In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.

For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).

Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.

When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.

To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = ⁴/₃c.

Now we can plug into the last equation. We plug in ⁴/₃c for a, 4c for b, and leave c as it is. We must find a common denominator (⁴/₃c – ¹²/₃c + ³/₃c) and add the numerators to find that our equation equals ^–5/₃c.

There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).

If x² + 5x – 24 = 0, 

(x – 3)(x + 8) = 0 or x = –8 or +3. 

y² – 9y + 20 = 0, then 

(y – 5)(y – 4) = 0, or y = +4 or +5. 

y is always greater than x.

We can factor the equation x² + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.

x - 3 = 0

x = 3

We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.

This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.

Quantity A is greater.

We know that Quantity A = y / 3 = 15 /3 = 5.

If we plug in 15 for y, we can solve for x, for Quantity B.

y = x² - 10

y = 15

15 = x² - 10 (Add 10 to both sides.)

25 = x²

x = 5 or -5

Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.

The point of intersection for two lines is the same as the values of x and y that mutually solve each equation.  Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:

3x + 4y = 6

15x - 4y = 3

18x = 9; x = 0.5

You can now plug x into the first equation:

3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125

Therefore, our point of intersection is (0.5, 1.125)

The cars have a distance from each other of 25 + 120t miles, where t is the number of hours, 25 is their initial distance and 120 is 50 + 70, or their combined speeds.  Solve this equation for 400:

25 + 120t = 400; 120t = 375; t = 3.125

However, the question asked for minutes, so we must multiply this by 60:

3.125 * 60 = 187.5 minutes.

Start by setting up an equation using Quantity A and Quantity B. In other words, you can solve an inequality where Quantity A > Quantity B. You would have one of four outcomes:

1. Quantity A = Quantity B: the two quantities are equal.
2. The inequality is always satisfied: Quantity A is always larger. 
3. The inequality is never satisfied (but the two are unequal): Quantity B is always larger. 
4. The inequality is not always correct or incorrect: the relationship cannot be determined. 

So solve:

 –5x + 4 > 8 – 2x (Quantity A > Quantity B)

    +2x      +2x

  –3x + 4 > 8

    –4      –4

  –3x > 4 or x < –4/3

*remember to switch the direction of the inequality when you divide by a negative number

As the inequality [x < –4/3] is always false for [x>0], Quantity B is always larger.

Set up a system of linear equations based on our data:

40,000P + 45,000A = 40,415,000

P/A = 4/3

To make things easiest, solve the second equation for A in terms of P:

A = (3/4) P

Replace this value into the first equation:

40,000P + 45,000 * (3/4)P = 40,415,000

Simplify:

40,000P + 33,750P = 40,415,000

73,750P = 40,415,000

P = 548 (The number of professors)