Intermediate Algebra Problems With Solutions - Sample 1

Problem: For what value of the constant \( K \) does the equation \( 2x + Kx = 3 \) have one solution?

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Solution:

Given the equation:

\[ 2x + Kx = 3 \]

Factor \( x \) out:

\[ x(2 + K) = 3 \]

Solve for \( x \) to obtain:

\[ x = \dfrac{3}{2 + K} \]

\( x \) is undefined if the denominator is equal to zero. The denominator equals zero if:

\[ K = -2 \]

Therefore, the given equation has one solution for all real values of \( K \neq -2 \).

Problem: For what value of the constant \( K \) does the equation \( Kx^2 + 2x = 1 \) have two real solutions?

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Solution:

Rewrite the given quadratic equation in standard form:

\[ Kx^2 + 2x - 1 = 0 \]

Calculate the discriminant \( \Delta \) of the quadratic equation:

\[ \Delta = b^2 - 4ac = 2^2 - 4(K)(-1) = 4 + 4K \]

For the equation to have two real solutions, the discriminant \( \Delta \) must be positive. Hence, we solve the inequality:

\[ 4 + 4K > 0 \]

The solution set to the inequality is:

\[ K > -1 \]

For any \( K > -1 \), the given equation has two real solutions.

Problem: For what value of the constant \( K \) does the system of equations

\[ \begin{cases} 2x - y = 4 \\ 6x - 3y = 3K \end{cases} \]

have an infinite number of solutions?

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Solution:

For the system of equations to have an infinite number of solutions, the two equations must be equivalent.

If we multiply all terms of the first equation by \( 3 \), we obtain:

\[ 6x - 3y = 12 \]

For the two equations to be equivalent, we must have:

\[ 12 = 3K \]

Solving for \( K \) gives:

\[ K = 4 \]

Problem: Find \( Y \) so that the points \( A(0, 0) \), \( B(2, 2) \), and \( C(-4, Y) \) are the vertices of a right triangle with hypotenuse \( \overline{AC} \).

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Solution:

Since the triangle is a right triangle with hypotenuse \( \overline{AC} \), we apply the Pythagorean Theorem:

\[ AC^2 = AB^2 + BC^2 \quad (I) \]

Compute the squared distances:

\[ AB^2 = (2 - 0)^2 + (2 - 0)^2 = 4 + 4 = 8 \] \[ BC^2 = (2 - (-4))^2 + (2 - Y)^2 = 36 + (2 - Y)^2 \] \[ AC^2 = (-4 - 0)^2 + (Y - 0)^2 = 16 + Y^2 \]

Substitute into equation \( (I) \):

\[ 16 + Y^2 = 8 + 36 + (2 - Y)^2 \] \[ 16 + Y^2 = 44 + (4 - 4Y + Y^2) \] \[ 16 + Y^2 = 48 - 4Y + Y^2 \]

Subtract \( 16 + Y^2 \) from both sides:

\[ 0 = 32 - 4Y \implies 4Y = 32 \implies Y = 8 \]

Final Answer: \[ \boxed{Y = 8} \]

Problem: Find \( M \) so that the lines with equations

\[ -2x + My = 5 \quad \text{and} \quad 4y + x = -9 \]

are perpendicular.

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Solution:

We first find the slopes of the two lines: rewriting them gives slopes of \( \dfrac{2}{M} \) and \( -\dfrac{1}{4} \).

For two lines to be perpendicular, the product of their slopes must equal \( -1 \). Hence:

\[ \left(\dfrac{2}{M}\right) \cdot \left(-\dfrac{1}{4}\right) = -1 \] \[ \dfrac{-2}{4M} = -1 \]

Multiply all terms by \( 4M \) and simplify:

\[ -2 = -4M \implies M = \dfrac{1}{2} \]

Problem: The length of a rectangular field is \( \dfrac{7}{5} \) its width. If the perimeter of the field is \( 240 \) meters, what are the length and width of the field?

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Solution:

Let \( L \) be the length and \( W \) be the width:

\[ L = \dfrac{7}{5}W \]

Perimeter equation:

\[ 2L + 2W = 240 \]

Substitute \( L = \dfrac{7}{5}W \) into the perimeter equation:

\[ 2\left(\dfrac{7}{5}W\right) + 2W = 240 \]

Multiply all terms by \( 5 \) and simplify:

\[ 14W + 10W = 1200 \implies 24W = 1200 \implies W = 50\,\text{m} \]

Solve for \( L \):

\[ L = \dfrac{7}{5}(50) = 70\,\text{m} \]

Problem: Peter wants to get \$200,000 for his house after commission. An agent charges \( 20\% \) of the selling price for selling the house for Peter.

a) What should be the selling price?

b) What will be the agent's commission?

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Solution:

Let \( x \) be the selling price. Peter wants to get \$200,000 after commission:

\[ x - 20\% \cdot x = 200,000 \] \[ x - 0.2x = 200,000 \implies 0.8x = 200,000 \]

a) Solve for \( x \):

\[ x = \$250,000 \]

b) Calculate the agent's commission:

\[ 20\% \text{ of } \$250,000 = 0.20 \cdot 250,000 = \$50,000 \]

Problem: John's annual salary after a raise of \( 15\% \) is \$45,000. What was his salary before the raise?

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Solution:

Let \( x \) be the salary before the increase. After the raise:

\[ x + 15\%x = 45,000 \] \[ x + 0.15x = 45,000 \implies 1.15x = 45,000 \]

Solve for \( x \):

\[ x = \dfrac{45,000}{1.15} \approx \$39,130 \]

Problem: It took Malcom \( 3.5 \) hours to drive from city A to city B. On his way back to city A, he increased his speed by \( 20 \) km per hour and it took him \( 3 \) hours. Find the average speed for the whole journey.

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Solution:

Let \( x \) and \( x + 20 \) be the speeds of the car from A to B and from B to A, respectively. The distance from A to B is \( 3.5x \) and from B to A is \( 3(x + 20) \).

Since the distances are equal:

\[ 3.5x = 3(x + 20) \implies 0.5x = 60 \implies x = 120\,\text{km/hr} \]

The formula for average speed is:

\[ \text{Average speed} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{3.5x + 3(x + 20)}{3.5 + 3} \]

Substitute \( x = 120 \):

\[ \text{Average speed} = \dfrac{3.5 \cdot 120 + 3(120 + 20)}{6.5} = \dfrac{420 + 420}{6.5} = \dfrac{840}{6.5} \approx 129.2\,\text{km/hr} \]

Problem: Let \( R \) be a relation defined by:

\[ R = \{(4, 3), (x^2, 2), (1, 6), (-4, 0)\} \]

Find all values of \( x \) so that \( R \) is NOT a function.

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Solution:

For \( R \) not to be a function, two ordered pairs must have the same first component (input) with different second components (outputs). The first components are \( 4, x^2, 1, -4 \).

Check when \( x^2 \) equals another input:

\[ x^2 = 4 \implies x = \pm 2 \]

This gives two pairs with input 4: \( (4, 3) \) and \( (4, 2) \). Since outputs differ, \( R \) is not a function.

Next:

\[ x^2 = 1 \implies x = \pm 1 \]

This gives two pairs with input 1: \( (1, 6) \) and \( (1, 2) \). Different outputs mean \( R \) is not a function.

\( x^2 = -4 \) has no real solution.

Final Answer: \[ \boxed{R \text{ is not a function when } x = \pm 2 \text{ or } x = \pm 1.} \]