Question 540520
Let {{{a}}} = the numerator
Let {{{b}}} = the denominator
given:
(1) {{{ ( a + 2 ) / ( b + 1 ) = 1/2 }}}
(2) {{{ ( a + 1 ) / ( b - 2 ) = 3/5 }}}
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Multiply both sides of (1) by {{{ 2*( b + 1 ) }}}
(1) {{{ 2*( a + 2 ) = b + 1 }}}
(1) {{{ 2a + 4 = b + 1 }}}
(1) {{{ 2a - b = -3 }}}
Multiply both sides of (2) by {{{ 5*( b - 2 ) }}}
(2) {{{ 5*( a + 1 ) = 3*( b - 2 ) }}}
(2) {{{ 5a + 5 = 3b - 6 }}}
(2) {{{ 5a - 3b = -11 }}}
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Multiply both sides of (1) by {{{3}}} and
subtract (1) from (2)
(1) {{{ 6a - 3b = -9 }}}
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(2) {{{ 5a - 3b = -11 }}}
(1) {{{ -6a + 3b = 9 }}}
{{{ -a = -2 }}}
{{{ a = 2 }}}
and, since
(1) {{{ 2a - b = -3 }}}
(1) {{{ 2*2 - b = -3 }}}
(1) {{{ -b = -7 }}}
(1) {{{ b = 7 }}}
the fraction is {{{a/b = 2/7 }}}
check answer:
(1) {{{ ( 2 + 2 ) / ( 7 + 1 ) = 1/2 }}}
(1) {{{ 4/8 = 1/2 }}}
(1) {{{ 1/2 = 1/2 }}}
OK
(2) {{{ ( 2 + 1 ) / ( 7 - 2 ) = 3/5 }}}
(2) {{{ 3/5 = 3/5 }}}
OK