Algebra Practice Quiz: Master Key Concepts

Subtract 5 from both sides of the equation to find x = 7. This linear equation demonstrates basic algebra skills.

In slope-intercept form, y = mx + b, the coefficient m is the slope. Therefore, the slope is 3.

The greatest common factor of 4x and 12 is 4, so factoring out 4 yields 4(x + 3). The other options do not factor using the greatest common factor.

2 raised to the 3rd power means 2 multiplied by itself three times: 2 x 2 x 2 = 8. This basic exponentiation demonstrates the concept of repeated multiplication.

Divide both sides by 5 to isolate x, resulting in x = 20/5 = 4. This problem reinforces the concept of solving simple linear equations.

The quadratic factors to (x - 2)(x - 3) = 0, so the solutions are x = 2 and x = 3. Factoring is a common method to solve quadratic equations.

Factor the numerator as (x - 3)(x + 3) and cancel the common term (x + 3), yielding x - 3 provided that x ≠ -3. This demonstrates the simplification of rational expressions.

Distribute 2 to get 2x - 6 = 4x + 1, then rearrange the equation to obtain -7 = 2x, so x = -7/2. The solution is obtained by isolating the variable x.

The vertex of a parabola in the form y = ax^2 + bx + c can be found using x = -b/(2a). Here, x = 2 and substituting back gives y = -3, so the vertex is (2, -3).

Taking the square root of both sides leads to x + 2 = 4 or x + 2 = -4, so x = 2 or x = -6. Both solutions arise from considering the positive and negative square roots.

The quadratic factors as (3x + 2)(x + 3) because multiplying these factors produces 3x^2 + 11x + 6. This factorization is confirmed by expansion.

Since 50 can be factored into 25 and 2, the square root becomes √25 * √2 which simplifies to 5√2. This is the simplest radical form, demonstrating proper radical simplification.

Cross multiplying the equation gives 1 = 3(x - 2). Solving for x results in x = 7/3, which is the correct solution.

By adding 5 to both sides, the inequality becomes 2x > 8. Dividing both sides by 2 yields x > 4, so the solution includes all numbers greater than 4.

To find the inverse, switch x and y in the equation y = 2x + 3 and solve for y, which gives f❻¹(x) = (x - 3)/2. This process reverses the operations of the function.

Using the logarithmic rule, the equation becomes log₂[(x - 1)(x + 3)] = 3, which implies (x - 1)(x + 3) = 8. Solving the resulting quadratic gives x = -1 ± 2√3, but only 2√3 - 1 meets the domain restrictions, making it the correct solution.

After isolating one square root and squaring both sides, the resulting equation leads to a quadratic with a negative discriminant, indicating no real solutions. Evaluating the original equation confirms that the left side always exceeds 1 within the domain.

The square root requires that x + 4 be nonnegative, so x must be at least -4, and the denominator cannot be zero, which means x ≠ 2. Together, these conditions establish the domain as x ≥ -4 with x ≠ 2.

Setting 2x - 3 equal to x² results in the equation x² - 2x + 3 = 0, which has a negative discriminant. This means the system has no real solution, so the graphs do not intersect in the real plane.

Substitute x = 3 and x = -2 into the function to obtain f(3) = 5/7 and f(-2) = -5/2. Their sum, 5/7 - 5/2, simplifies to -25/14 when expressed with a common denominator.