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Sagot :
Let x and y be the candy pounds that sells for $0.87 and $1.22 , respectively. Since they both must add up to 9 lb, we have
[tex]x+y=9...(A)[/tex]On the other hand, the mixture should sell for $0.91 per lib, so we can write,
[tex]0.87x+1.22y=9\times0.91[/tex]Or euivalently,
[tex]\begin{gathered} \frac{0.87}{0.91}x+\frac{1.22}{0.91}y=9 \\ that\text{ is, } \\ 0.95604x+1.340659y=9...(B) \end{gathered}[/tex]Then, we need to solve the following system of equations:
[tex]\begin{gathered} x+y=9...(A) \\ 0.95604x+1.340659y=9 \end{gathered}[/tex]Solving by elimination method.
In order to eliminate variable x, we can to multiply equation (A) by -0.95604 and get an equivalent system of equations:
[tex]\begin{gathered} -0.95604x-0.95604y=-8.60439 \\ 0.95604x+1.340659y=9 \end{gathered}[/tex]Then, by adding both equations, we get
[tex]0.384619y=0.39561[/tex]Then, y is given by
[tex]\begin{gathered} y=\frac{0.39561}{0.384619} \\ y=1.02857 \end{gathered}[/tex]Once we have obtained the result for y, we can substitute in into equation (A), that is,
[tex]x+1.02857=9[/tex]then, x is given as
[tex]\begin{gathered} x=9-1.02857 \\ x=7.9714 \end{gathered}[/tex]Therefore, by rounding to two decimal places, the answer is:
$ 0.87 per lb of candy: 7.97 lb
$1.22-per-lb of candy: 1.03 lb
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