Intermediate Algebra: Equations, Problem Solving
with Detailed Solutions — Sample 3

Problem: Solve the following system of equations:

\[ \begin{cases} - \dfrac{x}{2} + \dfrac{y}{3} = 0 \\ x + 6y = 16 \end{cases} \]

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Solution:

Multiply all terms of the first equation by the LCM of 2 and 3, which is 6:

\[ 6\left(-\dfrac{x}{2} + \dfrac{y}{3}\right) = 6(0) \implies -3x + 2y = 0 \]

Solve the equivalent system of equations:

\[ \begin{aligned} -3x + 2y &= 0 \\ x + 6y &= 16 \end{aligned} \]

Solving this system yields:

\[ x = \dfrac{8}{5}, \quad y = \dfrac{12}{5} \]

Problem: How many real solutions does each quadratic equation shown below have?

1. \( x^2 + \dfrac{4}{5}x = -\dfrac{1}{4} \)
2. \( x^2 - 7x + 10 = 0 \)
3. \( x^2 - \dfrac{2}{3}x + \dfrac{1}{9} = 0 \)

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Solution: Find the discriminant \( \Delta = b^2 - 4ac \) for each equation:

1. \( x^2 + \dfrac{4}{5}x + \dfrac{1}{4} = 0 \)
   \( \Delta = \left(\dfrac{4}{5}\right)^2 - 4(1)\left(\dfrac{1}{4}\right) = \dfrac{16}{25} - 1 = -\dfrac{9}{25} < 0 \)
   No real solutions.
2. \( x^2 - 7x + 10 = 0 \)
   \( \Delta = (-7)^2 - 4(1)(10) = 49 - 40 = 9 > 0 \)
   2 real solutions.
3. \( x^2 - \dfrac{2}{3}x + \dfrac{1}{9} = 0 \)
   \( \Delta = \left(-\dfrac{2}{3}\right)^2 - 4(1)\left(\dfrac{1}{9}\right) = \dfrac{4}{9} - \dfrac{4}{9} = 0 \)
   1 repeated real solution.

Problem: Does the data in the table below represent \( y \) as a function of \( x \)? Explain.

\[ \begin{array}{|c|c|} \hline \textbf{x} & \textbf{y} \\ \hline -2 & 4 \\ 3 & 6 \\ -5 & 4 \\ 6 & 12 \\ 7 & 16 \\ 0 & 34 \\ \hline \end{array} \]

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Solution:

Yes, this data represents \( y \) as a function of \( x \) because each input value \( x \) is paired with exactly one output value \( y \). Even though some \( y \)-values repeat (for example, \( y = 4 \) appears for both \( x = -2 \) and \( x = -5 \)), no \( x \)-value maps to more than one \( y \). Therefore, the relation satisfies the definition of a function.

Problem: Solve the following quadratic equation:

\[ 0.01x^2 - 0.1x - 0.3 = 0 \]

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Solution:

Multiply all terms of the equation by 100 to obtain integer coefficients:

\[ x^2 - 10x - 30 = 0 \]

Calculate the discriminant:

\[ \Delta = (-10)^2 - 4(1)(-30) = 100 + 120 = 220 \]

Apply the quadratic formula:

\[ x = \dfrac{-(-10) \pm \sqrt{220}}{2(1)} = \dfrac{10 \pm 2\sqrt{55}}{2} = 5 \pm \sqrt{55} \]

Problem: In a cafeteria, 3 coffees and 4 donuts cost \$10.05. In the same cafeteria, 5 coffees and 7 donuts cost \$17.15. How much do you have to pay for 4 coffees and 6 donuts?

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Solution:

Let \( x \) be the price of one coffee and \( y \) be the price of one donut.

From the first statement:

\[ 3x + 4y = 10.05 \]

From the second statement:

\[ 5x + 7y = 17.15 \]

Subtract the first equation from the second:

\[ (5x + 7y) - (3x + 4y) = 17.15 - 10.05 \implies 2x + 3y = 7.10 \]

Multiply all terms by 2 to find the cost of 4 coffees and 6 donuts:

\[ 4x + 6y = 2(7.10) = 14.20 \]

Final Answer: 4 coffees and 6 donuts cost \$14.20.

Problem: Find the slope of the lines through the given points and state whether each line is vertical, horizontal, or neither.

1. Line \( L_1 \): \( (-2, 3) \) and \( (8, 3) \)
2. Line \( L_2 \): \( (4, 3) \) and \( (4, -3) \)
3. Line \( L_3 \): \( (-1, 7) \) and \( (3, -3) \)

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Solution:

1. Slope of line \( L_1 = \dfrac{3 - 3}{8 - (-2)} = \dfrac{0}{10} = 0 \) → Horizontal line.
2. Slope of line \( L_2 = \dfrac{-3 - 3}{4 - 4} = \dfrac{-6}{0} \) (undefined) → Vertical line.
3. Slope of line \( L_3 = \dfrac{-3 - 7}{3 - (-1)} = \dfrac{-10}{4} = -\dfrac{5}{2} \) → Neither horizontal nor vertical.

Problem: Find four consecutive even integers whose sum is 388.

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Solution:

Let \( x, x + 2, x + 4, \text{ and } x + 6 \) be four consecutive even integers. Their sum is 388:

\[ x + (x + 2) + (x + 4) + (x + 6) = 388 \] \[ 4x + 12 = 388 \implies 4x = 376 \implies x = \dfrac{376}{4} = 94 \]

The four numbers are \( 94, 96, 98, \text{ and } 100 \).

Problem: Thomas drove for 2 hours and had lunch. After lunch he drove for 3 more hours at a speed that is \( 20\text{ km/hour} \) more than before lunch. The total trip distance was \( 460\text{ km} \). What was his speed after lunch?

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Solution:

Let \( x \) be the speed before lunch (in km/h). The distance driven before lunch is \( 2x \) km.

After lunch, the speed is \( x + 20 \), and the distance driven is \( 3(x + 20) \) km.

Total distance is 460 km:

\[ 2x + 3(x + 20) = 460 \] \[ 2x + 3x + 60 = 460 \implies 5x + 60 = 460 \implies 5x = 400 \implies x = 80 \]

His speed before lunch is 80 km/h, so his speed after lunch is:

\[ 80 + 20 = 100\text{ km/h} \]

Problem: Compare the following expressions:

1. \( 2^{-4} \)
2. \( (-2)^{-4} \)
3. \( \left(-\dfrac{1}{2}\right)^4 \)

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Solution: Simplify each expression:

1. \( 2^{-4} = \dfrac{1}{2^4} = \dfrac{1}{16} \)
2. \( (-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16} \)
3. \( \left(-\dfrac{1}{2}\right)^4 = \dfrac{(-1)^4}{2^4} = \dfrac{1}{16} \)

All three expressions simplify to the exact same value: \( \dfrac{1}{16} \).

Problem: Evaluate and convert to scientific notation:

\[ (3.4 \times 10^{11})(5.4 \times 10^{-3}) \]

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Solution:

\[ (3.4 \times 5.4) \times 10^{11} \times 10^{-3} = 18.36 \times 10^8 = 1.836 \times 10^9 \]