Quiz

I think it’s time for my readers to show how well I have presented the concepts. Below are some problems that cover some of the concepts presented on this blog. Below are several problems you can try. Further below, are the worked solutions.

1. Evaluate (which means – do the math): ${10}\hspace{0.33em}{+}\hspace{0.33em}{6}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{+}\hspace{0.33em}{10}$
2. Evaluate: ${9}\hspace{0.33em}\times\hspace{0.33em}{(}{4}\hspace{0.33em}\times\hspace{0.33em}{10}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\div\hspace{0.33em}{4}$
3. Solve for all values of x that make the following equation true: ${(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{2}{)(}{3}{x}\hspace{0.33em}{-}\hspace{0.33em}{9}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$
4. Find the two values of x that make the following equation true: ${x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{9}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}$
5. Find the two values of x that make the following equation true: ${x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{121}$
6. Simplify (that is use algebra rules to make the expression shorter): $\frac{4{x}^{7}{y}^{8}}{{x}^{3}{y}^{7}}$

SOLUTIONS:

1. ${10}\hspace{0.33em}{+}\hspace{0.33em}{6}\hspace{0.33em}\times\hspace{0.33em}{1}\hspace{0.33em}{+}\hspace{0.33em}{10}$. You do the multiplication first to get 10 + 6 + 10. Then add all the numbers to get 26.
2.  ${9}\hspace{0.33em}\times\hspace{0.33em}{(}{4}\hspace{0.33em}\times\hspace{0.33em}{10}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\div\hspace{0.33em}{4}$. This is a BODMAS within a BODMAS. You evaluate the expression in the brackets first, but within the brackets, you do the multiplication first:

${9}\hspace{0.33em}\times\hspace{0.33em}{(}{4}\hspace{0.33em}\times\hspace{0.33em}{10}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\div\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{9}\hspace{0.33em}\times\hspace{0.33em}{(}{40}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\div\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{9}\hspace{0.33em}\times\hspace{0.33em}{(}{36}{)}\hspace{0.33em}\div\hspace{0.33em}{4}$. Now do the multiplication and division left to right:

${9}\hspace{0.33em}\times\hspace{0.33em}{(}{36}{)}\hspace{0.33em}\div\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{324}\hspace{0.33em}\div\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{81}$

1. ${(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{2}{)(}{3}{x}\hspace{0.33em}{-}\hspace{0.33em}{9}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$. Since the only way for three things multiplied together can equal zero is if any or all of the things are zero. So to solve set each factor equal to zero:
$\begin{array}{l} {{x}\hspace{0.33em}{-}\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\ {{x}\hspace{0.33em}{+}\hspace{0.33em}{2}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{-}{2}}\\ {{3}{x}\hspace{0.33em}{-}\hspace{0.33em}{9}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{3}} \end{array}$

1. ${x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{9}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}$. Using the distributive property to un-distribute the x you get ${x}{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{9}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$. Now setting each factor equal to zero results in
$\begin{array}{l} {{x}\hspace{0.33em}{=}\hspace{0.33em}{0}}\\ {{x}\hspace{0.33em}{-}\hspace{0.33em}{9}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{9}} \end{array}$
1.  ${x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{121}$. Taking the square root of both sides and remembering that this action introduces two plus or minus answers results in $\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{121}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{11}$
2. $\frac{4{x}^{7}{y}^{8}}{{x}^{3}{y}^{7}}$. You need to remember the rule that when you have the same base with exponents in the numerator and denominator, you subtract the exponents. So

$\frac{4{x}^{7}{y}^{8}}{{x}^{3}{y}^{7}}\hspace{0.33em}{=}\hspace{0.33em}{4}{x}^{{7}{-}{3}}{y}^{{8}{-}{7}}\hspace{0.33em}{=}\hspace{0.33em}{4}{x}^{4}{y}$ where the “1” exponent of y can be left off.