Department of Mathematics



Algebra



Self-Test:


1)       $\displaystyle{\frac{3[2+(-5)-(-8)]+2^2-1}{\sqrt{3^0 - (-2)^3}} = }$

      $\displaystyle{\frac{18}{2\sqrt{2}} }$
      $\displaystyle{\frac{16}{3} }$
      $4$
      $6$
      None of the Above

Hint 1 Use the Order of Operations (BEDMAS – Brackets, Exponents, Division, Multiplication, Addition, and Subtraction), which actually allows you to solve the numerator and denominator separately.
Hint 2 $a^0 = 1$ for $a \neq 0$ and $\sqrt{a}=k$ if $k^2 =a$

2)       $\displaystyle{ \left[ \left( \frac{4}{9}\right)^{-3/2} - \frac{2}{3} \right] \div \frac{1}{2} }$

      $\displaystyle{\frac{-13}{27} }$
      $\displaystyle{\frac{65}{12} }$
      $\displaystyle{\frac{-52}{27} }$
      $\displaystyle{\frac{-65}{48} }$
      None of the Above

Hint $a^{-n} = \frac{1}{a^n}$ and $a^{m/n} = (\sqrt[n]{a})^m$

3)       $2\sqrt{3}+5\sqrt{12}-9\sqrt{3}-\sqrt{120}$

      $3\sqrt{3}-2\sqrt{30}$
      $13\sqrt{3}-2\sqrt{30}$
      $-7\sqrt{3}-5\sqrt{12}$
      $-17\sqrt{3}$
      $0$

Hint $\sqrt{ab} = \sqrt{a}\sqrt{b}$, and see Hint 2 from #1

4)       $\displaystyle{ \frac{3b^2d(14a^2b^4c^3 - 21a^{-1}b^2c^7)}{7a^3b^2c^3} = }$

      $-3a^6b^{-1}c^{-3}d$
      $-3b^{-2}cd$
      $6a^{-1}b^{4}d - 21a^{-4}c^{4}$
      $6a^{-1}b^{4}d - 9a^{-4}b^2c^{4}d$
      Cannot be simplified

Hint $a^m a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$

5)       $\displaystyle{ \frac{x+1}{x-2}+\frac{3-x}{2x+1} \times \frac{2}{x-2} = }$

      $\displaystyle{ \frac{2x^2+x+7}{(x-2)(2x+1)} }$
      $\displaystyle{ \frac{2x^2-2x+7}{(x-2)(2x+1)} }$
      $\displaystyle{ \frac{4x^2+4x+8}{(x-2)(2x+1)} }$
      $\displaystyle{ \frac{7}{2x+1} }$
      $\displaystyle{ \frac{8}{(x-2)(2x+1)} }$

Hint 1 $$\frac{a}{b}\cdot \frac{c}{d} = \frac{ac}{bd} $$
Hint 2 Before adding two fractions, use the factored forms of the denominators to determine the Least Common Denominator

6)       $\displaystyle{ \frac{(x+-1)(x+3)}{(x^2-1)(x-2)} \times \frac{(x-2)^2}{(x+3)} \div \frac{(x+1)}{(x-2)} = }$

      $\displaystyle{ \frac{x^2+2x+1}{x^2-1} }$
      $0$
      $1$
      $x+1$
      $\displaystyle{ \left( \frac{x-2}{x+1} \right)^2 }$

Hint 1 Express everything in fully factored form and simplify where possible
Hint 2 $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b}\cdot \frac{d}{c}$ (multiply the first fraction by the reciprocal of the second)

7)       When $x=-1, \; \sqrt{(3x)^2+4x}-2x^5 =$
      $\sqrt{13}-2$
      $\sqrt{5}+2$
      $\sqrt{7}-2$
     $\sqrt{7}+2$
      None of the Above

Hint Everywhere you see an $x,$ replace it with a $(-1),$ then perform the required operations

8)       When $x=2$ and $y=3, \; (xy+3)^{1/4} \left( (x+y)^{1/2} - (y-1)x^{-1} \right) = $

      $\sqrt{15}-1$
      $\sqrt{15}$
      $\sqrt{12}$
      $\sqrt{15}-\sqrt{3}$
      None of the Above

Hint 1 Everywhere you see an $x,$ replace it with a $(2)$ and everywhere you see a $y,$ replace it with a $(3).$ Then perform the required operations
Hint 2 See the hint for #2 for help with exponents

9)       The solution $(x,y)$ to the system: $y=2x+4$ and $6x+3y=12$ is

      $(0,4)$
      $(4,0)$
      $(\frac{-12}{5},\frac{-4}{5})$
      $(\frac{-4}{5},\frac{-12}{5})$
      None of the Above

Hint Try substituting the expression for $y$ (in terms of $x$) from the first equation into the second equation

10)       The $y$ value of the solution $(x,y)$ to the system: $2x+3y=5$ and $3x+7y=1$ is

      greater than $2$
      between $1$ and $2$ inclusive
      between $-2$ and $1$ inclusive
      less than $-2$
      None of the Above

Hint Multiply each equation by a constant such that when you add them, one of the variables cancels out, leaving you with one equation and one unknown.





Worked Examples and Practice Problems