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(-2) = x + 4 is a linear equation, not a quadratic.
.
x+4 = -2
subtracting 4 from both sides
x = -6

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The given inequality is sqrt(x-3) is equal to or greater than 7.
.
We first create the equivalent equation:
sqrt(x-3) = 7
.
Then we solve the equation:
.
Squaring both sides of the equation, we have
(x-3) = 49
.
Adding 3 to both sides
x = 52
.
Now we substitute back into the inequality:
x is any real number equal to or greater than 52.

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We are told the bus travels at 4/5 speed of the car. So the car travels at 5/4 speed of bus.
b = 4/5*c
c = 5/4*b
.
We are told the car drives for 3 hr at the speed c = 5/4*b.
d = rt
d = 5/4b*3 = 15/4b
.
We are told that at that time the car had traveled 85 km more than the bus.
.
d = 3b + 85
.
If a=b and b=c, then a=c.
.
15/4*b = 3b+ 85
or
15/4*b = 12/4*b + 85
.
Subtract 12/4*b from both sides
.
3/4b = 85
.
b = (4*85)/3 = 340/3 = 113 1/3 = 113.3333333...
.
So at the time the car broke down, the bus had traveled
d = 3*113.333 = 340 km
.
340 km is 85 less than the car travelled, so the car had travelled 340+85 = 425.
.
The bus will cover the 85 km in t time
d = rt
t = d/r = 85/113.333
t = 0.75 hr = 45 min
.
So the bus will pass the car at 12:45 pm.
.
We can check this solution by checking to see if the bus can travel 425 km in 3.75 hr.
3.75*113.333333333333 = 425.
.
Done.

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By theorem we know that to maximize the area of a quadrilateral we define a square.
Given 8 ft of fencing is to enclose 2 sides, then each side will be 4 ft.
The volume will be l*w*h = 4 * 4 * 3 = 48 cu ft.
You can experiment with other dimensions to ensure this is the maximum.

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Given
2x + 3y = 12
2x - y = 4
.
Change both in y = mx+b format
.
2x + 3y = 12
3y = -2x + 12
y = -(2/3)X + 4
.
2x -y = 4
-y = -2x + 4
y = 2x - 4
.

.
We can see they intersect at (3,2), so they are consistent and independent.
.
We can check our work using x=3 and y=2.
.
Do these equations produce the correct answer when you substitute for x & y?
2x + 3y = 12
2(3) + 3(2) = 6 + 6 = 12
Check.
.
2x - y = 4
2(3) - 2 = 6 -2 = 4
Check.

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Given the setup:

.
We can multiply through:

.
Collecting terms

.
Subtracting 27 from both sides

.
Can we factor 32 into two terms that are 4 apart? 4 * 8 = 32, so we have:

.
So we have to solutions: y = -8 and y = 4.
.
Checking , we have:



Check.
.
Checking for , we have:



Check.
Done.

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The Pythagorean Theorem is fundamental to the solution:
, where
c = hypotenuse
a = longer leg
b = other leg
.
The problem setup gives you 3 important relationships.

and

.
So,

.
Of course,

.
Going back to the Pythagorean Theorem:

.
By substitution, we can cast the entire problem in terms of b, the shorter leg.

.

.
Subtract b^2 from both sides

.
Subtract 32b from both sides

.
Subtract 256 from both sides

so,

.
Can we factor 60 such that the factors are 4 apart? 2*30? no 3*20? no. 4*15? no. 5*12? no. 6*10? yes!

.
Thus we have two candidate values for b: 10 and -6. Since the leg of a triangle cannot be negative, the proposed solution is b = 10.
.
Substituting back into what we were told in the problem setup:


.
We always check our proposed solution. Here we just use the Pythagorean Theorem again.




That checks, so we have the correct answer.
Done.

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The club has x members now, so the average cost = 200,000/x.
If they add 5 members, the average cost = 200,000/(x+5) - 2000.
So, we can express this as:

.

.
Multiply through by to eliminate the denominators.
.

.

.

.
Divide by 2000

.
The 100x cancel, so collect terms and rearrange

.
Multiply by -1

.
Factor

.
So x can = either -25 or 20. -25 members does not make sense, so x = 20 and x+5 = 25.
.
Checking our answer.
average cost with 20 members

.
average cost with 25 members

.
So the difference in average cost is 2000, as the problem requires.

