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Sagot :
[tex]x-weight\ of\ candy\ cost\ \$6\ per\ pound\\y-weight\ of\ candy\ cost\ \$4.50\ per\ pound\\\\ \left\{\begin{array}{ccc}x+y=2\\2(6x+4.5y)=15\end{array}\right\\\left\{\begin{array}{ccc}x+y=2&|multiply\ both\ sides\ by\ (-3)\\12x+9y=15&|divide\ both\ sides\ by\ 3\end{array}\right\\+\left\{\begin{array}{ccc}-3x-3y=-6\\4x+3y=5\end{array}\right\ \ \ \ |add\ sides\ of\ the\ equations\\---------\\.\ \ \ \ \ \ \ \ \ x=1\\\\1+y=2\to y=1[/tex]
[tex]Answer:Fifty Fifty (1\ pounds\ of\ candy\ cost\ \$6\ and\ 1\ pound\ of\ candy\ cost\ \$4.50[/tex]
[tex]Answer:Fifty Fifty (1\ pounds\ of\ candy\ cost\ \$6\ and\ 1\ pound\ of\ candy\ cost\ \$4.50[/tex]
[tex]c-the \ 6\$ \ candies \\ y-the \ 4.5-\$ \ candis \\ \\ c+y=2 \\ \\2(6+4.50)=15 \\\\ c+y=2 \ |(-3) \\12c+9y=15 \ |:(+3) \\\\ -3c-3y=-6\\ {4c+5y=5}\\...................... \\\\ -3c-3y=-6 \\ 4c+3y=5 \\ ................. \\\\ \boxed{c=1} \\\\ y=2-1 \\\\ \boxed{y=1}[/tex]
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