Question

Solve the equation and check. (n-5)/(6)+(n+3)/(7)=(11)/(21)

Answer 1

Hello!

Answer:

`\Large \boxed{\sf n = 3}`

Explanation:

→ We want to solve this equation:

`\sf (n-5)/(6) +(n+3)/(7) =(11)/(21)`

◼ Find the common denominator:

→ Find the LCM (Least Common Multiple) of 6; 7; 21:

6: 6, 12, 18, 24, 30, 36, 42, 48, 64, 70...

7: 7, 14, 21, 29, 36, 42, 49, 56, 63, 70...

21: 21, 42, 63, 84, 95...

→ Therefore, the LCM of 6; 7; 21 is 42.

`\sf 42 = 6 * 7 \\42 = 2 * 21`

◼ Now let's convert each denominator to 42:

`\sf ((n-5)* 7)/(6* 7) +((n+3)* 6)/(7* 6) =(11* 2)/(21 * 2 )`

◼ Simplify the fractions:

`\sf (7n - 35)/(42) +(6n+18)/(42) =(22)/(42 )`

◼ Add the fractions:

`\sf (7n - 35+6n+18)/(42) =(22)/(42 )`

◼ Simplify the fraction:

`\sf (13n - 17)/(42) =(22)/(42 )`

◼ Multiply both sides by 42:

`\sf (13n - 17)/(42)} * 42 =(22)/(42 ) * 42`

◼ Simplify both sides:

`\sf 13n - 17 =22}`

◼ Add 17 to both sides:

`\sf 13n - 17 +17 =22+17`

◼ Simplify both sides:

`\sf 13n =39`

◼ Divide both sides by 13:

`\sf (13n)/(13) = (39)/(13)`

◼ Simplify both sides:

`\boxed{\sf n = 3}`

Check:

→ Let's replace n by 3 in the equation:

`\sf (3-5)/(6) +(3+3)/(7) =(11)/(21)`

→ Verify if the egality is true:

`\sf (-2)/(6) +(6)/(7) \\\\\\= (-2* 7)/(6* 7 ) +(6 * 6)/(7* 6) \\\\\\= (-14)/(42 ) +(36)/(42)\\\\\\= (-14+36)/(42)\\\\\\= (22)/(42)\\\\\\= (22 / 2 )/(42 / 2)\\\\\\= (11)/(21)`

Therefore, the egality is true if n = 3.

Conclusion:

The solution of the equation
`\sf (n-5)/(6) +(n+3)/(7) =(11)/(21)` is
`\sf 3`.