Homework 4

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Directions:

1. Show each step of your work and fully simplify each expression.
2. Turn in your answers in class on a physical piece of paper.
3. Staple multiple sheets together.
4. Feel free to use Desmos for graphing.

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Answer the following:

1. Show why $x = 2$ is a solution to the equation \[\dfrac{1}{x} - \dfrac{1}{x-4} = 1\]
2. Isolate the given variable in the following equations:
   1. $4x + 2 = 6x - w, \ $ for $x$
   2. $\dfrac{x + 1}{x - 3} = 1, \ $ for $x$
   3. $4xy - w(2xz - 3yz) + 3 = -4x - y, \ $ for $x$
   4. $3x^2 - 2(y' + xy') - 3y^2y' = 4, \ $ for $y'$
3. Find all solutions for the following equations that are real numbers.
   1. $3x + 4 = 7$
   2. $2x + 3 = 7 - 3x$
   3. $2x^2 - 5x = -2$
   4. $2x^2 = 8$
   5. $4x^2 - x = 0$
   6. $x^2 - 2x = -1$
4. Solve the following equations. Remember to check your work if necessary!
   1. $\dfrac{1}{x} = \dfrac{4}{3x} + 1$
   2. $\dfrac{1}{x}-\dfrac{2}{3\left(x-3\right)}=-\dfrac{4}{x^{2}-9}$
   3. $\dfrac{\dfrac{4}{x^2} - 1}{x} = 0$
5. Isolate $y'$ in the equation \[2y^2y' - 2(xy + y') = 0\]
6. Solve the following equations. Fully simplify all complex solutions and remember to check solutions.
   1. $\dfrac{1}{x-1} - \dfrac{2}{x^2} = 0$
   2. $\sqrt{8x - 1} = 3$
   3. $\sqrt{2x + 1} + 1 = x$
   4. $\dfrac{2}{\sqrt{4x - 1}} - \dfrac{1}{x} = 0$
   5. $3x^2 + 1 = 0$
   6. $x^2 + 2x + 2 = 0$
7. Using the discriminant $b^2 - 4ac$, determine whether the following equations have two real solutions, one real solution, or two distinct complex solutions.
   1. $3x^2 - 5x = -2$
   2. $x^2 + 1 = 0$
   3. $x^2 - 2x - 1 = 0$
   4. $2x^2 - x = x^2 + 1$
8. Simplify the following:
   1. $\sqrt{-4}$
   2. $(-25)^\frac{1}{2}$
   3. $\sqrt{-2}\cdot\sqrt{6}\cdot\sqrt{-3}$
   4. $\dfrac{\sqrt{-49}}{7}$
   5. $\dfrac{\sqrt{-4}}{3}\cdot \dfrac{\sqrt{-3}}{2}$
9. Solve the following inequalities:
   1. $2x \geq 10$
   2. $5 - 2x < -15$
   3. $-1 < 2x - 5 < 7$
   4. $2 \leq -x + 5 < 4$
   5. $2(7x - 3) < 12x + 16$