Question 137785
Let x, y  =  amounts  invested
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(1) x+y = $30,000,,,,,x=30,000-y
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(2) .03x +.04y =1020
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substitute  (1)  into  (2)
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.03(30,000-y) + .04y = 1020
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900 - .03y +.04y = 1020
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.01y = 120
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y= $ 12,000,,,,,,,,x=30,000-12,000=  $ 18,000
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check,,,,.03(18,000) + .04(12,000) = 540 +480 = 1020,,,,,,ok