Bill's Two Investments 33) With \$35,000, Bill made two investments. Since one had paid 8% interest, and the other 12%, his total intererest-earning from both was \$3,920, for the year. How much had he initially invested at each percent? x = 8% investment y = 12% investment x + y = 35,000 (The two investments and their total amount.) .08(x) + .12(y) = \$3,920 (The interest-earning of each investment, and their combined annual yield.) 8x + 12y = 392,000 (We eliminated the decimal places by multiplying by 100. x + y = 35,000 8x + 12y = 392,000 -8x -8y = -280,000 (We multiplied the above equation by -8.) 8x +12y = 392,000 ------------------ 4y = 112,000 y = 28,000 (We divided both sides by 4 to find the value of y.) Plugging the value of y (28,000) into the above formula, we can now solve for the value of x: x + 28,000 = 35,000 x = 7,000 So, now we know that Bill had invested as follows: \$28,000 (at 12%) = \$3,360 (interest earned) \$7,000 (at 8%) = \$560 (interest earned) ------------------------- \$35,000 \$3,920 (combined interest earned) We can also divide \$3,920 by 35,000 to find his interest average for both investments: 11.2%