Solve algebraically for x: 3600 + 1.02x < 2000 + 1.04x

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asked Aug 3, 2021 187k views

5 votes

Solve algebraically for x:
3600 + 1.02x < 2000 + 1.04x

Mathematics
high-school

LoG asked

by LoG

7.7k points

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1 Answer

3 votes

Answer:

x < 80000

Explanation:

$3600 + 1.02x < 2000 + 1.04x \\ ( \: \: \: \: \: \: \: \: \: \: \: \: - 1.04x < \: \: \: \: \: \: \: \: \: \: \: \: \: - 1.04x)/(3600 - 0.02x < 2000) \\ \\ 3600 - 0.02x < 2000 \\ ( - 3600 \: \: \: \: \: \: \: \: \: < - 3600)/( - 0.02x < - 1600) \\ \\ ( - 0.02x)/( - 0.02) < ( - 1600)/( - 0.02) \\ \\ x < 80000$

Kaan Taze answered Aug 7, 2021

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