Intermediate Algebra Questions and Answers
Lines, Equations, Systems, Expressions, Factoring

Problem: Write \( 1.5 \times 10^{-5} \) in standard form.

Solution:

\[ 1.5 \times 10^{-5} = \dfrac{1.5}{10^5} = \dfrac{1.5}{100000} = 0.000015 \]

Problem: Evaluate: \( 30 - |-x + 6| \) for \( x = 10 \).

Solution:

Substitute \( x = 10 \) into the expression: \[ 30 - |-(10) + 6| = 30 - |-10 + 6| = 30 - |-4| = 30 - 4 = 26 \]

Problem: Evaluate: \( 2xy^3 + x - 2y \) for \( x = 2 \) and \( y = -2 \).

Solution:

Substitute \( x = 2 \) and \( y = -2 \): \[ 2(2)(-2)^3 + 2 - 2(-2) = 4(-8) + 2 + 4 = -32 + 6 = -26 \]

Problem: What is the slope of the line perpendicular to the line \( y = -4 \)?

Solution:

The equation \( y = -4 \) is a horizontal line. A line perpendicular to a horizontal line is a vertical line, and its slope is undefined.

Problem: Write an equation of the line with slope 2 and x-intercept \( (-4, 0) \).

Solution:

Use point-slope form \( y - y_1 = m(x - x_1) \): \[ y - 0 = 2(x - (-4)) \implies y = 2(x + 4) \implies y = 2x + 8 \]

Problem: Solve the equation: \( -3(-x + 5) + 20 = -10(x - 3) + 4 \).

Solution:

Expand both sides: \[ 3x - 15 + 20 = -10x + 30 + 4 \] \[ 3x + 5 = -10x + 34 \] Add \( 10x \) and subtract 5: \[ 13x = 29 \implies x = \dfrac{29}{13} \]

Problem: Solve the inequality: \( 4(x - 6) + 4 < 8(x - 4) \).

Solution:

Expand both sides: \[ 4x - 24 + 4 < 8x - 32 \] \[ 4x - 20 < 8x - 32 \] Rearrange terms: \[ -4x < -12 \implies x > 3 \]

Problem: Solve the equation: \( 3(x - 2)^2 - 12 = 0 \).

Solution:

Add 12 and divide by 3: \[ 3(x - 2)^2 = 12 \implies (x - 2)^2 = 4 \] Take the square root of both sides: \[ x - 2 = \pm 2 \implies x = 2 \pm 2 \] The solutions are \( x = 0 \) and \( x = 4 \).

Problem: Solve the equation: \( \dfrac{x}{3} + \dfrac{2}{7} = \dfrac{x}{7} - 5 \).

Solution:

Multiply all terms by the LCD (21): \[ 21\left(\dfrac{x}{3}\right) + 21\left(\dfrac{2}{7}\right) = 21\left(\dfrac{x}{7}\right) - 21(5) \] \[ 7x + 6 = 3x - 105 \] Subtract \( 3x \) and 6: \[ 4x = -111 \implies x = -\dfrac{111}{4} \]

Problem: Line \( L \) passes through \( (2, 7) \) and is perpendicular to \( x + y = 0 \). What is the point of intersection of \( L \) and the line \( x + y = 0 \)?

Solution:

The line \( x + y = 0 \) can be written as \( y = -x \), which has a slope of \( -1 \).
The perpendicular slope is the negative reciprocal: \( m = 1 \).
Using point-slope form for line \( L \) through \( (2, 7) \): \[ y - 7 = 1(x - 2) \implies y = x + 5 \] To find the intersection with \( y = -x \), set them equal: \[ x + 5 = -x \implies 2x = -5 \implies x = -\dfrac{5}{2} \] Substitute back to find \( y \): \( y = \dfrac{5}{2} \).
The point of intersection is \( \left(-\dfrac{5}{2}, \dfrac{5}{2}\right) \).