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I will solve the candy problem. Start with what you know.
x = candy @ $1.80/lb
y = candy @ $2.40/lb
.
She wants to produce 48 lbs valued at $2.00/lb.
.
So we have x(1.80) + y(2.40) = 48(2.00) = 96.
.
But with only one equation, having two unknowns is not solvable. But since we know the total, we define x in terms of y or y in terms of x.
x lb = 48 lb - y lb
y lb = 48 lb = x lb
.
Rewriting the equation we have:
1.8x + 2.4y = 96
.
Substitute y = 48-x
1.8x + 2.4(48-x) = 96
.
Multiply thru
1.8x + 115.2 - 2.4x = 96
.
Collect terms
-0.6x + 115.2 = 96
.
Subtract 115.2 from both sides
-0.6x = 96 - 115.2 = -19.2
.
Multiply both sides by -10
6x = 192
.
Divide both sides by 6
x = 32
.
Looking back to our setup, we know that y = 48-x = 48-32 = 16.
So we have 32 lb @ $1.80/lb = $57.60
And we have 16 lb @ $2.40/lb = $38.40
Thus we have a total of 48 lb and a total cost of $57.60 + $38.40 = $96.00.
96/48 = $2/lb.
48 lb @ $2/lb = $96.
So our solution checks.
.

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P = 240 = 2L + 2W
A = L*W
.
By theorem (or from calculus), we know the maximum area is a square, so: L=W.
.
P = 2L + 2W = 4L = 240
L = 60
.
So a 60 by 60 room is 3600 sq ft in area.
.
We can test the area by testing another room with dimensions close to 60 by 60 and P = 240.
.
Let's try 59 by 61. P = 2(59)+2(61) = 118+122 = 240. OK.
A = 59 * 61 = 3599, which is less than 3600.
.
Let's try 58 by 62. P = 2(58) + 2(62) = 116 + 124 = 240. OK
A = 58 * 63 = 3596, which is less than 3600 and less than 3599.
.
Let's try 57 by 63. P = 240. OK
A = 57 * 63 = 3591, which is less than 3600 and less than 3599 and less than 3596. The area is decreasing as the values get further apart.
.
How about 1 by 119? P = 2(1) + 2(119) = 2 + 238 = 240.
A = 1*119 = 119, which is way less...
.
Another way to look at this problem is the define the perimeter = 2C, where C = L+W.
.
That way L = C-W, so the area will be (C-W) * W = CW - W^2.
.
The maximum value of CW - W^2 is C^2/4, which occurs at W = C/2.
.
So, if W = C/2, and C=L+W, then L=C/2, so we have a square.

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What is the area of the island in square meters? Call it A.
What is its average elevation in meters? Call it E.
A*E = volume of the island above sea level in cubic meters.
Divide by 300,000 cubic meters/second to yield the number of seconds.
Convert the seconds to minutes or to hours or to days or to years.

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A sketch of the line would help a lot!
.
Start of Line: __________ Anil __________ Darsh ___________:Front of line
.
The number of people from the start of the line to Darsh is said to be: x
.
The number of people from Anil to Darsh is given: y
.
The number of people from Anil to the front of the line is given as: z
.
So we can separate the line into 3 segments:
.
Start of line to Anil = x - y
.
From Anil to Darsh = y
.
From Darsh to Front of the line = z - y
.
So we add the three segments and get: x-y + y + z-y = x + z - y
.
We can check with a concrete example. Assume 10 people in the line behind Anil, 5 between Anil and Darsh, and 5 from Darsh to the front of the line. So there are 22 people in the line: 10+5+5+Anil+Darsh.
.
||||||||| Anil ||||| Darsh |||||, where each vertical bar represents a person.
.
x = 10 + Anil + 5 = 16
.
From Anil to the front of the line is z:
.
z = 5 + Darsh + 5 = 11
.
We know that y = 5 because that is the number of people we put between Anil and Darsh in our setup.
.
So, we have 16 + 11 - 5 = 22, which is the number of people we know are in the line: x + z - y.

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When you factor you find that it =

.
To check, multiply through:
=

.
We observe that:
is equivalent to:

.
So, equals:

.
Note: I had to assume the original equation is as I wrote it, which has a in the middle, not .

