# Chapter 6 - Algebra: Equations and Inequalities - 6.3 Applications of Linear Equations - Exercise Set 6.3 - Page 373: 4

The number is $40$.

#### Work Step by Step

step 1 Let x represent the unknown number in the problem. step 2 "$30\%$ of itself" = $\displaystyle \frac{30}{100}\cdot x=0.3x$ " ... x is decreased by $30\%$ of itself" = $x-0.3x$ " the result is $28$" is written as " = $28$" step 3 The equation is $x-0.3x=28$ step 4 $x-0.3x=28\qquad$/simplify LHS $0.7x=28\qquad/\div 0.7$ $x=\displaystyle \frac{28}{0.7}=\frac{280}{7}=40$ The number is $40$. step 5 Check the solution in the original wording of the problem. The number ($40$), decreased by 30% of 40, ($\displaystyle \frac{30}{100}\cdot 40=12),$ ($40-12$) gives the result $28$. OK.

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