Finding the Unknown · Lesson 5 of 13
Unknowns on Both Sides
“Bring related terms together while keeping both sides equal.”
• Solve equations with the same unknown on both sides. • Expand brackets before combining terms when useful. • Distinguish a zero solution from an equation with no solution. • Avoid dividing by an unknown that may equal zero.
Collect copies of the unknown
An unknown may appear on both branches of a balance, and it may appear on both sides of an equation. Removing equal groups from both sides still works. In algebra, subtracting the same unknown term from both sides removes those copies together. The aim is to leave unknown terms on one side and ordinary numbers on the other.
In 6y + 7 = 4y + 21, subtracting 4y from both sides leaves 2y + 7 = 21. This is the same reasoning as removing four matching sacks from each branch. The number y is still unknown, but subtracting the same quantity on both sides does not require knowing its weight first.
Problem
Solve 6y + 7 = 4y + 21.
- 1.Subtract 4y from both sides: 6y − 4y + 7 = 21, so 2y + 7 = 21.
- 2.Subtract 7 from both sides: 2y = 14. Divide by 2: y = 7.
- 3.Check the original: LHS = 6 × 7 + 7 = 49; RHS = 4 × 7 + 21 = 49.
Problem
Solve 3u − 7 = 2u + 3.
- 1.Subtract 2u from both sides: u − 7 = 3.
- 2.Add 7 to both sides: u = 10.
- 3.Check: 3 × 10 − 7 = 23 and 2 × 10 + 3 = 23.
Brackets describe a whole group
A multiplier outside brackets multiplies every term inside them. For example, 4(m + 6) means four copies of m + 6, so it equals 4m + 24. Expanding can expose the unknown terms and constant numbers so that like terms can be combined. Like terms have the same letter part: 4m and 2m combine, but 4m and 24 do not.
Like terms have the same letter part and can be combined, such as 4m and 2m. A constant term is a known number without the unknown, such as 24.
The number multiplying the unknown. In 4m, the coefficient of m is 4.
Problem
Solve 4(m + 6) − 8 = 2m − 4.
- 1.Expand the bracket: 4m + 24 − 8 = 2m − 4. Combine the constants: 4m + 16 = 2m − 4.
- 2.Subtract 2m from both sides: 2m + 16 = −4. Subtract 16: 2m = −20.
- 3.Divide by 2: m = −10.
- 4.Check: left = 4(−10 + 6) − 8 = −16 − 8 = −24; right = 2(−10) − 4 = −24.
Problem
Solve 7m = m − 3.
- 1.Subtract m from both sides: 6m = −3.
- 2.Divide both sides by 6: m = −3/6 = −1/2.
- 3.Check: LHS = −7/2; RHS = −1/2 − 3 = −7/2. The fraction balances both sides.
Zero is a solution; a contradiction is not
Sometimes removing unknown terms leaves a multiple of the unknown equal to zero. That gives a zero solution. In other cases all the unknown terms cancel and two unequal ordinary numbers remain. Such a contradiction says that no value of the unknown could have made the starting equation true. Always interpret the simplified statement rather than forcing a familiar-looking answer.
Problem
Solve 5s = 3s.
- 1.Subtract 3s from both sides: 2s = 0.
- 2.Divide by the non-zero number 2: s = 0.
- 3.Check: 5 × 0 = 3 × 0 = 0. Dividing the original by s would wrongly assume s is non-zero and would lose this answer.
Problem
Solve x + 4 = x + 5.
- 1.Subtract x from both sides: 4 = 5.
- 2.This is false, independently of x. Therefore the equation has no solution.
- 3.Adding 4 and adding 5 to the same number always gives results differing by 1. Choosing x = 0, a negative number, or a fraction cannot remove that difference.
Do not divide by an unknown unless you have established that it is non-zero. In 5s = 3s, division by s removes the only solution. Also, no solution is different from the solution 0: zero must still make the original sides equal.
Quiz
What remains after subtracting 4y from 6y + 7 = 4y + 21?
How should 4(m + 6) be expanded?
Which equation has solution zero?
What does 4 = 5 at the end of valid solving steps mean?
What solves k + 8 = 12 − k?
Practice Problems
- Solve k + 8 = 12 − k, checking both sides.
- Solve 3n = 10 + n and verify your answer.
- Solve 8a − 6 = 5a + 9.
- Solve 2(b + 3) = b − 4 and check the brackets in the original.
- Decide whether 9t = 4t has a zero solution or no solution. Explain why dividing by t is unsafe.
- Decide whether z − 7 = z + 2 can have any solution. Explain the constant statement left after subtraction.
- Solve 2x + 4 = 5x − 14. Show an operation on both sides for every line.
- Create an equation with no solution. Explain the contradiction left after the unknown terms cancel.
Key Takeaways
• Unknown terms can be subtracted equally from both sides. • Expand a bracket by multiplying every term inside it. • Combine like terms before isolating the unknown. • A zero solution makes the original equality true. • A false constant statement means no solution; division by an unknown can lose zero.