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In the setup to the problem you are told:

.
So we can cross multiply

.
Multiplying through the numerators
.

.
Subtracting 48 from both sides
2x = 3y
.
So, by inspection, the ratio x:y = 2:3 using their coefficients.

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You can start this problem thinking in terms of
K amount of Kenyan coffee @ $10 per kilogram
+ V amount of Venezuelan coffee @ $12 per kilogram
= 360 kg @ $11/kg
.
The problem asks how many kg of Kenyan coffee in the mixture.
Well, it can be said the amount is 360 - V.
That is, the amount of Kenyan = known total - unknown amount of Venezuelan coffee.
.
At first, this may not seem like much help. But in fact, that is one of the key ways to reduce the number of unknown variables in algebra word problems.
.
K = 360 - V
V = V
.
Substituting in the total cost equation:
.
10*(360-V) + 12*(V) = 360*11
.
Multiplying through:
3600-10V + 12V = 3960
.
Collecting like terms
3600 + 2V = 3960
.
Subtracting 3600 from both sides
2V = 360
.
Dividing both sides by 2
V = 180
.
So, V = 180, which means K = 360-180 = 180.
.
We have equal amount of K and V in the total, which makes sense because we want the mixture to cost $11/kg and the two components cost $10/kg and $12/kg. $11/kg is the average.
.
We can check our work by substitution back into the equation to see if our result produces the given result.
180*10 + 180*12 = 36*11
1800 + 2160 = 3960 = 36*11
.
Done.

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The jet is going 1800 miles/hr. You want to know how much time it takes to go 15 miles.
Looking at the units, you can see that you need to divide 15 by 1800 to end up with hours.
.
15/1800 = 0.008333 hr
.
Since 1 hr = 60 minutes, we can multiply by 60 to get how many minutes to fly 15 miles.
.
0.4999 min (or about 1/2 min)
.
Multiplying by 60 seconds, we will get how many seconds to fly 15 miles.
.
29.9999 seconds (or about 1/2 min).
.
Done.

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We always want to use the fewest unknowns, which usually means defining as many variables in terms of another variable as possible. Instead of defining two consecutive integers as x & y, define them as x and x+1.
.
We are told the sum (x + x+1) is 29 less than the product (x * (x+1)).
.
x * (x+1) = x + (x+1) + 29
.
x^2 + x = 2x + 30
.
Subtracting 2x from both sides
x^2 - x = 30
Subtracting 30 from both sides
x^2 - x - 30 = 0
.
We need to determine the factors of 30 that are 1 apart to solve this without using the quadratic equation. We see 5 * 6 = 30 probably will work.
.
(x - 6)(x + 5) = 0
.
That gives us two values of 'x' to test: x=6 and x=-5.
.
Testing x=6.
x = 6
x+1 = 7
Product = 6*7 = 42
Sum = 6+7 = 13
Does 42 = 13 + 29?
Yes. So this solution works.
.
Testing x=-5.
x = -5
x+1 =-4
Product = -5*-4 = 20
Sum = -5 + -4 = -9
Are these 29 apart?
Yes, so this checks, too.
.
Our solutions are:
5, 6
and
-5,-4
.
Done.

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For most problems like this, your best interest is served by using as few variables as possible. For example, the setup says "3 consecutive EVEN integers," which is a temptation to call x, y, and z.
But from what we know about consecutive numbers, they're just 1 apart. Consecutive EVEN integers are 2 apart. So we can define the 3 unknown numbers in terms of x.
a = x
b = x+2
c = x+4
.
We're all told the square of the largest number is 12 more than the sum of the squares of the other two numbers. We can write this fact as: . Combining we have:
.

.
Resolving the squares.



.
Collecting like terms
Subtract x^2 from both sides

Subtracting 8x from both sides

Subtracting 16 from both sides


.
We can factor an 'x' from the left-hand.

.
So we have two possible solutions:
x=0
x=4
.
Testing x=0.
0^2 = 0
2^2 = 4
4^2 = 16
Is 4^2 equal to the sum of the other two squares + 12?
Yes.
.
Testing x=4.
4^2 = 16
6^2 = 36
8^2 = 64
Is 8^2 equal to the sum of the other two squares + 12?
16+36=52, with is 12 less than 64.
So this works, too.
.
The EVEN consecutive digits are:
0, 2 & 4
and
4, 6 & 8
.
Done.

