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Sagot :
[tex]\begin{cases}2x+5y=\frac{4}{5} \\ 6x-5y=\frac{5}{6} \end{cases}\\\\\begin{cases} 5y=-2x+\frac{4}{5} \ \ / *\frac{1}{5} \\ 6x-5y=\frac{5}{6} \end{cases}\\\\\begin{cases} y=-\frac{2}{5}x+4 \\ 6x-5y=\frac{5}{6} \end{cases}[/tex]
[tex]substitution : \\\\ 6x-5 *(-\frac{2}{5}x+\frac{4}{25})=\frac{5}{6}\\\\6x+2x- \frac{4}{5}=\frac{5}{6} \\\\8x=\frac{5}{6} +\frac{4}{5}\\\\8x=\frac{25}{30}+\frac{24}{30}[/tex]
[tex]8x=\frac{49}{30} \ \ /*\frac{1}{8}\\\\x=\frac{49}{240}\\\\\\2*\frac{49}{240}+5y=\frac{4}{5}\\\\\frac{49}{120}+5y=\frac{4}{5}[/tex]
[tex]5y=\frac{4}{5}-\frac{49}{120} \\\\5y=\frac{96}{120}-\frac{49}{120} \\\\5y=\frac{47}{120}\ \ /*\frac{1}{5}\\\\y=\frac{47}{600} \\\\Answer : \ \begin{cases} x= \frac{49}{240}\\ y=\frac{47}{600} \end{cases}[/tex]
[tex]substitution : \\\\ 6x-5 *(-\frac{2}{5}x+\frac{4}{25})=\frac{5}{6}\\\\6x+2x- \frac{4}{5}=\frac{5}{6} \\\\8x=\frac{5}{6} +\frac{4}{5}\\\\8x=\frac{25}{30}+\frac{24}{30}[/tex]
[tex]8x=\frac{49}{30} \ \ /*\frac{1}{8}\\\\x=\frac{49}{240}\\\\\\2*\frac{49}{240}+5y=\frac{4}{5}\\\\\frac{49}{120}+5y=\frac{4}{5}[/tex]
[tex]5y=\frac{4}{5}-\frac{49}{120} \\\\5y=\frac{96}{120}-\frac{49}{120} \\\\5y=\frac{47}{120}\ \ /*\frac{1}{5}\\\\y=\frac{47}{600} \\\\Answer : \ \begin{cases} x= \frac{49}{240}\\ y=\frac{47}{600} \end{cases}[/tex]
2x + 5y = 4/5 ---- equation 1
6x - 5y =5/6 ----equation 2
since you are to use substitution method then first thing to do is to have one equation be modified such that one variable could have a value in terms of the other..
Let's have equation 1
2x + 5y = 4/5
2x = 4/5 - 5y
[tex]x = ( \frac{4}{5} - 5y)( \frac{1}{2}) [/tex]
[tex]x = (\frac{4}{5} )( \frac{1}{2} ) - \frac{5y}{2} [/tex]
[tex]x = \frac{2}{5} - \frac{5y}{2} [/tex]
substituting this to equation 2 you'll have:
6x - 5y = 5/6
[tex]6( \frac{2}{5} - \frac{5y}{2}) - 5y = \frac{5}{6} [/tex]
[tex]6( \frac{2}{5}) - \frac{6(5y)}{2} - 5y = \frac{5}{6} [/tex]
[tex] \frac{12}{5} - 15y - 5y = \frac{5}{6} [/tex]
[tex] \frac{12}{5} - \frac{5}{6} = 15y + 5y [/tex]
[tex]20y = \frac{12}{5} - \frac{5}{6} [/tex]
[tex]20y = \frac{12(6)-5(5)}{30} [/tex]
[tex]20y = \frac{71-25}{30} [/tex]
[tex]20y = \frac{47}{30} [/tex]
[tex]y = \frac{47}{30(20)} [/tex]
[tex]y = \frac{47}{600} [/tex]
substituting the value for y to equation 1
2x + 5y = 4/5
2x + 5(47/600) = 4/5
[tex]2x = \frac{4}{5} - \frac{5(47)}{600} [/tex]
[tex]2x = \frac{4}{5} - \frac{47}{120} [/tex]
[tex]2x = \frac{49}{120} [/tex]
[tex]x = \frac{49}{240} [/tex]
6x - 5y =5/6 ----equation 2
since you are to use substitution method then first thing to do is to have one equation be modified such that one variable could have a value in terms of the other..
Let's have equation 1
2x + 5y = 4/5
2x = 4/5 - 5y
[tex]x = ( \frac{4}{5} - 5y)( \frac{1}{2}) [/tex]
[tex]x = (\frac{4}{5} )( \frac{1}{2} ) - \frac{5y}{2} [/tex]
[tex]x = \frac{2}{5} - \frac{5y}{2} [/tex]
substituting this to equation 2 you'll have:
6x - 5y = 5/6
[tex]6( \frac{2}{5} - \frac{5y}{2}) - 5y = \frac{5}{6} [/tex]
[tex]6( \frac{2}{5}) - \frac{6(5y)}{2} - 5y = \frac{5}{6} [/tex]
[tex] \frac{12}{5} - 15y - 5y = \frac{5}{6} [/tex]
[tex] \frac{12}{5} - \frac{5}{6} = 15y + 5y [/tex]
[tex]20y = \frac{12}{5} - \frac{5}{6} [/tex]
[tex]20y = \frac{12(6)-5(5)}{30} [/tex]
[tex]20y = \frac{71-25}{30} [/tex]
[tex]20y = \frac{47}{30} [/tex]
[tex]y = \frac{47}{30(20)} [/tex]
[tex]y = \frac{47}{600} [/tex]
substituting the value for y to equation 1
2x + 5y = 4/5
2x + 5(47/600) = 4/5
[tex]2x = \frac{4}{5} - \frac{5(47)}{600} [/tex]
[tex]2x = \frac{4}{5} - \frac{47}{120} [/tex]
[tex]2x = \frac{49}{120} [/tex]
[tex]x = \frac{49}{240} [/tex]
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