Question 178279
Let {{{a}}}= pounds of soybean meal needed
Let {{{b}}}= pounds of cornmeal needed
In words:
(pounds of protein in soybean meal + pounds of protein in cornmeal) /
(total pounds of soybean and cornmeal) = % protein
given:
(1) {{{a + b = 320}}}
(2) {{{(.16a + .08b)/(a + b) = .13}}}
{{{(.16a + .08b)/320 = .13}}}
{{{.16a + .08b = .13*320}}}
(3){{{.16a + .08b = 41.6}}}
Multiply both sides of (1) by {{{.08}}} and subtract from (3)
(3) {{{.16a + .08b = 41.6}}}
(1) {{{-.08a - .08b = -25.6}}}
{{{.08a = 16}}}
{{{a = 200}}}
And, since
(1) {{{a + b = 320}}}
{{{200 + b = 320}}}
{{{b = 120}}}
200 pounds of soybean meal are needed and
120 pounds of cornmeal are needed
check:
(2) {{{(.16a + .08b)/(a + b) = .13}}}
{{{(.16*200 + .08*120)/320 = .13}}}
{{{(32 + 9.6)/320 = .13}}}
{{{41.6/320 = .13}}}
{{{41.6 = 41.6}}}
OK