Question Video: Creating Linear Equations with One Variable Mathematics • 7th Grade

Video Transcript

The ages of four siblings are consecutive multiples of three, and the sum of their ages is 78. Write an equation that can be used to find 𝑛, the age of the oldest sibling.

We’re told in the question to let the age of the oldest sibling equal 𝑛. As the ages of the four siblings are consecutive multiples of three, we know that the second oldest sibling is three years younger than the oldest. Likewise, the third oldest is three years younger than the second oldest. And finally, the youngest sibling is three years younger than the third oldest.

Subtracting three from 𝑛 gives us 𝑛 minus three. Therefore, the age of the second oldest sibling is 𝑛 minus three. Subtracting three from this gives us 𝑛 minus six as negative three minus three is negative six. The age of the third oldest sibling is 𝑛 minus six.

Finally, the age of the youngest sibling is 𝑛 minus nine as 𝑛 minus six minus three equals 𝑛 minus nine. We were also told that the sum of their ages is 78. This means that when we add 𝑛, 𝑛 minus three, 𝑛 minus six, and 𝑛 minus nine, we get a total of 78. Therefore, the equation that can be used to find 𝑛 — the age of the oldest sibling — is 𝑛 plus 𝑛 minus three plus 𝑛 minus six plus 𝑛 minus nine is equal to 78.

Whilst we weren’t asked in this question to solve the equation, we will show you how this can be done to work out the ages of the four children. Simplifying the left-hand side of the equation by collecting or grouping like terms gives us four 𝑛 minus 18. 𝑛 plus 𝑛 plus 𝑛 plus 𝑛 is equal to four 𝑛. And negative three minus six minus nine is equal to negative 18.

The equation can, therefore, be simplified to four 𝑛 minus 18 equals 78. Adding 18 to both sides of this equation gives us four 𝑛 is equal to 96 as 78 plus 18 equals 96. Dividing both sides of this equation by four gives us a value of 𝑛 equal to 24. 96 divided by four equals 24.

As 𝑛 was equal to the age of the oldest sibling, the oldest sibling must be 24 years of age. The second oldest sibling must be 21 as 24 minus three equals 21. 24 minus six is equal to 18. Therefore, the third oldest sibling is 18. Finally, 24 minus nine is equal to 15. Therefore, the youngest sibling is 15 years of age.

These four numbers are consecutive multiples of three. 15, 18, 21, and 24 are consecutive numbers in the three times table. Adding the four numbers gives us a total of 78. Therefore, the sum of their ages is indeed 78.

Using these checks prove that our equation 𝑛 plus 𝑛 minus three plus 𝑛 minus six plus 𝑛 minus nine equals 78 was correct.