Which ordered pair is a solution to the system of linear equations (1)/(2)x-(3)/(4)y=(11)/(60) and (2)/(5)x+(1)/(8)y=(3)/(10) 7 ((1)/(5),-(2)/(3)) ((1)/(5),(2)/(3)) ((2)/(3),-(1)/(5)) o c ((2)/(3),(1)/(5))

To determine which ordered pair is a solution to the system of linear equations, we need to substitute each pair into both equations and see if they satisfy both equations.<br /><br />Given system of linear equations:<br />$\frac {1}{2}x-\frac {3}{4}y=\frac {11}{60}$<br />$\frac {2}{5}x+\frac {1}{8}y=\frac {3}{10}$<br /><br />Let's test each ordered pair:<br /><br />1. $(\frac {1}{5},-\frac {2}{3})$<br /> Substituting into the first equation:<br /> $\frac {1}{2}(\frac {1}{5})-\frac {3}{4}(-\frac {2}{3})=\frac {11}{60}$<br /> $\frac {1}{10}+\frac {1}{2}=\frac {11}{60}$<br /> $\frac {3}{10}=\frac {11}{60}$ (False)<br /><br /> Substituting into the second equation:<br /> $\frac {2}{5}(\frac {1}{5})+\frac {1}{8}(-\frac {2}{3})=\frac {3}{10}$<br /> $\frac {2}{25}-\frac {1}{12}=\frac {3}{10}$<br /> $\frac {24}{300}-\frac {25}{300}=\frac {3}{10}$<br /> $-\frac {1}{300}=\frac {3}{10}$ (False)<br /><br /> Therefore, $(\frac {1}{5},-\frac {2}{3})$ is not a solution.<br /><br />2. $(\frac {1}{5},\frac {2}{3})$<br /> Substituting into the first equation:<br /> $\frac {1}{2}(\frac {1}{5})-\frac {3}{4}(\frac {2}{3})=\frac {11}{60}$<br /> $\frac {1}{10}-\frac {1}{2}=\frac {11}{60}$<br /> $-\frac {1}{5}=\frac {11}{60}$ (False)<br /><br /> Substituting into the second equation:<br /> $\frac {2}{5}(\frac {1}{5})+\frac {1}{8}(\frac {2}{3})=\frac {3}{10}$<br /> $\frac {2}{25}+\frac {1}{12}=\frac {3}{10}$<br /> $\frac {24}{300}+\frac {25}{300}=\frac {3}{10}$<br /> $\frac {49}{300}=\frac {3}{10}$ (False)<br /><br /> Therefore, $(\frac {1}{5},\frac {2}{3})$ is not a solution.<br /><br />3. $(\frac {2}{3},-\frac {1}{5})$<br /> Substituting into the first equation:<br /> $\frac {1}{2}(\frac {2}{3})-\frac {3}{4}(-\frac {1}{5})=\frac {11}{60}$<br /> $\frac {1}{3}+\frac {3}{20}=\frac {11}{60}$<br /> $\frac {20}{60}+\frac {9}{60}=\frac {11}{60}$<br /> $\frac {29}{60}=\frac {11}{60}$ (False)<br /><br /> Substituting into the second equation:<br /> $\frac {2}{5}(\frac {2}{3})+\frac {1}{8}(-\frac {1}{5})=\frac {3}{10}$<br /> $\frac {4}{15}-\frac {1}{40}=\frac {3}{10}$<br /> $\frac {64}{240}-\frac {6}{240}=\frac {3}{10}$<br /> $\frac {58}{240}=\frac {3}{10}$ (False)<br /><br /> Therefore, $(\frac {2}{3},-\frac {1}{5})$ is not a solution.<br /><br />4. $(\frac {2}{3},\frac {1}{5})$<br /> Substituting into the first equation:<br /> $\frac {1}{2}(\frac {2}{3})-\frac {3}{4}(\frac {1}{5})=\frac {11}{60}$<br /> $\frac {1}{3}-\frac {3}{20}=\frac {11}{60}$<br /> $\frac {20}{60}-\frac {9}{60}=\frac {11}{60}$<br /> $\frac {11}{60}=\frac {11}{60}$ (True)<br /><br /> Substituting into the second equation:<br /> $\frac {2}{5}(\frac {2}{3})+\frac {1}{8}(\frac {1}{5})=\frac {3}{10}$<br /> $\frac {4}{15}+\frac {