Intermediate Algebra: Linear Equations, Slopes,
and Practical Applications with Step-by-Step Solutions

Problem: Find the slopes of the lines given by the following equations:

1. \( 2x + 3y = 2 \)
2. \( y = 2 \)
3. \( x = 4 \)

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Solution:

1. Solve equation for \( y \): \[ y = -\dfrac{2}{3}x + \dfrac{2}{3} \] Slope: \( -\dfrac{2}{3} \)
2. The slope of a horizontal line such as \( y = 2 \) is equal to \( 0 \).
3. The slope of a vertical line such as \( x = 4 \) is undefined.

Problem: Find the slopes of the lines through points A and B given by:

1. \( A(2, 1) \), \( B(3, 4) \)
2. \( A(3, -4) \), \( B(5, -4) \)
3. \( A(2, -6) \), \( B(2, -7) \)

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Solution:

1. Slope: \( \dfrac{4 - 1}{3 - 2} = \dfrac{3}{1} = 3 \)
2. Slope: \( \dfrac{-4 - (-4)}{5 - 3} = \dfrac{0}{2} = 0 \) (horizontal line).
3. Slope: \( \dfrac{-7 - (-6)}{2 - 2} = \dfrac{-1}{0} \) (undefined, vertical line).

Problem: Find the slopes of the lines:

1. parallel to the line whose equation is given by \( 5x - 3y = -3 \)
2. perpendicular to the line whose equation is given by \( -4x - 8y = -3 \)
3. parallel to the line whose equation is given by \( x = 2 \)
4. perpendicular to the line whose equation is given by \( x = -9 \)
5. parallel to the line whose equation is given by \( y = 3 \)
6. perpendicular to the line whose equation is given by \( x = 0 \)

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Solution:

1. Rewrite in slope-intercept form: \( -3y = -5x - 3 \implies y = \dfrac{5}{3}x + 1 \).
   Slope of given line: \( m = \dfrac{5}{3} \). Slope of parallel lines is \( \dfrac{5}{3} \).
2. Rewrite equation: \( -8y = 4x - 3 \implies y = -\dfrac{1}{2}x + \dfrac{3}{8} \).
   Slope of given line: \( m = -\dfrac{1}{2} \). Perpendicular slope is the negative reciprocal: \( m_{\perp} = 2 \).
3. \( x = 2 \) is a vertical line (undefined slope). Parallel lines are also vertical, so slope is undefined.
4. \( x = -9 \) is a vertical line. Lines perpendicular to vertical lines are horizontal, so slope = 0.
5. \( y = 3 \) is a horizontal line (slope = 0). Parallel lines have slope = 0.
6. \( x = 0 \) is the y-axis (vertical line, undefined slope). Lines perpendicular to it are horizontal, so slope = 0.

Problem: Find the slope of each of the lines graphed below.

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Solution:

1. \( L_1 \) is a vertical line and its slope is undefined.
2. \( L_2 \) is a horizontal line and its slope is equal to 0.
3. Using the x-intercept at \( (-4, 0) \) and y-intercept at \( (0, 4) \):
   \[ \text{slope} = \dfrac{4 - 0}{0 - (-4)} = 1 \]

Problem: Find the value of \( k \) so that the slope of the line through the points \( (4, 2) \) and \( (k, 6) \) is equal to \( -2 \).

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Solution:

Slope formula:

\[ \dfrac{6 - 2}{k - 4} = -2 \implies \dfrac{4}{k - 4} = -2 \]

Cross multiply:

\[ 4 = -2(k - 4) \implies 4 = -2k + 8 \implies 2k = 4 \implies k = 2 \]

Problem: Find the value of \( k \) so that the lines with equations \( 2y + kx = 2 \) and \( 3x - 2y = 6 \) have equal slopes.