Problem: What is the point of intersection of the lines: \( x + 2y = 4 \) and \( -x - 3y = -7 \)?

Solution:

Add the two equations together: \[ (x + 2y) + (-x - 3y) = 4 + (-7) \] \[ -y = -3 \implies y = 3 \] Substitute \( y = 3 \) into the first equation: \[ x + 2(3) = 4 \implies x + 6 = 4 \implies x = -2 \] The intersection point is \( (-2, 3) \).

Problem: How many solutions do the system of equations \( 2x - 3y = 4 \) and \( 4x - 6y = -7 \) have?

Solution:

Multiply the first equation by 2: \[ 4x - 6y = 8 \] Compare with the second equation \( 4x - 6y = -7 \). Since the left sides are identical but the right sides differ ($8 \neq -7$), the lines are parallel.
Therefore, there are no solutions.

Problem: For what value(s) of \( A \) does the system of equations \( Ax + 6y = 0 \) and \( 2x - 7y = 3 \) have no solutions?

Solution:

A linear system has no solutions when the lines are parallel (equal slopes, different y-intercepts).
Slope of first equation \( Ax + 6y = 0 \) is \( -\dfrac{A}{6} \).
Slope of second equation \( 2x - 7y = 3 \) is \( \dfrac{2}{7} \).
Set slopes equal: \[ -\dfrac{A}{6} = \dfrac{2}{7} \implies A = -\dfrac{12}{7} \]

Problem: Solve \( |2x - 4| - 2 = 6 \).

Solution:

Add 2 to both sides: \[ |2x - 4| = 8 \] Split into two cases: \[ 2x - 4 = 8 \implies 2x = 12 \implies x = 6 \] \[ 2x - 4 = -8 \implies 2x = -4 \implies x = -2 \] The solutions are \( x = -2 \) and \( x = 6 \).

Problem: How many solutions does the equation \( 2x^2 + 3x = 8 \) have?

Solution:

Rewrite in standard form: \( 2x^2 + 3x - 8 = 0 \).
Calculate the discriminant \( \Delta = b^2 - 4ac \): \[ \Delta = 3^2 - 4(2)(-8) = 9 + 64 = 73 \] Since the discriminant is positive (\( 73 > 0 \)), the equation has 2 real solutions.

Problem: Solve the equation \( 3x^2 + 6x - 1 = 8 \).

Solution:

Rearrange into standard quadratic form: \[ 3x^2 + 6x - 9 = 0 \] Divide by 3: \[ x^2 + 2x - 3 = 0 \] Factor the quadratic: \[ (x + 3)(x - 1) = 0 \implies x = -3 \text{ or } x = 1 \]

Problem: Solve the system of equations: \( 2x + 5y = 18 \) and \( -3x - y = -1 \).

Solution:

Multiply the second equation by 5: \[ -15x - 5y = -5 \] Add this to the first equation: \[ (2x + 5y) + (-15x - 5y) = 18 + (-5) \implies -13x = 13 \implies x = -1 \] Substitute \( x = -1 \) into the second equation: \[ -3(-1) - y = -1 \implies 3 - y = -1 \implies y = 4 \] The solution is \( x = -1, y = 4 \).

Problem: What is the range of the function \( f \) defined by: \( f = \{(2, 3), (1, 4), (5, 4), (0, 3)\} \)?

Solution:

The range of a function is the set of all second coordinates (outputs): \[ \{3, 4, 4, 3\} \] Removing duplicates gives the range: \( \{3, 4\} \).

Problem: Factor the expression \( 2x^2 + 3x + 1 \).

Solution:

Split the middle term: \[ 2x^2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1) \]

Problem: Factor the expression \( 10x^2 + 20x - 80 \).

Solution:

Factor out the GCF (10): \[ 10(x^2 + 2x - 8) \] Factor the remaining quadratic expression: \[ 10(x - 2)(x + 4) \]