Question

A two-digit number is 4 more than 6 times the sum of its

digit. If 8 is subtracted from the number the digits are

reversed. Find the number​

Answer 1

Let x= ones digit and y=tens digit

According to the question

∴x+10y=6(x+y)+4

⇒10y=6x+6y+4−x

⇒4y=5x+4

And , from if 18 is substracted from the number , the digits are reversed ,

∴x+10y−18=y+10x

⇒10y−18=y+9x

⇒9y−18=9x

⇒y−2=x

Substitute x value into :

4y=5x+4

4y=5(y−2)+4

4y=5y−6

y=6

Substitute y into :

∴y−2=x

⇒6−2=x

⇒4=x

∴ The number is 64.

Answer 2

Answer

Let x= ones digit and y=tens digit

According to the question ,

$∴x+10y=6(x+y)+4 \\ </p><p>⇒10y=6x+6y+4−x \\ </p><p>⇒4y=5x+4$

And , from if 18 is substracted from the number , the digits are reversed ,

$∴x+10y−18=y+10x \\ </p><p>⇒10y−18=y+9x \\ </p><p>⇒9y−18=9x \\ </p><p>⇒y−2=x$

Substitute x value into :

$4y=5x+4 \\ </p><p>4y=5(y−2)+4 \\ </p><p>4y=5y−6 \\ </p><p>y=6</p><p>$

Substitute y into :

$∴y−2=x \\ </p><p>⇒6−2=x \\ </p><p>⇒4=x$

∴ The number is 64.

$hope \: its \: help \: u$