = 45, we know y^9 = 9.
x^2 + 9 = 45
.
Subtract 9 from both sides
x^2 = 36
.
Take the square root.
x = 6.
.
So our second candidate answer is x = 6.
.
Checking our work.
.
What does x^2 + y^2 equal 45?
3^2 + 6^2 = 9 + 36 = 45
.
Does x - y = 3?
6-3 = 3
.
OK

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The slope-intercept formula is: y = mx+b.
The slope = m = change in y divided by change x.
.
We have an equation y = 2x -3.
m = 2 = slope
It means that for every change in y = 2, there is a change in x = 1.
This extends indefinitely in both directions. change in y = 4, means change in x = 2.
Etc.
.
The y-intercept = -3, so the related point is (0, -3)
.
Adding 4 to y means we add 2 to x, which produces the point (2, 1).
.
Plot these two points and draw the line connecting them.
.

.
The same logic applies to the second equation.
y = 2/3x - 4.
.
When x =0, y = -4, so we have the point (0, -4)
.
The slope = 2/3, which means for every change of 2 in y, there is a change of 3 in x.
So we can use this information to calculate a second point.
Add 2 to y and add 3 to x: (3, -2)
Plot these point and draw a line.
.

.
You can plot both on the same graph to check if they intercept one another.
.

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5r - 4s = -6
4r + 5s = 69
.
Subtract the second equation from the first
.
r - 9s = -75
.
Add 9s to both sides.
r = 9s - 75
.
Substitute
.
4r + 5s = 69
4(9s - 75) + 5s = 69
36s - 300 + 5s = 69
.
Add 300 to both sides
36s + 5s = 300 + 69
.
Collect terms.
41s = 369
.
Divide both sides by 41.
s = 369/41 = 9
.
Substitute into the other equation.
4r + 5(9) = 69
4r + 45 = 69
.
Subtract 45 from both sides.
4r = 24
.
Divide both sides by 4.
r = 6
.
So, we have the candidate answer is:
s = 9
r = 6
.
Check by substituting the values into the original equations.
.
5r - 4s = -6
5r = 30
4s = 36
5r - 4s = -6
OK
.
4r + 5s = 69
4r = 24
5s = 45
4r + 5s = 69
OK

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Since they are similar, you can use ratios to find the missing value.
.
4/15 = 9/x
.
Cross multiply.
4x = 135
.
Divide both sides by 4.
x=33.75
.
You can check by computing the ratios.
4/15 = .266667
.
9/33.75 = .266667
.
OK

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P = 2L + 2W
L = 3W + 1 (length is 1 more than 3*width)
.
substituting
P = 2(3W+1) + 2W
.
P = 106 (given)
.
substituting for P
2(3W+1) + 2W = 106
.
multiply thru
6W + 2 + 2W = 106
.
collect terms and subtract 2 from both sides
8W = 104
.
divide both sides by 8
W = 13
.
substitute back into the formula to solve for L
L = 3W+1 = 3(13)+1 = 40
.
Checking to see if P = 2L + 2W
2L = 80
2W = 26
P = 106
OK
.
Check to see if L = 3W + 1
3W = 39
+ 1 = 40
OK

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There are 10 people, so there are 10 ways you could pick the first person. Those are the possibilities. Two of them are married, so the chance that you pick either one of the married couple is 2/10 = 1/5 or 20%.
.
Now that you have picked one member of the married couple, there are 9 ways you can pick the next person. But only 1 of those is the spouse. So the probability is 1/9.
.
So the chance you pick the couple = 1/5 * 1/9 = 1/45
.
As a check, you can determine the number of combinations of 2 people from a set of 10. That defines the number of possibilities.

.

.
The 8! terms cancel. So we have . There is only 1 way you pick the couple. 1/45.
.

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.
Multiply both sides by

.
Subtract 4 from both sides

or

.
Factor

.
So, or

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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=20 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 0.23606797749979, -4.23606797749979. Here's your graph:

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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=28 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.64575131106459, -3.64575131106459. Here's your graph:

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The given equation is y = 4x + 1.
It is slope-intercept form: y = mx + b.
So any line that has slope 4 will be parallel to the first line.
For example, y = 4x + 3

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Put the equations in slope-intercept form.
.
Solve: x-7y=6
subtract x from both sides...
-7y = -x +6
divide both sides by -7...
y = 1/7*x + 6/7
.
Solve: 5x+5y=110
divide both sides by 5...
x + y = 22
Subtract x from both sides...
y = -x + 22
.
Graph the two lines...

.