Respuesta :
so hmmm let's say the cost for 1lb of walnut is "w" and the cost for 1lb of chocolate chips is "c" [tex]\bf 5x+6c=23\\
3w+2c=11[/tex]
so let's use the elimination method then, let's multiply the second equation by -3, so we end up with -3*2c is -6c and thus getting vertically 6c-6c =0, thus "eliminating" the "c" variable
[tex]\bf \begin{array}{llll} 5w+6c=23&\implies &\quad 5w+6c=23\\ 3w+2c=11&\boxed{\times -3}\implies &-9w-6c=-33\\ &&--------\\ &&-4w+0=-10 \end{array} \\\\\\ -4w=-10\implies w=\cfrac{-10}{-4}\implies \boxed{w=\cfrac{5}{2}}\\\\ -------------------------------\\\\ \textit{now, let's plug that in the 1st equation} \\\\\\ 5\cdot \boxed{\cfrac{5}{2}}+6c=23\implies \cfrac{25}{2}+6c=23\implies 6c=23-\cfrac{25}{2} \\\\\\ 6c=\cfrac{21}{2}\implies c=\cfrac{21}{12}\implies \boxed{c=\cfrac{7}{4}}[/tex]
so, "w" is 5/2 or 2 bucks and 50 cents per pound
and "c" is 7/4 or 1 buck and 75 cents per pound
so let's use the elimination method then, let's multiply the second equation by -3, so we end up with -3*2c is -6c and thus getting vertically 6c-6c =0, thus "eliminating" the "c" variable
[tex]\bf \begin{array}{llll} 5w+6c=23&\implies &\quad 5w+6c=23\\ 3w+2c=11&\boxed{\times -3}\implies &-9w-6c=-33\\ &&--------\\ &&-4w+0=-10 \end{array} \\\\\\ -4w=-10\implies w=\cfrac{-10}{-4}\implies \boxed{w=\cfrac{5}{2}}\\\\ -------------------------------\\\\ \textit{now, let's plug that in the 1st equation} \\\\\\ 5\cdot \boxed{\cfrac{5}{2}}+6c=23\implies \cfrac{25}{2}+6c=23\implies 6c=23-\cfrac{25}{2} \\\\\\ 6c=\cfrac{21}{2}\implies c=\cfrac{21}{12}\implies \boxed{c=\cfrac{7}{4}}[/tex]
so, "w" is 5/2 or 2 bucks and 50 cents per pound
and "c" is 7/4 or 1 buck and 75 cents per pound
