ryleighhancock
Answered

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x-2y=16 and x-y=8 its a system of equation. I don't know how to do this please help.

Sagot :

[tex]\left\{\begin{array}{ccc}x-2y=16\\x-y=8\end{array}\right\\\\\#1\ elimination\ method:\\\left\{\begin{array}{ccc}x-2y=16\\x-y=8&|change\ signs\end{array}\right\\\underline{+\left\{\begin{array}{ccc}x-2y=16\\-x+y=-8\end{array}\right}\ \ \ \ |add\ both\ sides\ of\ the\ equations\\.\ \ \ \ \ -y=8\to\boxed{y=-8}\\\\subtitute\ value\ of\ y=-8\ to\ the\ second\ equation\\\\x-(-8)=8\\x+8=8\ \ \ \ |subtract\ 8\ from\ both\ sides\\\boxed{x=0}\\\\Answer:\boxed{\boxed{\left\{\begin{array}{ccc}x=0\\y=-8\end{array}}}[/tex]

[tex]\#\ subtitute\ method:\\\left\{\begin{array}{ccc}x-2y=16\\x-y=8&|add\ "y"\ to\ both\ sides\end{array}\right\\\left\{\begin{array}{ccc}x-2y=16&(1)\\x=y+8&(2)\end{array}\right\\\\subtitute\ (2)\ to\ (1)\\\\(y+8)-2y=16\\y+8-2y=16\\-y+8=16\ \ \ \ |subtract\ 8\ from\ both\ sides\\-y=8\to \boxed{y=-8}\\\\subtitute\ value\ of\ y=-8\ to\ (2)\\x=-8+8\\\boxed{x=0}\\\\Answer:\boxed{\boxed{\left\{\begin{array}{ccc}x=0\\y=-8\end{array}\right}}[/tex]
crisforp
x = 2y + 16 and x = 8 + y ;
2y + 16 = 8 + y ;
2y - y = 8 - 16 ;
y = - 8 ;
x = 8 + ( - 8 ) ;
x = 0;
checking ( 0 , - 8 ) ;
0 - 2 × ( - 8 ) = 16 ( correct !
0 - ( - 8 ) = 8 correct !

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