In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
long boards (wood): 6, 7, 9, and 10 feet
2/3 foot ===> 5 pieces
5/7 foot ===> some pieces
Step 02:
only one long board:
5 pieces * 2/3 ft = 10/3 ft
If long board = 6 ft:
[tex]6\text{ ft - }\frac{10}{3}ft\text{ = }\frac{18\text{ - 10}}{3}ft\text{ = }\frac{8}{3}\text{ ft}[/tex]no waste (whole number of pieces):
[tex]\frac{8}{3}\text{ ft / }\frac{5}{7}\text{ ft = }\frac{8\cdot7\text{ }}{5\cdot3}\text{ = }\frac{56}{15}\text{ = }3.73\text{ pieces }[/tex]It is not the solution
If long board = 7 ft:
[tex]7\text{ ft - }\frac{10}{3}ft\text{ = }\frac{21\text{ - 10}}{3}ft\text{ = }\frac{11}{3}\text{ ft}[/tex]no waste (whole number of pieces):
[tex]\frac{11}{3}\text{ ft / }\frac{5}{7}\text{ ft = }\frac{11\cdot7\text{ }}{5\cdot3}\text{ = }\frac{77}{15}=5.13\text{ pieces }[/tex]It is not the solution
If long board = 9 ft:
[tex]9\text{ ft - }\frac{10}{3}ft\text{ = }\frac{27\text{ - 10}}{3}ft\text{ = }\frac{17}{3}\text{ ft}[/tex]no waste (whole number of pieces):
[tex]\frac{17}{3}\text{ ft / }\frac{5}{7}\text{ ft = }\frac{17\cdot7\text{ }}{5\cdot3}\text{ = }\frac{119}{15}=7.93\text{ pieces }[/tex]7.93 pieces ≅ 8 pieces
If long board = 10 ft:
[tex]10\text{ ft - }\frac{10}{3}ft\text{ = }\frac{30\text{ - 10}}{3}ft\text{ = }\frac{20}{3}\text{ ft}[/tex]no waste (whole number of pieces):
[tex]\frac{20}{3}\text{ ft / }\frac{5}{7}\text{ ft = }\frac{20\cdot7\text{ }}{5\cdot3}\text{ = }\frac{140}{15}=9.33\text{ pieces }[/tex]It is not the solution
The answer is:
long board = 9 ft