10 Practice Problems with Step-by-Step Solutions
Test your knowledge and solve algebra questions involving linear equations, simplifying algebraic and absolute value expressions, finding the distance between two points, determining x-intercepts, evaluating functions, and calculating the slope of a line.
Solution:
Expand the terms on both sides:
\[ -15x - 10 - x + 3 = -16x - 20 + 13 \]Group like terms:
\[ -16x - 7 = -16x - 7 \]Add \( 16x + 7 \) to both sides:
\[ -16x - 7 + 16x + 7 = -16x - 7 + 16x + 7 \]Simplify to obtain:
\[ 0 = 0 \]Conclusion: The statement is true for all values of \( x \), therefore all real numbers are solutions. This equation is an identity.
Solution:
Expand the factors:
\[ = 2a - 6 + 4b - 2a + 2b + 6 + 5 \]Group and combine like terms:
\[ = (2a - 2a) + (4b + 2b) + (-6 + 6 + 5) \]Answer:
\[ = 6b + 5 \]If \( x \lt 2 \), simplify the expression:
\[ |x - 2| - 4|-6| \]Solution:
If \( x \lt 2 \), then \( x - 2 \lt 0 \).
According to the definition of absolute value, since \( (x - 2) \lt 0 \):
\[ |x - 2| = -(x - 2) \]And we know that \( |-6| = 6 \).
Substitute these back into the given expression:
\[ -(x - 2) - 4(6) = -x + 2 - 24 \]Group like terms:
Answer: \( -x - 22 \)
Solution:
Add 3 to both sides to isolate the absolute value:
\[ |-2x + 2| = 0 \]According to the definition of absolute value, if the absolute value equals 0, the expression inside must be 0:
\[ -2x + 2 = 0 \]Solve for \( x \):
Answer: \( x = 1 \)
Find the distance between the points \( (-4, -5) \) and \( (-1, -1) \).
Solution:
According to the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), substitute the coordinates:
\[ d = \sqrt{(-1 - (-4))^2 + (-1 - (-5))^2} \]Simplify the terms inside the parentheses:
\[ d = \sqrt{(-1 + 4)^2 + (-1 + 5)^2} \] \[ d = \sqrt{(3)^2 + (4)^2} \] \[ d = \sqrt{9 + 16} \]Answer: \( d = \sqrt{25} = 5 \)
Find the x-intercept of the graph of the equation: \( 2x - 4y = 9 \)
Solution:
To find the x-intercept, set \( y = 0 \) in the equation and solve for \( x \):
\[ 2x - 4(0) = 9 \] \[ 2x = 9 \]Answer: \( x = \frac{9}{2} \). The x-intercept is at the point \( (\frac{9}{2}, 0) \).
Find the slope of the line passing through the points \( (-1, -1) \) and \( (2, 2) \).
Solution:
Use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\[ m = \frac{2 - (-1)}{2 - (-1)} \] \[ m = \frac{2 + 1}{2 + 1} \]Answer: \( m = \frac{3}{3} = 1 \)
Find the slope of the line: \( 5x - 10y = 7 \)
Solution:
Rewrite the equation in slope-intercept form (\( y = mx + b \)):
\[ -10y = -5x + 7 \]Divide all terms by \( -10 \):
\[ \frac{-10y}{-10} = \frac{-5x}{-10} + \frac{7}{-10} \] \[ y = \frac{1}{2}x - \frac{7}{10} \]Answer: The slope is the coefficient of \( x \), which is \( \frac{1}{2} \).
Find the equation of the line that passes through the points \( (-1, -1) \) and \( (-1, 2) \).
Solution:
First, find the slope \( m \):
\[ m = \frac{2 - (-1)}{-1 - (-1)} = \frac{3}{0} \]Division by zero is undefined, which means the line is vertical and its equation takes the form \( x = \text{constant} \).
Answer: Since both points have an x-coordinate of \( -1 \), the equation of the line is \( x = -1 \).
Evaluate \( f(2) - f(1) \) given \( f(x) = 6x + 1 \).
Solution:
\( f(2) \) is found by substituting \( x = 2 \) into \( f(x) \), and \( f(1) \) is found by substituting \( x = 1 \).
\[ f(2) - f(1) = (6(2) + 1) - (6(1) + 1) \] \[ f(2) - f(1) = (12 + 1) - (6 + 1) \] \[ f(2) - f(1) = 13 - 7 \]Answer: \( 6 \)