Algebra 1 Practice Worksheets

Algebra 1 Practice Worksheets are ideal resources for students who wish to enhance their understanding and mastery of algebraic concepts. These worksheets provide carefully crafted exercises and problems, allowing students to practice and reinforce their skills in solving equations, working with variables, and manipulating algebraic expressions. With clear explanations and comprehensive examples, these worksheets serve as invaluable tools for intermediate level mathematics learners seeking to solidify their grasp of algebraic principles.

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  Two-Step Equations Worksheet

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What is the value of x in the equation 3x + 7 = 22?

The value of x in the equation 3x + 7 = 22 is x = 5.

Solve the inequality 4x - 9 > 7.

To solve the inequality 4x - 9 > 7, we need to isolate x. Adding 9 to both sides gives 4x > 16. Then, dividing by 4 on both sides gives x > 4. Therefore, the solution to the inequality is x > 4.

Factor the expression 4x^2 - 9.

The expression 4x^2 - 9 can be factored into the difference of squares as (2x + 3)(2x - 3).

Simplify the expression (3x^2 + 5x)(2x - 8).

To simplify the expression (3x^2 + 5x)(2x - 8), you use the distributive property to multiply each term in the first expression by each term in the second expression. This results in 6x^3 - 24x^2 + 10x^2 - 40x, which simplifies to 6x^3 - 14x^2 - 40x.

Solve the system of equations: 2x + y = 10 and x - y = 4.

By solving the system of equations using substitution or elimination method, we find x = 7 and y = 3. Thus, the solution to the system of equations is x = 7 and y = 3.

Graph the equation y = 2x + 3.

The graph of the equation y = 2x + 3 is a straight line with a slope of 2 and y-intercept of 3. To graph it, start at the point (0,3) and move up 2 units and right 1 unit to find another point on the line. Continue this process to plot more points and connect them to form a straight line. This line will have a positive slope and extend infinitely in both directions.

Find the slope of the line passing through the points (4, 2) and (-3, 5).

To find the slope of the line passing through the points (4, 2) and (-3, 5), we use the formula for slope: (y2 - y1) / (x2 - x1). Substituting the coordinates, we get (5 - 2) / (-3 - 4) which simplifies to 3 / -7. Therefore, the slope of the line passing through the points is -3/7.

Solve the quadratic equation 2x^2 + 5x - 3 = 0 using the quadratic formula.

The quadratic formula is x = (-b ± ?(b^2 - 4ac)) / 2a. Plugging in the values from the equation 2x^2 + 5x - 3 = 0, we have a = 2, b = 5, and c = -3. Substituting these values into the formula, we get x = (-5 ± ?(5^2 - 4*2*-3)) / 2*2 = (-5 ± ?(25 + 24)) / 4 = (-5 ± ?49) / 4 = (-5 ± 7) / 4. This gives us two solutions: x = (2/4) = 0.5 and x = (-12/4) = -3. Thus, the solutions to the equation 2x^2 + 5x - 3 = 0 are x = 0.5 and x = -3.

Simplify the expression (x^2 + 3x - 5) / (x - 2).

The simplified expression is x + 5.

Find the domain of the function f(x) = sqrt(x + 4).

To find the domain of the function f(x) = sqrt(x + 4), we look for values of x that make the expression inside the square root non-negative. In this case, x + 4 must be greater than or equal to 0 for the square root to be defined. Solving x + 4 ? 0 gives x ? -4. Therefore, the domain of the function f(x) = sqrt(x + 4) is all real numbers greater than or equal to -4, or in interval notation, it is [-4,?).