IDNLearn.com: Your trusted source for accurate and reliable answers. Get comprehensive answers to all your questions from our network of experienced experts.
Sagot :
[tex]x+\frac{4}{5}=\frac{2}{3}-x-\frac{\not3^1}{\not6_2}\\\\x+\frac{4}{5}=\frac{2}{3}-x-\frac{1}{2}\\\\find\ LCD(\frac{4}{5};\ \frac{2}{3};\ \frac{1}{2})=2\times3\times5=30\\\\multiply\ both\ sides\ by\ LCD=30\\\\30x+30\cdot\frac{4}{5}=30\cdot\frac{2}{3}-30x-30\cdot\frac{1}{2}\\\\30x+6\cdot4=10\cdot2-30x-15\cdot1\\\\30x+24=20-30x-15[/tex]
[tex]30x+24=5-30x\ \ \ \ |subtract\ 24\ from\ both\ sides\\\\30x=-19-30x\ \ \ \ \ |add\ 30x\ to\ both\ sides\\\\60x=-19\ \ \ \ \ |divide\ both\ sides\ by\ 60\\\\\boxed{x=-\frac{19}{60}}[/tex]
[tex]\frac{x+4}{5}=\frac{2}{3}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{2\cdot2}{3\cdot2}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4}{6}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4-(x-3)}{6}\\\\\frac{x+4}{5}=\frac{4-x+3}{6}\\\\\frac{x+4}{5}=\frac{7-x}{6}\ \ \ \ |cross\ multiply[/tex]
[tex]6(x+4)=5(7-x)\\\\6(x)+6(4)=5(7)-5(x)\\\\6x+24=35-5x\ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\6x=11-5x\ \ \ \ \ |add\ 5x\ to\ both\ sides\\\\11x=11\ \ \ \ \ |divide\ both\ sides\ by\ 11\\\\\boxed{x=1}[/tex]
[tex]30x+24=5-30x\ \ \ \ |subtract\ 24\ from\ both\ sides\\\\30x=-19-30x\ \ \ \ \ |add\ 30x\ to\ both\ sides\\\\60x=-19\ \ \ \ \ |divide\ both\ sides\ by\ 60\\\\\boxed{x=-\frac{19}{60}}[/tex]
[tex]\frac{x+4}{5}=\frac{2}{3}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{2\cdot2}{3\cdot2}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4}{6}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4-(x-3)}{6}\\\\\frac{x+4}{5}=\frac{4-x+3}{6}\\\\\frac{x+4}{5}=\frac{7-x}{6}\ \ \ \ |cross\ multiply[/tex]
[tex]6(x+4)=5(7-x)\\\\6(x)+6(4)=5(7)-5(x)\\\\6x+24=35-5x\ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\6x=11-5x\ \ \ \ \ |add\ 5x\ to\ both\ sides\\\\11x=11\ \ \ \ \ |divide\ both\ sides\ by\ 11\\\\\boxed{x=1}[/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.