Black Card Revoked Practice Quiz

To isolate x, add 7 to both sides to get 3x = 18, then divide by 3 to obtain x = 6. This straightforward process demonstrates the basic method of solving linear equations.

Distribute 2 over (3x + 4) to get 6x + 8, and then subtract 5 to obtain 6x + 3. This problem reinforces distribution and combining like terms.

Using the exponent rule, 2^3 * 2^2 = 2^(3+2) = 2^5, which equals 32. This demonstrates the property of multiplying powers with the same base.

The greatest common factor of 6 and 9 is 3. Factoring 3 out of the expression gives 3(2x + 3), which is the simplest form.

Recognize that x^2 - 9 is a difference of squares and factors as (x - 3)(x + 3). Setting each factor equal to zero gives the solutions x = 3 and x = -3.

Expanding 2(x - 3) gives 2x - 6, and setting the equation 2x - 6 = 4x - 10 leads to 2x = 4, so x = 2. This problem emphasizes rearranging equations to isolate the variable.

The quadratic factors as (x + 2)(x + 3) = 0, which results in the solutions x = -2 or x = -3. Factoring is an efficient method for solving such quadratic equations.

Substitute x = 5 into the function to get f(5) = 2(5) - 3 = 10 - 3 = 7. This question reinforces evaluation of a linear function.

Break 50 into 25 à - 2, and since √25 = 5, the expression simplifies to 5√2. This process demonstrates the simplification of radical expressions through factoring.

Raise each factor to the power of 2: (x^3)^2 becomes x^6 and (y^2)^2 becomes y^4, resulting in x^6y^4. This exercise reinforces the rules of exponents in algebra.

Multiply both sides by 6 to eliminate the fraction, resulting in 2x - 4 = 12. Solving for x by adding 4 and then dividing by 2 yields x = 8.

The quadratic is already in vertex form, y = a(x - h)^2 + k, where the vertex is (h, k). Here, h = 1 and k = 3, so the vertex is (1, 3).

Find two numbers that multiply to 6 and add to -5; these are -2 and -3. Thus, x^2 - 5x + 6 factors as (x - 2)(x - 3).

Adding the equations eliminates y, yielding 2x = 10 so that x = 5. Substituting back into one of the equations gives y = 2, solving the system.

Setting up the system x + y = 12 and x - y = 4, adding the two equations gives 2x = 16, so x = 8 and then y = 4. This problem demonstrates solving word problems using systems of equations.

Using the quadratic formula with a = 2, b = -3, and c = -2 gives a discriminant of 25. This leads to x = (3 ± 5) / 4, yielding solutions x = 2 and x = -1/2.

Factor the numerator as 2(x^2 - 4) = 2(x - 2)(x + 2) and then cancel the common factor with the denominator to obtain (x - 2)(x + 2) / (2x). The restriction x ≠0 is included because division by zero is undefined.

Recognize that 16 can be written as 2^4. Equate the exponents to get x + 1 = 4, which simplifies to x = 3. This problem illustrates solving exponential equations by matching bases.

First, expand the inequality to get 3 - 2x + 2 < 7, which simplifies to 5 - 2x < 7. After subtracting 5 and dividing by -2 (remembering to flip the inequality sign), the solution is x > -1.

Square both sides of the equation to obtain x + 5 = (x - 1)^2. This simplifies to x^2 - 3x - 4 = 0, which factors as (x - 4)(x + 1) = 0. Testing the potential solutions in the original equation shows that only x = 4 is valid.