if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 38. if $ 2,000 is the total amount invested, how much is invested at 8 percent?
Choices.
A) $ 700	B) $ 800	C) $ 900	D) $ 1100	E) $ 1200
<gadget id="calculator">100 * 10</gadget>
<output>1_000</output>

<gadget id="calculator">1_000 * 2</gadget>
<output>2_000</output>

<gadget id="calculator">10 * 8</gadget>
<output>80</output>

<gadget id="calculator">80 * 2</gadget>
<output>160</output>

<gadget id="calculator">160 + 38</gadget>
<output>198</output>

<gadget id="calculator">10 / 100</gadget>
<output>1/10 = around 0.1</output>

<gadget id="calculator">8 / 100</gadget>
<output>2/25 = around 0.08</output>

<gadget id="calculator">(1/10) + (2/25)</gadget>
<output>9/50 = around 0.18</output>

<gadget id="calculator">198 / (9/50)</gadget>
<output>1_100</output>

<gadget id="calculator">2_000 - 1_100</gadget>
<output>900</output>

<result>C</result>
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 38. if $ 2,000 is the total amount invested, how much is invested at 8 percent?
subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 38), add(divide(10, const_100), divide(8, const_100))))
"0.1 x = 0.08 ( 2000 - x ) + 38 0.18 x = 198 x = 1100 then the amount invested at 8 % is $ 2000 - $ 1100 = $ 900 the answer is c ."