Solutions to Algebra Problems
Detailed step-by-step solutions to common algebra problems are presented below to help students understand each concept clearly.
Problem 1: Solving a Linear Equation
Given the equation:
\[ 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13 \]Multiply the factors:
\[ -15x - 10 - x + 3 = -16x - 20 + 13 \]Group like terms:
\[ -16x - 7 = -16x - 7 \]Since both sides of the equation are identical, subtracting the same terms from both sides leads to:
\[ 0 = 0 \]This statement is true for all real values of \(x\). Therefore, the solution set is all real numbers.
Problem 2: Simplifying an Algebraic Expression
Given the expression:
\[ 2(a - 3) + 4b - 2(a - b - 3) + 5 \]Expand each term:
\[ 2a - 6 + 4b - 2a + 2b + 6 + 5 \]Combine like terms:
\[ 6b + 5 \]Final simplified expression: \(6b + 5\)
Problem 3: Simplifying an Expression with Absolute Value
Given the expression:
\[ |x - 2| - 4|-6| \]If \(x < 2\), then \(x - 2 < 0\) and: \[ |x - 2| = -(x - 2) \]
Also, since absolute value is always non-negative: \[ |-6| = 6 \]
Substitute and simplify:
\[ -(x - 2) - 4(6) = -x - 22 \]Problem 4: Distance Between Two Points
Find the distance between the points \((-4,-5)\) and \((-1,-1)\).
Distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Substitute the values:
\[ d = \sqrt{(-1 + 4)^2 + (-1 + 5)^2} = \sqrt{9 + 16} = 5 \]Problem 5: Finding the x-Intercept
Given the equation:
\[ 2x - 4y = 9 \]To find the x-intercept, set \(y = 0\) and solve for \(x\):
\[ 2x = 9 \quad \Rightarrow \quad x = \frac{9}{2} \]The x-intercept is the point: \[ \left(\frac{9}{2}, 0\right) \]
Problem 6: Evaluating a Function
Given the function:
\[ f(x) = 6x + 1 \]Compute \(f(2) - f(1)\):
\[ (6 \cdot 2 + 1) - (6 \cdot 1 + 1) = 13 - 7 = 6 \]Problem 7: Finding the Slope of a Line
Given the points \((-1,-1)\) and \((2,2)\), the slope is:
\[ m = \frac{2 - (-1)}{2 - (-1)} = 1 \]Problem 8: Slope-Intercept Form
Given the equation:
\[ 5x - 5y = 7 \]Rewrite in slope-intercept form:
\[ y = x - \frac{7}{5} \]The slope \(m\) is the coefficient of \(x\), so: \[ m = 1 \]
Problem 9: Equation of a Vertical Line
The slope between \((-1,-1)\) and \((-1,2)\) is:
\[ m = \frac{3}{0} \]Since the slope is undefined, the line is vertical. Both points have the same x-coordinate, so the equation is:
\[ x = -1 \]Problem 10: Solving an Absolute Value Equation
Given the equation:
\[ |-2x + 2| - 3 = -3 \]Add 3 to both sides:
\[ |-2x + 2| = 0 \]An absolute value equals zero only when its expression equals zero:
\[ -2x + 2 = 0 \quad \Rightarrow \quad x = 1 \]