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We know the area of a rectangle is width times the length.

.
We are told A = 80
.
We also are told L is 2 feet longer than the width.

or

You pick.
.
Substituting what we know:


Multiplying through the right-hand:

Equal things are equal, so we can rewrite:

Subracting 80 from both sides

.
We now consider whether it can be factored. The factors of 80 are 2*40, 4*20, 5*18, 8*10...
We have to have different signs to get them to multiply to -80.
And we need them to be 2 apart to have the solve the L term. 8 & 10 is the way to go.
.

.
So we have two candidate solutions: L=10 and L=-8.
Since we are talking about the area of a rectangle, a length of -8 is not reasonable. So we'll test 10.
.
Looking back at the equations we established:

.
So we propose the L=10 and W=8.
.
Checking the Area, we test:

.
That checks. So we're done.

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Given the two points, you can draw a line. But any line will be perpendicular to any other line that has a slope = negative inverse of the first line; and it will be parallel to any line that has the same slope. So you need to give us more information in your problem statement.
.
With two points, you can determine the slope = change in y divided by change in x, which is also known as the rise over the run.
.
(x1,y1) = (3,1)
(x2,y2) = (6,3)
.
m = slope = (y2 - y1) / (x2 - x1)
m = (3-1) / (6-3)
m = 2/3
.
The slope-intercept formula for a line is: y = mx + b.
We only have a value for m, so we cannot determine the y-intercept (that is, where it crosses the y-axis when x=0).
.
We can 'assume' b = 0 to be able to draw a line to show you what the line looks like.
.
y = 2/3x + 0.
Graph:

.
As I mentioned above, a line with negative inverse slope will be perpendicular.
Given y = 2/3x, then the perpendicular line will be y = -3/2x.
Graph:

.
A parallel line will have the same slope, but a different y-intercept. So, we'll 'assume' the second intercept is 3 so that the graph will show the parallel.
y = 2/3x + 3
Graph:

.
To finish this lesson, we can plot a line parallel to the first perpendicular. Let's assume a y-intercept of -4, so y = -3/2x - 4.
Graph:

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This problem hinges on your remembering that the sum of the angles in a triangle is always 180.
.
In this case we have 3 unknown angles: a, b, and c.
However, we are told they have a specific relationship: 1 to 3 to 5. So the second angle is 3 times the first and the third angle is 5 times the first.
.
a = a
b = 3a
c = 5a
.
a + b + c = 180
Substituting,
a + 3a + 5a = 180
9a = 180
Divide both sides by 9
a = 20
.
Substituting back into the formulas we defined.
a = 20
b = 3a = 3(20) = 60
c = 5a = 5(20) = 100
.
Check the work.
Does 20 + 60 + 100 = 180?
Yes.
.
Done.

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A linear equation in standard form is: Ax + By = C
The A, B, and C are numeric values. Traditionally, A is a + number.
.
The other common form is slope-intercept form: y = mx + b
I use slope-intercept to solve most problems.
.
You will need to be more specific about what your problem with positive and negative numbers.

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With the substitution method, you isolate one unknown and then substitute it into the other equation to find a solution. The equations are:
EQ1: 4m + n = 23
EQ2: m - 4n= 10
.
Working with EQ2.
Add 4n to both sides.
m = 4n + 10
.
Now we can substitute it back into EQ1.
4(4n+10) + n = 23
Multiply through the left-hand pararentheses
16n + 40 + n = 23
Collect like terms
17n + 40 = 23
Subtract 40 from both sides
17n = - 17
Divide both sides by 17
n = -1
.
Now we take that value and substitute back into EQ2.
m = 4(-1) + 10
m = 6
.
So our proposed solution is:
n = -1
m = 6
.
We can check by putting these values back into the equations.
4(6) + (-1) = 24 - 1 = 23. Check.
6 - 4(-1) = 6 + 4 = 10. Check.

