Question 374126
Let {{{a}}} = amount invested at 9%
Let {{{b}}} = mount invested at 10%
given:
{{{150*12 = 1800}}} interest received during
the year from both investments
(1) {{{a = b + 1000}}}
(2) {{{.09a + .1b = 1800}}}
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Multiply both sides of (2) by {{{100}}}
(2) {{{.09a + .1b = 1800}}}
(2) {{{9a + 10b = 180000}}}
By substitution:
(2) {{{9*(b + 1000) + 10b = 180000}}}
{{{9b + 9000 + 10b = 180000}}}
{{{19b = 180000 - 9000}}}
{{{19b = 171000}}}
{{{b = 9000}}}
And, from (1)
{{{a = 9000 + 1000}}}
{{{a = 10000}}}
$10,000 was invested at 9% and
$9,000 was invested at 10%