Value of [tex]x=\frac{2}{3}\,\,and\,\,value\,\,of\,\,y=\frac{1}{5}[/tex]
Solution set is [tex](\frac{2}{3},\frac{1}{5})[/tex]
Step-by-step explanation:
We need to solve the following system of equations:
[tex]\frac{1}{2}x-\frac{3}{4}y=\frac{11}{60}\,\,\,eq(1)\\\frac{2}{5}x+\frac{1}{6}y=\frac{3}{10}\,\,\,eq(2)[/tex]
We can solve this using Substitution.
Isolate x in [tex]\frac{1}{2}x-\frac{3}{4}y=\frac{11}{60}[/tex]
[tex]\frac{1}{2}x=\frac{11}{60}+\frac{3}{4}y\\ x=\frac{11}{30}+\frac{3}{2}y[/tex]
Putting value of x in equation(2)
[tex]\frac{2}{5}(\frac{11}{30}+\frac{3}{2}y)+\frac{1}{6}y=\frac{3}{10}\\Solving:\\Taking\,\,LCM\,\,of\,\,5,6,10 \,\,we\,\,get: 30\\Multiply\,\,by\,\,LCM\\\frac{2}{5}(\frac{11}{30}+\frac{3}{2}y)\times30+\frac{1}{6}y\times30=\frac{3}{10}\times30\\Simplifying:\\12(\frac{11}{30}+\frac{3}{2}y)+5y=9\\\frac{22}{5}+18y+5y=9\\23y=9-\frac{22}{5}\\23y=\frac{45-22}{5}\\23y=\frac{23}{5}\\y=\frac{\frac{23}{5} }{23}\\y=\frac{1}{5}[/tex]
Putting value of y=1/5 in [tex]x=\frac{11}{30}+\frac{3}{2}y[/tex]
[tex]x=\frac{11}{30}+\frac{3}{2}\times\frac{1}{5} \\x=\frac{11}{30}+\frac{3}{10}\\x=\frac{11+3*3}{30}\\x=\frac{11+9}{30} \\x=\frac{20}{30}\\x=\frac{2}{3}[/tex]
So, value of [tex]x=\frac{2}{3}\,\,and\,\,value\,\,of\,\,y=\frac{1}{5}[/tex]
Solution set is [tex](\frac{2}{3},\frac{1}{5})[/tex]
Keywords: System of linear Equations
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