All ACT Math Resources
Example Questions
Example Question #201 : Equations / Inequalities
Given the follow inequality, which of the following presents a range of possible answers for the inequality: –3 < 3x + 2 ≤ 3.5
(–1, ½)
( –2, 2)
(½, 1)
(–1,1)
(–3, 1/2)
(–1, ½)
If you plug in the outer limits of the given ranges, (–1, ½) is the only combination that fits within the given equation. It is important to remember that "<" means “less than,” and "≤" means “less than or equal.” For example, if you answered (–2,2), plugging in 2 would make the the expression equal 8, which is greater than 3.5. And plugging in –2 for x would make the expression equal –4, which is less than –3, not greater. However, plugging in the correct answer (–1, ½) gives you –1 as your lower limit and 3.5 as your upper limit, which satisfies the equation. Both limits of the data set must satisfy the equation.
Example Question #202 : Equations / Inequalities
If , what is the product of the largest and smallest integers that satisfy the inequality?
0
–5
7
5
–10
0
The inequality in the question possesses an absolute value; therefore, we most solve for the variable being less than positive 6 and greater than negative 6. Let's start with the positive solution.
Add 4 to both sides of the inequality.
Divide both sides of the inequality by 2.
Now, let's solve for the negative solution
Add 4 to both sides of the inequality.
Divide both sides of the inequality by 2.
Using these solutions we can write the following statement:
The smallest integer that satisfies this equation is 0, and the largest is 4. Their product is 0.
Example Question #201 : Equations / Inequalities
Solve 8x2 – 2x – 15 = 0
x = 3/2 or 5/4
x = 3/2 or -5/4
x = -3/2 or 5/4
x = -3/2 or -5/4
x = 3/2 or -5/4
The equation is in standard form, so a = 8, b = -2, and c = -15. We are looking for two factors that multiply to ac or -120 and add to b or -2. The two factors are -12 and 10.
So you get (2x -3)(4x +5) = 0. Set each factor equal to zero and solve.
Example Question #202 : Equations / Inequalities
If (x2 + 2) / 2 = (x2 - 6x - 1) / 5, then what is the value of x?
4
3
2
-3
-2
-2
(x2 + 2) / 2 = (x2 - 6x - 1) / 5. We first cross-multiply to get rid of the denominators on both sides.
5(x2 + 2) = 2(x2 - 6x - 1)
5x2 + 10 = 2x2 - 12x - 2 (Subtract 2x2, and add 12x and 2 to both sides.)
3x2 + 12x + 12 = 0 (Factor out 3 from the left side of the equation.)
3(x2 + 4x + 4) = 0 (Factor the equation, knowing that 2 + 2 = 4 and 2*2 = 4.)
3(x + 2)(x + 2) = 0
x + 2 = 0
x = -2
Example Question #203 : Equations / Inequalities
Which of the following is a factor of the polynomial x2 – 6x + 5?
x + 2
x – 5
x – 8
x – 6
x + 1
x – 5
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.
x2 – 6x + 5 = (x – 1)(x – 5)
Because only (x – 5) is one of the choices listed, we choose it.
Example Question #204 : Equations / Inequalities
7 times a number is 30 less than that same number squared. What is one possible value of the number?
Either:
or:
Example Question #205 : Equations / Inequalities
Which of the following is equivalent to ?
The answer is .
To determine the answer, must be distrbuted,
. After multiplying the terms, the expression simplifies to .
Example Question #206 : Equations / Inequalities
For what value of b is the equation b2 + 6b + 9 = 0 true?
3
0
5
–3
–3
Factoring leads to (b+3)(b+3)=0. Therefore, solving for b leads to -3.
Example Question #207 : Equations / Inequalities
What is the solution to:
2
1
4
0
6
4
First you want to factor the numerator from x2 – 6x + 8 to (x – 4)(x – 2)
Input the denominator (x – 4)(x – 2)/(x – 2) = (x – 4) = 0, so x = 4.
Example Question #208 : Equations / Inequalities
What is the value of where:
The question asks us to find the value of , because it is in a closed equation, we can simply put all of the whole numbers on one side of the equation, and all of the containing numbers on the other side.
We utilize opposite operations to both sides by adding to each side of the equation and get
Next, we subtract from both sides, yielding
Then we divide both sides by to get rid of that on
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