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The perimeter of a triangle = a + b + c.
where
a = first side
b = second side
c = third side
.
We are told P = 110.
.
To solve word problems, it usually is best to define these variables in terms of one another, which the problem tells you how to do so. For example,
.
a = a
b = 2a
c = b+30 = 2a+30
.
110 = a + b + c
Substituting for 'b' and 'c' defined in terms of 'a'.
110 = a + 2a + 2a+30
Collecting terms
110 = 5a + 30
Subtract 30 from both sides
80 = 5a
Equals are equal, so
5a = 80
Divide both sides by 5
a = 16
.
Substituting back into the formulas we set up above.
b = 2a = 2(16) = 32
c = b + 30 = 32 + 30 = 62
.
Checking our work:
a + b + c = 16 + 32 + 62 = 110
Done.

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Given the perimeter of any triangle is a + b + c.
Given your triangle is an isosceles, it's perimeter will be 2*sides + base.
So, given P = 54, then s cannot be 30.

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To say that a value doubles every year means that it will go up in value by 2, 4, 8, 16, 32, etc. Notice that those are 2 raised to the 't', where 't' = years numbered 1, 2, 3, 4, 5, etc.
.
f(x) = future value
P = present value = $40
t = year - 2001, which means t will take on values 0, 1, 2, 3, 4, 5, etc.
Note that an exponent of 0 means the value is 1.
.
f(t) = P*2^t
.
year...t...f(t)
2001...0...40
2002...1...80
2003...2...160
2004...3...320
2005...4...640
2006...5...1280
2007...6...2560
2008...7...5120
.
So the card first reaches $5000 value in 2008.

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Convert Jake's total time into minutes: 2*60 + 22 = 120 + 22 = 144
.
We are told Jill can do the job 27 minutes quicker, so her speed is 144-27 = 117.

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(1) Since I don't know the dimensions of the rectangle, I will assume it is a square. Given the area = 4850, then the sides are the square root of 4850, which = 69.64. Since the perimeter of a square is simply 4s, s = length of a side, then YES your 280 ft of fencing can do the job.
.
(2) This one requires you to set up the given equation and then substitute the values of t, where

year   t
1987   0
1988   1
1989   2
1990   3
etc.

.
The equation is

.
Using Excel, I created the following table for the years 1987-2007. Notice that the sales go up for a while and then start back down.

yr	t	Sales
1987	0	0.500
1988	1	1.265
1989	2	1.960
1990	3	2.585
1991	4	3.140
1992	5	3.625
1993	6	4.040
1994	7	4.385
1995	8	4.660
1996	9	4.865
1997	10	5.000
1998	11	5.065
1999	12	5.060
2000	13	4.985
2001	14	4.840
2002	15	4.625
2003	16	4.340
2004	17	3.985
2005	18	3.560
2006	19	3.065
2007	20	2.500

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.

.
The division can be replaced by multiplication by the inverse.
So instead of:

.
we can substitute:
.

.
The 'xy' terms will cancel, so we have:
.

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First, get the equation into y = mx + b format.
.
6y + 9x + 4 = 16 + 9x
Subtract 4 from both sides
6y + 9x = 12 + 9x = 9x + 12
Subtract 9x from both sides
6y = 12
Divide both sides by 6
y = 12/6 = 2
.
Note that the 'x' term seems to have 'disappeared.' Of course, the x-axis exists, so it cannot 'disappear.' So logically it has to be there, but with a coefficient = 0.
.
y = 0m + 2
.
So, the slope = m = 0, which means it is a horizontal line.
.
When x = 0, y = 2. (Of course, when x=anything, y=2.)
.
Graph:

.
Done.

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CORRECTION! I read the problem too fast the first time. (sorry)
.
The perimeter of a rectangle = 2L + 2W, where L=length and W=width.
We are told L is 50 ft less than HALF the W, so L = .5W -50.
We are told the perimeter = 800.
.
Substituting we have:
800 = 2L + 2W
800 = 2(.5W-50) + 2W
Multiplying the factor
800 = W - 100 + 2W
Collecting like terms
800 = 3W - 100
Adding 100 to both sides
900 = 3W
3W = 900
Dividing both sides by 3
W = 900/3 = 300
.
Looking back, we know L = .5W-50, so L = 150 - 50 = 100.
.
Check the perimeter.
2W + 2L = 2(300) + 2(100) = 800.
This is what it's supposed to be.
Done.