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The line with equation ax+by=20 passes through the points(-2,10) and (1,5)
find a and b

Sagot :

Lilith
[tex]ax+by=20 \\ \\(-2,10) \\ \\ a*(-2)+b*10 = 20 \\ \\-2a +10b = 20 \\ \\ (1,5)\\ \\ a*1+b*5 = 20 \\ \\ a +5b = 20[/tex]
 
[tex]\begin{cases} -2a+10b =20\\ a+5b =20 \ /*2\end{cases} \\ \\\begin{cases} -2a+10b =20\\ 2a+10b =40 \end{cases}\\--------\\ 20b=60 \ / :20 \\ \\b=\frac{60}{20} =3[/tex]

[tex]-2a+10b=20 \\ \\-2a +10*3=20 \\ \\-2a +30=20\\ \\-2a=20-30\\ \\-2a=-10 \ /: (-2)[/tex]

[tex]a=\frac{10}{2}=5\\ \\\begin{cases} a= 5 \\b=3\end{cases}[/tex]
[tex]ax+by=20\\\\for\ (-2;\ 10)\to x=-2;\ y=10\ substitute\\\\-2a+10b=20\\\\for\ (1;\ 5)\to x=1;\ y=5\ substitute\\\\1a+5b=20\\\\\\ \left\{\begin{array}{ccc}-2a+10b=20&/:2\\a+5b=20\end{array}\right[/tex]
[tex]+\left\{\begin{array}{ccc}-a+5b=10\\a+5b=20\end{array}\right\\------------\\.\ \ \ \ \ \ \ 10b=30\ \ \ \ /:10\\.\ \ \ \ \ \ \ \ b=3\\\\\left\{\begin{array}{ccc}b=3\\a+5b=20\end{array}\right\\\left\{\begin{array}{ccc}b=3\\a+5\cdot3=20\end{array}\right\\\left\{\begin{array}{ccc}b=3\\a+15=20\end{array}\right\\\left\{\begin{array}{ccc}b=3\\a=20-15\end{array}\right\\\left\{\begin{array}{ccc}b=3\\a=5\end{array}\right[/tex]


[tex]Answer:a=5\ and\ b=3\to5x+3y=20.[/tex]
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