Question 1092323
Let {{{ a }}} = ounces of 6% peanuts needed
Let {{{ b }}} = ounces of 10% peanuts needed
Let {{{ c }}} = ounces of 50% peanuts needed
-----------------------------------------
(1) {{{ a + b + c = 16 }}}
(2) {{{ ( .06a + .1b + .5c ) / 16 = .135 }}}
(3) {{{ c = (1/4)*b }}}
-------------------------------------
There are 3 equations and 3 unknowns, so it's solvable
(2) {{{ .06a + .1b + .5c = 2.16 }}}
(2) {{{ 6a + 10b + 50c = 216 }}}
(2) {{{ 3a + 5b + 25c = 108 }}}
-------------------------------
Multiply both sides of (1) by {{{3} and
subtract (1) from (2)
(2) {{{ 3a + 5b + 25c = 108 }}}
(1) {{{ -3a - 3b - 3c = -48 }}}
------------------------------
{{{ 2b + 22c = 60 }}}
{{{ b + 11c = 30 }}}
and
(3) {{{ b = 4c }}}
{{{ 4c + 11c = 30 }}}
{{{ 15c = 30 }}}
{{{ c = 2 }}}
and
{{{ b = 4c }}}
{{{ b = 4*2 }}}
{{{ b = 8 }}}
and
(1) {{{ a + b + c = 16 }}}
(1) {{{ a + 8 + 2 = 16 }}}
(1) {{{ a = 16 - 10 }}}
(1) {{{ a = 6 }}}
----------------------
6 ounces of 6% peanuts are needed
8 ounces of 10% peanuts are needed
2 ounces of 50% peanuts are needed
----------------------------------
check:
(2) {{{ ( .06a + .1b + .5c ) / 16 = .135 }}}
(2) {{{ ( .06*6 + .1*8 + .5*2 ) / 16 = .135 }}}
(2) {{{ ( .36 + .8 + 1 ) / 16 = .135 }}}
(2) {{{ 2.16 / 16 = .135 }}}
(2) {{{ 2.16 = 2.16 }}}
OK