Often times algebraic concepts have a geometric meaning too. You've now done enough arithmetic with pronumerals to be able to learn this secret of the universe.

Consider \(3(2x+5)\). You can expand this by distributing the \(3\) to each term in the brackets like so: 

\vspace{0.5cm}
\begin{equation} (\source{3}\source{})(\target{2x}+\target{5})=6x + 15\mbox{\drawarrowssingle} \end{equation}
\vspace{0.5cm}

Or you can think about this as having some kind of original rectangle with dimensions $ 3 $ by $ 2x $ and then extending the width by $ 5 $.

\begin{center}
    \begin{tikzpicture}[scale=0.5]
        \draw (0,0) rectangle (7,3);
        \draw (2,0) rectangle (7,3);

        \node at (1,3) [above] {$2x$};
        \node at (0, 1.5) [left] {$3$};
        \node at (3.5, 3) [above] {$5$};
    \end{tikzpicture}
\end{center}

Now finding the area of the enlarged shape is algebraically equivalent to $ 3(2x + 5) $ and often times expanding this will make substitution easier if you know what the value of $ x $ is!

These kinds of expansions are the backbone of mathematics and becoming proficient at these will help you simplify harder problems. Let's get better at expanding:

 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     \begin{questions}                                                                                          
         \Question[3] Expand:                                                                                   
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(2(a+3)=\)                                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(2a+6\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(3(x-2)=\)                                                                              
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(3x-6\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(4(2m-7)=\)                                                                             
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(8m-28\)                                                                                      
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[3] Now try:                                                                                  
             \begin{multicols}{3}                                                                               
             \begin{parts}                                                                                      
                 \part \(5(a+1)+6=\)                                                                            
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(5a+5+6\)                                                                                     
                 \(5a+11\)                                                                                      
                 \end{solutionordottedlines}                                                                    
                 \part \(4(2b-1)+7=\)                                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(8b-4+7\)                                                                                     
                 \(8b+3\)                                                                                       
                 \end{solutionordottedlines}                                                                    
                 \part \(6(d+5)-3d=\)                                                                           
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(6d+30-3d\)                                                                                   
                 \(3d+30\)                                                                                      
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[2] Can you handle some more terms?                                                           
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \(2(b+5)+3(b+2)=\)                                                                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(2b+10+3b+6\)                                                                                 
                 \(5b+16\)                                                                                      
                 \end{solutionordottedlines}                                                                    
                 \part \(3(x-2)-2(x+1)=\)                                                                       
                 \begin{solutionordottedlines}[1cm]                                                             
                 \(3x-6-2x-2\)                                                                                  
                 \(x-8\)                                                                                        
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               


 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[10] Have a go at these ones yourselves:                                                      
         \begin{parts}                                                                                          
         \begin{multicols}{2}                                                                                   
             \part \(\frac{3}{5}(6x+\frac{7}{3})=\)\dotfill                                                     
             \( \frac{18}{5}x + \frac{7}{5} \)                                                                  
             \part \(\frac{4}{3}(6x+11)+\frac{2}{3}=\)\dotfill                                                  
             \( 8x + \frac{44}{3} + \frac{2}{3} = 8x + \frac{46}{3} \)                                          
             \part \(-12(4y-5)=\)\dotfill                                                                       
             \( -48y + 60 \)                                                                                    
             \part \(\frac{2}{3}(12p+6)=\)\dotfill                                                              
             \( 8p + 4 \)                                                                                       
             \part \(-\frac{1}{2}(10d-6)=\)\dotfill                                                             
             \( -5d + 3 \)                                                                                      
             \part \(-\frac{4}{5}(25m-100)=\)\dotfill                                                           
             \( -20m + 80 \)                                                                                    
             \part \(\frac{3}{5}(\frac{x}{6}+\frac{1}{3})=\)\dotfill                                            
             \( \frac{1}{10}x + \frac{1}{5} \)                                                                  
             \part \(-\frac{3}{5}(\frac{a}{3}-\frac{2}{3})=\)\dotfill                                           
             \( -\frac{1}{5}a + \frac{2}{5} \)                                                                  
             \part \(c(c-5)=\)\dotfill                                                                          
             \( c^2 - 5c \)                                                                                     
             \part \(2i(5i+7)=\)\dotfill                                                                        
             \( 10i^2 + 14i \)                                                                                  
         \end{multicols}                                                                                        
         \end{parts}                                                                                            
     \end{questions}                                                                                            
 \end{exercisebox}