# Algebra

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an investment of 72,000 was made by a business club. the investment was split into three parts and lasted for one year. the first part of the investment earned 8% interest. the second 6% and the third 9%. total interest from the investments was \$5640. the interest from the first investment was 3 times the interest from the second. find the amounts of three parts of the investment

• Algebra -

\$X-in 1st inv.
\$Y-in 2nd inv.
\$Z-in 3rd inv.

0.08X + 0.06Y + 0.09Z = \$5640.

Eq1: 0.08X + 0.06Y + 0.09Z = 5640.
Eq2: X + Y + Z = 72000.
Multiply Eq1 by 100 and Eq2 by -9:
8X + 6Y + 9Z = 564000,
-9X - 9Y - 9Z = - 648000,
-X - 3Y = -84,000,

0.08X = 3*0.06Y,
X = 2.25Y
Substitute 2.25Y for X:
-2.25Y - 3Y = -84000,
-5.25Y = -84000,

Y = \$16000.

X = 2.25Y = 2.25 * 16000 = \$36,000.

X + Y + Z = \$72000,
36000 + 16000 + Z = 72000,

Z = \$20,000.

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