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Solution:

Write \( 2y + kx = 2 \) in slope-intercept form:

\[ 2y = -kx + 2 \implies y = -\dfrac{k}{2}x + 1 \implies m = -\dfrac{k}{2} \]

Write \( 3x - 2y = 6 \) in slope-intercept form:

\[ -2y = -3x + 6 \implies y = \dfrac{3}{2}x - 3 \implies s = \dfrac{3}{2} \]

Set slopes equal:

\[ -\dfrac{k}{2} = \dfrac{3}{2} \implies k = -3 \]

Problem: Find the value of \( k \) so that the lines with equations \( -3y + 2x = 4 \) and \( kx + 2y = 3 \) are perpendicular.

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Solution:

Find the slope \( m \) of \( -3y + 2x = 4 \):

\[ -3y = -2x + 4 \implies y = \dfrac{2}{3}x - \dfrac{4}{3} \implies m = \dfrac{2}{3} \]

Find the slope \( s \) of \( kx + 2y = 3 \):

\[ 2y = -kx + 3 \implies y = -\dfrac{k}{2}x + \dfrac{3}{2} \implies s = -\dfrac{k}{2} \]

Perpendicular lines satisfy \( m \times s = -1 \):

\[ \left(\dfrac{2}{3}\right)\left(-\dfrac{k}{2}\right) = -1 \implies -\dfrac{k}{3} = -1 \implies k = 3 \]

Problem: The oil consumption of a certain country was \( 330 \) thousand barrels per day in 2004 and \( 450 \) thousand barrels per day in 2006. Assume that the oil consumption in this country increases linearly and estimate the oil consumption in 2015.

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Solution:

Let \( t \) be the number of years after 2004 (so \( t = 0 \) in 2004). Linear consumption model: \( C = mt + C_0 \).

In 2004 (\( t = 0 \)), \( C = 330 \), so \( C_0 = 330 \).

In 2006 (\( t = 2 \)), \( C = 450 \). Substitute to find slope \( m \):

\[ 450 = m(2) + 330 \implies 2m = 120 \implies m = 60 \]

For 2015 (\( t = 2015 - 2004 = 11 \)):

\[ C = 60(11) + 330 = 660 + 330 = 990 \text{ thousand barrels per day} \]

Problem: A family spent \$3600 on food last year and \$3000 the year before last. Assume that the spending on food of this family increases linearly and estimate their spending on food for this year.

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Solution:

Spending two years ago: \( S_0 = \$3000 \)

Spending last year: \( S_1 = \$3600 \)

Yearly increase: \( \Delta S = S_1 - S_0 = 3600 - 3000 = \$600 \)

Assuming a linear increase, spending this year (\( S_2 \)) is:

\[ S_2 = S_1 + \Delta S = 3600 + 600 = \$4200 \]

Estimated spending this year: \$4200

Problem: To convert the measure of temperature given in degrees Fahrenheit \( T_f \) into degrees Celsius \( T_c \), use the formula \( T_c = \dfrac{5}{9}(T_f - 32) \). If the temperature of an item increases by 9 degrees Fahrenheit, by how many degrees Celsius has the temperature of this item increased?

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Solution:

A change of \( \Delta T_f \) in Fahrenheit corresponds to a change in Celsius \( \Delta T_c \) given by:

\[ \Delta T_c = \dfrac{5}{9}\Delta T_f \]

Substitute \( \Delta T_f = 9 \):

\[ \Delta T_c = \dfrac{5}{9} \times 9 = 5 \]

The temperature increases by 5 degrees Celsius.

Problem: To convert degrees Celsius \( T_c \) into Fahrenheit, use the formula \( T_f = \dfrac{9}{5}T_c + 32 \). If the temperature of an item increases by 10 degrees Celsius, by how many degrees Fahrenheit has the temperature increased?

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Solution:

A change \( \Delta T_c \) in Celsius corresponds to a change in Fahrenheit \( \Delta T_f \):

\[ \Delta T_f = \dfrac{9}{5}\Delta T_c \]

Substitute \( \Delta T_c = 10 \):

\[ \Delta T_f = \dfrac{9}{5} \times 10 = 18 \]

The temperature increases by 18 degrees Fahrenheit.