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

This page features a carefully selected set of intermediate algebra problems designed to strengthen your understanding of key concepts such as equations, scientific notation, and problem solving. Each problem comes with detailed, step-by-step solutions to help students and educators master intermediate algebra skills effectively.

Questions 1

Solve the following system of equations: \[ \begin{cases} - \dfrac{x}{2} + \dfrac{y}{3} = 0 \\ x + 6y = 16 \end{cases} \]

Solution:

We first multiply all terms of the first equation by the LCM of 2 and 3, which is 6. \[ 6\left(-\dfrac{x}{2} + \dfrac{y}{3}\right) = 6(0) \] Simplify \[ -3x + 2y = 0 \]

We then solve the following equivalent system of equations:

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

Solving this system gives the solution: \[ x = \dfrac{8}{5}, \quad y = \dfrac{12}{5} \]

Questions 2

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

Solution:

Find the discriminant of each equation.

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

Questions 3

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

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

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 twice. What matters is that no \( x \) value maps to more than one \( y \). Therefore, the relation satisfies the definition of a function.

Questions 4

Solve the following quadratic equation. \[ 0.01 x^{2} - 0.1 x - 0.3 = 0 \]

Solution:

Multiply all terms of the equation by 100 to obtain an equivalent equation with integer coefficients. \[ x^2 - 10x - 30 = 0 \] Discriminant: \[ \Delta = (-10)^2 - 4(1)(-30) = 100 + 120 = 220 \] Solutions: \[ x = \dfrac{-(-10) \pm \sqrt{220}}{2(1)} = \dfrac{10 \pm 2\sqrt{55}}{2} = 5 \pm \sqrt{55} \]

Questions 5

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?

Solution:

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

We now use "3 coffees and 4 donuts cost \$10.05" to write the equation:

\[ 3x + 4y = 10.05 \]

And use "5 coffees and 7 donuts cost \$17.15" to write the equation:

\[ 5x + 7y = 17.15 \]

Subtracting the first equation from the second:

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

Multiply all terms in the equation by 2:

\[ 4x + 6y = 14.2 \]

So, 4 coffees and 6 donuts cost \$14.20.

Questions 6

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

  1. Line L1 : \( (-2 , 3) \) and \( (8 , 3) \)
  2. Line L2 : \( (4 , 3) \) and \( (4 , -3) \)
  3. Line L3 : \( (-1 , 7) \) and \( (3 , -3) \)

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 \( L_3 = \dfrac{-3 - 7}{3 - (-1}) = \dfrac{-10}{4} = \dfrac{-5}{2} \), \( L_3 \) is neither horizontal nor vertical.

Questions 7

Find four consecutive even integer numbers whose sum is \( 388 \).

Solution:

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

Their sum is \( 388 \), hence the equation: \[ x + (x + 2) + (x + 4) + (x + 6) = 388 \]

Group like terms: \[ 4x + 12 = 388 \] \[ 4x = 388 - 12 = 376 \]

Solve the equation: \[ x = \dfrac{376}{4} = 94 \]

The four numbers are: \( 94,\, 96,\, 98,\, 100 \).

Check the sum:

\[ 94 + 96 + 98 + 100 = 388 \]

Questions 8

Going for a long trip, 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 distance of the total trip was \( 460 \; \text{km} \).
What was his speed after lunch?

Solution:

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

After lunch, the speed is 20 km/h more than before lunch, so it is \( x + 20 \). The distance driven after lunch is \( 3(x + 20) \) km.

The total distance driven is 460 km. Therefore, the equation is:

\[ 2x + 3(x + 20) = 460 \]

Solve the equation:

\[ \begin{align*} 2x + 3(x + 20) &= 460 \\ 2x + 3x + 60 &= 460 \\ 5x + 60 &= 460 \\ 5x &= 400 \\ x &= 80 \end{align*} \]

So, the speed before lunch is \( {80} \) km/h and his speed after lunch \( \boxed{80+20} = 100 \) km/h .

Questions 9

Compare the following expressions:

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

Solution:

We first simplify each expression.

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

The three expressions simplify to the same value.

Questions 10

Evaluate and convert to scientific notation. \[ (3.4 \times 10^{11})(5.4 \times 10^{-3}) \]

Solution:

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

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