Question 512137
Let {{{a}}} = liters of 3% solution needed
Let {{{b}}} = liters of 7% solution needed
Let {{{c}}} = liters of 9% solution needed
given:
(1) {{{ a + b + c = 15 }}}
(2) {{{ (  .03a + .07b + .09c ) / 15 = .06 }}}
(3) {{{ a = 2c }}}
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There are 3 equations and 3 unknowns, so it's solvable
(2) {{{ .03a + .07b + .09c  = .06*15 }}}
(2) {{{ .03a + .07b + .09c  = .9 }}}
(2) {{{ 3a + 7b + 9c  = 90 }}}
Multiply both sides of (1) by {{{7}}}
and subtract (2) from (1)
(1) {{{ 7a + 7b + 7c = 105 }}}
(2) {{{ -3a - 7b - 9c  = -90 }}}
(4) {{{ 4a - 2c = 15 }}}
Substitute (3) into (4)
(4) {{{ 8c - 2c = 15 }}}
(4) {{{ 6c = 15 }}}
(4) {{{ c = 2.5 }}}
and, since
(3) {{{ a = 2c }}}
(3) {{{ a = 5 }}}
and
(1) {{{ a + b + c = 15 }}}
(1) {{{ 5 + b + 2.5 = 15 }}}
(1) {{{ b = 15 - 7.5 }}}
(1) {{{ b = 7.5 }}}
5 liters of 3% solution are needed
7.5 liters of 7% solution are needed
2.5 liters of 9% solution are needed
check answer:
(2) {{{ (  .03*5 + .07*7.5 + .09*2.5 ) / 15 = .06 }}}
(2) {{{ ( .15 + .525 + .225 ) / 15 = .06 }}}
(2) {{{  .15 + .525 + .225  = .06*15 }}}
(2) {{{ .9 = .9 }}}
OK