p = price for one Peony.
d = price for one Daylily.
[tex]\bf \begin{cases} \stackrel{\textit{Mr Hanson}}{4p+7d}~~ = ~~\stackrel{\textit{total cost}}{21.25}\\\\ \stackrel{\textit{Mrs. Florez}}{5p+3d}~~ = ~~\stackrel{\textit{total cost}}{16.50} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{solving for "d" on the 1st equation}}{4p+7d=21.25\implies 7d=21.25-4p}\implies d = \cfrac{21.25-4p}{7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{5p+3d=16.5}\implies 5p+3\left( \cfrac{21.25-4p}{7} \right)=16.5[/tex]
[tex]\bf \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{7}}{7\left[ 5p+3\left( \cfrac{21.25-4p}{7} \right) \right]=7(16.5)}\implies 35p+3(21.25-4p)=115.5 \\\\\\ 35p+63.75-12p=115.5\implies 23p+63.75=115.5\implies 23p=51.75 \\\\\\ p=\cfrac{51.75}{23}\implies \blacktriangleright p = 2.25 \blacktriangleleft[/tex]
[tex]\bf \stackrel{\textit{we know that}}{d = \cfrac{21.25-4p}{7} }\implies d = \cfrac{21.25-4(2.25)}{7} \\\\\\ d = \cfrac{21.25-9}{7}\implies d = \cfrac{12.25}{7}\implies \blacktriangleright d = 1.75 \blacktriangleleft[/tex]
