Question

a number 84 is divided into two parts. if the difference between half of the first part and one-third of the second part is 12, find two parts of number.

Answer 1

The first part is 48 and the second part is 36.

Step-by-step explanation:

Let,

The first half = x

The second half = y

According to given statement;

x+y=84       Eqn 1

$\frac{1}{2}x-\frac{1}{3}y=12\ \ \ Eqn\ 2$

Multiplying Eqn 1 by $\frac{1}{2}$

$\frac{1}{2}(x+y=84)\\\\\frac{1}{2}x+\frac{1}{2}y=42\ \ \ Eqn\ 3$

Subtracting Eqn 2 from Eqn 3

$(\frac{1}{2}x+\frac{1}{2}y)-(\frac{1}{2}x-\frac{1}{3}y=42-12\\\\\frac{1}{2}x+\frac{1}{2}y-\frac{1}{2}x+\frac{1}{3}y=30\\\\\frac{1}{2}y+\frac{1}{3}y=30\\\\\frac{3y+2y}{6}=30\\\\\frac{5y}{6}=30$

Multiplying both sides by $\frac{6}{5}$

$\frac{6}{5}*\frac{5}{6}y=30*\frac{6}{5}\\\\y=\frac{180}{5}\\\\y=36$

Putting y=36 in Eqn 1

$x+36=84\\x=84-36\\x=48$

The first part is 48 and the second part is 36.

Keywords: linear equation, elimination method

Learn more about elimination method at:

• brainly.com/question/10046556
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