### Math Notes

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#### Algebra Solutions

##### Topics || Problems

Divide 80 into two parts such that the ratio of one part, decreased by 4, to the other part, decreased by 8, is 1:3.

Solution:

Let $$x$$ be the first part and $$y$$ be the second part.

$$x+y = 80$$

$$\frac{x-4}{y-8}=\frac{1}{3}$$

$$3(x-4) = y-8$$

$$y = 80-x$$

$$3(x-4) =80-x-8$$

$$3x+x = 72 +12$$

$$x = 21$$

$$y = 80-21 = 59$$

Thus, the parts are 59 and 21