Question 1005763
Let {{{ b }}} = amount that Bill has
Let {{{ j }}} = amount that Jim has
Let {{{ s }}} = amount that Sue has
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(1) {{{ b = 2j }}}
(2) {{{ b = s/2 }}}
(3) {{{ b + j + s = 120 }}}
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There are 3 unknowns and 3 equations,
so it is solvable
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From (1):
(1) {{{ j = b/2 }}}
From (2):
(2) {{{ s = 2b }}}
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Substitute (1) and (2) into (3)
(3) {{{ b + j + s = 120 }}}
(3) {{{ b +b/2 + 2b = 120 }}}
(3) {{{ (7/2)*b = 120 }}}
(3) {{{ b = (2/7)*120 }}}
(3) {{{ b = 34.29 }}}
and, since:
(1) {{{ j = b/2 }}}
(1) {{{ j = 34.29/2 }}}
(1) {{{ j = 17.15 }}}
also:
(2) {{{ s = 2b }}}
(2) {{{ s = 2*34.29 }}}
(2) {{{ s = 68.58 }}}
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Bill has $34.29
Jim has $17.15
Sue has $68.58
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check:
(3) {{{ b + j + s = 120 }}}
(3) {{{ 34.29 +17.15 + 68.58 = 120 }}}
(3) {{{ 120.02 = 120 }}} ( error due to rounding off )