Equations / Inequalities - ACT Math

Set each component of this equation equal to zero and solve for d and c.

(x-d)(x+c)=0

x-d=0

d=0

x+c=0

x=-c

Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.

Solving for P will give the answer.

The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.

Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81

Solving for x gives Claire’s age, which can be used to find Nick’s age.

If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.

If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).

The y’s cancel leaving us with an answer of 5.

We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.

Factor the polynomial by choosing values that when FOIL'ed will add to equal the middle coefficient, 3, and multiply to equal the constant, 1.

x_2 – 6_x + 5 = (x – 1)(x – 5)

Because only (x – 5) is one of the choices listed, we choose it.

Begin by multiplying both sides by (x – 1):

x2 – x = x – 1

Solve as a quadratic equation: x2 – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

Generally, quadratic equations have two answers.

First, the equations must be put in standard form: 3x2 + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.

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.

To solve, remove the radical by squaring both sides

(√3x) 2 = 92

3x = 81

x = 81/3 = 27

First, simplify the equation:

√(8y) + 18 = 4

√(8y) = -14

Then square both sides

(√8y) 2 = -142

8y = 196

y = 196/8 = 24.5

140 minutes is 7/3 of an hour. Multiplied by the speed of 78mph, we obtain 182 miles traveled.

Adding 4x to both sides of the equation and dividing by 7 yields a slope of 4/7.

Plugging in **–**1 into the equation for x and solving for c yields **–**2.

For a line to be parallel with another, the slopes must be equal. All of the equations have a slope of 4/3 except 8x +6y=34.

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)

1. Substitute y in the equation for 4.
2. You now have 6 * 4 = 10z + 4
3. Simplify the equation: 24 = 10z + 4
4. Subtract 4 from both sides: 24 – 4 = 10z + 4 – 4
5. You now have 20 = 10z
6. Divde both sides by 10 to solve for z.
7. z = 2.

This goes up at a constant number between values, making it an arthmetic sequence. The first number is 2, with a difference of 3. Plugging this into the arithmetic equation you get An = 2 + 3 (n – 1). Plugging in 20 for n, you get a value of 59.

Looking at the sequence you can see that it doubles each term, making it a geometric sequence. Since it doubles r = 2 and the first term is 5. Plugging this into the geometric equation you get An = 5(2)n–1. Setting n = 6, you get 160 as the 6th term.