A set of intermediate algebra problems are presented with detailed solutions. Topics include the Pythagorean theorem, distance, speed and time calculations, systems of equations, slopes of lines, percentages, and rectangle dimensions. Each problem is explained clearly, with math expressions formatted using LaTeX for easy understanding.
For what value of the constant \( K \) does the equation \( 2x + K x = 3 \) have one solution?.
Given the equation \[ 2x + K x = 3 \] 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 is equal to zero if \[ k = - 2 \] The given equation has one solution for all real values of \( k \neq -2 \).
For what value of the constant \( K \) does the equation \( K x^2 + 2x = 1 \) have two real solutions?
Rewrite the given quadratic equation in standard form: \[ Kx^2 + 2x - 1 = 0 \] Calculate the discriminant \( \Delta \) of the above quadratic equation: \[ \Delta = b^2 - 4 a c = 4 - 4(K)(-1) = 4 + 4K \] For the equation to have two real solutions, the discriminant \( \Delta \) must be positive. Hence we need to solve the inequality: \[ 4 + 4K > 0 \] The solution set to the above inequality is given by: \[ K > -1 \] for which the given equation has two real solutions.
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?
For the system of equations \begin{cases} 2x - y = 4 \\\\ 6x - 3y = 3K \end{cases} 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
\[ 12 = 3K \]which, when solved, gives
\[ K = 4 \]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 \( \overline{AC} \).
We are given three points: \( A(0, 0) \) , \( B(2, 2) \) and \( C(-4, Y) \).
We are told the triangle is a right triangle with hypotenuse \( \overline{AC} \), so we use 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 \( (I) \): \[ 16 + Y^2 = 8 + 36 + (2 - Y)^2 \] \[ 16 + Y^2 = 44 + (4 - 4Y + Y^2) \] \[ 16 + Y^2 = 44 + 4 - 4Y + Y^2 \] \[ 16 + Y^2 = 48 - 4Y + Y^2 \] Subtract \( 16 + Y^2 \) from both sides: \[ 0 = 48 - 4Y + Y^2 - 16 - Y^2 \] \[ 0 = 32 - 4Y \] \[ 4Y = 32 \] \[ Y = 8 \] \[ \boxed{Y = 8} \]
Find \( M \) so that the lines with equations \[ -2x + My = 5 \quad \text{and} \quad 4y + x = -9 \] are perpendicular.
We first find the slopes of the two lines: \(\dfrac{2}{M}\) and \(-\dfrac{1}{4}\)
For two lines to be perpendicular, the product of their slopes must be equal to \(-1\). Hence the equation: \[ \left(\dfrac{2}{M}\right) \cdot \left(-\dfrac{1}{4}\right) = -1 \] which may be written as \[ \left(\dfrac{-2}{4 M}\right) = -1 \] multiply all terms by \( 4 M \) and simplify \[ -2 = - 4 M \] Solve for \(M\) to find: \[ M = \dfrac{1}{2} \]
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?
Let \( L \) be the length and \( W \) be the width. \[ L = \dfrac{7}{5}W \] Perimeter: \[ 2L + 2W = 240 \] Substitute \( L = \dfrac{7}{5}W \) in the above equation \[ 2\left(\dfrac{7}{5}W\right) + 2W = 240 \] Mutliply all terms by \( 5 \) and simplify \[ 14 W + 10W = 1200 \] Group \[ 24 W = 1200 \] Solve for \( W \) \[ W = 50 \,\text{m} \] Solve for \( L \) \[ L = \dfrac{7}{5}W = \dfrac{7}{5} 50 = 70\,\text{m} \]
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?
Let \( x \) be the selling price. Peter wants to get $\(200,000\) for his house after commission, hence \[ x - 20\% \cdot x = 200,\!000 \] which may be written \[ x - 0.2 x = 200,\!000 \Rightarrow 0.8 x = 200,\!000 \] a) Solve for \( x \) to find \[ x = \$ 250,\!000 \] b) \[ 20\% \; \text{of} \; x = 20\% \cdot 250,\!000 = \$ 50,\!000 \]
John's annual salary after a raise of \( 15\% \) is $\(45,000\). What was his salary before the raise?
Let \( x \) be the raise before the increase. Hence after the raise, we have \[ x + 15\% x = 45,000 \] which may be written as \[ x + 0.15 x = 45,000 \Rightarrow 1.15 x = 45,000 \] Solve for \( x \) to find \[ x = \$ 39,130 \]
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.
a) Find the average speed for the whole journey.
Let \( x \) and \( x + 20 \) be the speeds of the car from A to B and then from B to A respectively.
Hence the distance from A to B may be expressed as \( 3.5x \) and the distance from B to A as \( 3(x + 20) \).
\[ \text{The average speed} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{3.5x + 3(x + 20)}{3.5 + 3} \] The distance from A to B is equal to the distance from B to A, hence: \[ 3.5x = 3(x + 20) \] Rearrange the above equation \[0.5 x = 60 \] Solve for \( x \) to obtain \[ x = 120 \; \text{km/hr} \] . We now substitute \( x = 120 \) in the formula for the average speed to obtain: \[ \text{Average speed} = \dfrac{3.5 \cdot 120 + 3(120 + 20)}{3.5 + 3} = 129.2 \) km/hr \]
Let \( R \) be a relation defined by \[ R = \{(4,3), (x^2, 2), (1,6), (-4,0)\}. \] a) Find all values of \( x \) so that \( R \) is NOT a function.
To determine when \( R \) is not a function, we examine whether any two ordered pairs have the same first component (input) with different second components (outputs). The first components are: \[ 4,\ x^2,\ 1,\ -4 \] We consider values of \( x \) such that \( x^2 \) equals another input: \[ x^2 = 4 \Rightarrow x = \pm 2 \] This leads to two pairs with input 4: \[ (4,3) \quad \text{and} \quad (4,2) \] Since the outputs differ, \( R \) is not a function. Next: \[ x^2 = 1 \Rightarrow x = \pm 1 \] This gives two pairs with input 1: \[ (1,6) \quad \text{and} \quad (1,2) \] Again, different outputs, \( R \) is not a function. Finally: \[ x^2 = -4 \] has no real solution. \[ \boxed{ \text{Therefore, } R \text{ is not a function when } x = \pm 2 \text{ or } x = \pm 1. } \]