All ACT Math Resources
Example Questions
Example Question #211 : Equations / Inequalities
Factor the following equation:
First we factor out an x then we can factor the
Example Question #10 : How To Factor An Equation
Which of the following equations is NOT equivalent to the following equation?
The equation presented in the problem is:
We know that:
Therefore we can see that the answer choice is equivalent to .
is equivalent to . You can see this by first combining like terms on the right side of the equation:
Multiplying everything by , we get back to:
We know from our previous work that this is equivalent to .
is also equivalent since both sides were just multiplied by . Dividing both sides by , we also get back to:
.
We know from our previous work that this is equivalent to .
is also equivalent to since
Only is NOT equivalent to
because
Example Question #11 : How To Factor An Equation
A certain number squared, plus four times itself is equal to zero.
Which of the following could be that number?
The sentence should first be translated into an equation.
We will call the "certain number" .
This gives us .
So solve this we can factor out an to get .
This makes the zeros of the equation evident. We know that when the entire expression will equal zero, making the equation true. We also know that when , the quantity will be , which also satisfies the equation.
Therefore, the two possible solutions are and .
Example Question #11 : How To Factor An Equation
If , and , which of the following is a possible value of ?
-6
-12
-8
-4
-2
-12
The given expression is a quadratic equation; therefore, we can factor the equation
Use the format of the standard quadratic equation:
Since , we know that the quadratic's roots will resemble the following:
We also know that one of those signs has to be negative, since our two last terms multiply to equal the variable , and is negative in our quadratic. Now, we need to find two numbers that when multiplied together equal -24, and equal 10 when they are added together. Let's start by finding the factors of 24. The factors of 24 are 24 and 1, 12 and 2, 8 and 3, 6 and 4. Since one of those factors will be negative in our factored equation, we need to find the two factors whose difference is 10.
This means that the numbers in the factored equation are 12 and -2; thus, we may write the following:
.
By the zero multiplication rule, either portion of that equation can equal 0 for the result to be 0; thus, we have the following two expressions:
Subtract 12 from both sides of the equation:
Calculate the value of the variable in the second equation.
Add 2 to both sides of the equation.
Since we want a negative answer for our variable, the correct answer is:
Example Question #1 : Systems Of Equations
What is the sum of the x, y, and z coordinates of the point that satisfies this system of equations:
2x + 3y + 4z = 20
x = y – z
–3x + 2y + 2z = 23
0
–6
–10
4
4
Multiplying the second equation by 2 and subtracting it from the first gives the equation:
5y + 2z = 20
Multiplying the first equation by 3, multiplying the third equation by 2, and then adding those equations gives:
13y + 16z = 106
Now we have a system of two equations and two unknowns.
Multiply the equation 5y + 2z = 20 by 8, and then subtract this from 13y + 16z = 106
This yields –27y = –54, so y = 2. Then using the equation 5y + 2z = 20, z = 5. Then using any of the original three equations, x = –3.
The point of intersection is (-3, 2, 5), and the sum of these coordinates is 4.
Example Question #2 : Systems Of Equations
The cost of a movie ticket and a candy bar is $5. The cost of two tickets and a candy bar is $8.75. How much is a candy bar?
$2.50
$1.25
$3.75
$7.50
$0.75
$1.25
We start by setting up a system of equations. The price of one ticket and one candy bar is $5, so t+c=5. The price of two tickets and one candy bar is $8.75, so 2t+c=8.75. We can use the first equation to find out that c=5-t. We then substitute that value into the second equation, giving us 2t+(5-t)=8.75. This simplifies to 2t-t+5=8.75, so t+5=8.75, and finally t=3.75. We use the final value for t in the first equation, so 3.75+c=5. We solve for c, and get c=1.25.
Example Question #3 : Systems Of Equations
Jacob is 3 years older than Sarah, and Caroline is twice as old as Sarah. If Caroline is 28 years old, how many years old is Jacob?
One can describe the ages of Jacob, Sarah, and Caroline with the letters J, S, and C, respectively. From the information in the problem, J = S + 3, and C = 2S. Since C = 28, S = 28/2 = 14, and J = 14 + 3 = 17. Jacob is 17 years old.
Example Question #4 : Systems Of Equations
If ab = 24, a + b = 10, and a < b, what is the value of a – b?
–2
4
6
2
–4
–2
Solving the second equation for b and substituting b = 10 – a into the equation ab = 24 gives us
a(10 – a) = 10a – a2 = 24
which can be set up and solved as the following quadratic equation:
a1 – 10a +24 = 0
(a – 6) (a – 4) = 0
a = 6, a = 4
a must be 4 and b must be 6, since a < b; therefore, 4 – 6 = –2.
Example Question #4 : Systems Of Equations
Josh is counting his money. He has only quarters and nickels. He has two more quarters than nickels. He counts his money to discover he has $1.40. How many total coins does he have?
8
7
None of the answers are correct
5
3
8
The general formula for money problems is V1 x N1 + V2 x N2 = $Value where V is the value of the coin and N is the number of coins for each separate type of coin involved. $Value is the total value of all the money when counted. With this problem N = number of nickels and
Q = N + 2 is the number of quarters. Substituting into the general formula we get
0.25(N + 2) + 0.05N = 1.40. Solving for N yields N = 3, therefore Q = 5. So the answer to the question is actually N + Q = 3 + 5 = 8 total coins.
Example Question #1 : Systems Of Equations
A business makes $5 million less the second year than the first year. In the third year the business makes twice as much as in the second year. If the business makes $15 million in the third year how much did it make in the first?
$15 milion
$10 million
None of the other answers
$5 million
$12.5 million
$12.5 million
This problem is probably best solved by developing a series of equations. To relate the second and third years we can set up the equation 15 = 2x where x is the amount of money made in the second year, which is two times less than the amount of money made in the third. By dividing each side by two we see that the business made 7.5 million dollars in the second year. To relate the first and second years we can set up the equation x – 5 = 7.5 where x is the amount of money made in the first year. By adding 5 to both sides of the equation we are able to see that the business made 12.5 million dollars in the first year.