Solving Linear Equations

Subtracting 7 from both sides gives \( 3x = \;-6 \). Dividing both sides by 3 gives \( x = \;-2 \). We can verify: \( 3(-2) + 7 = \;-6 + 7 = 1 \) ✓. The solution is negative because the constant on the right (1) is smaller than the constant being added on the left (7), so \( x \) must compensate by being negative.

A useful way to think about it: if \( x \) were 0, the left side would be 7, which is already too big. To bring \( 3x + 7 \) down to just 1, we need \( x \) to be negative so that \( 3x \) pulls the total below 7.

Expanding the bracket in \( 2(x + 3) \) means multiplying each term inside by 2: \( 2 \times x + 2 \times 3 = 2x + 6 \). So the left-hand side of the first equation is identical to the left-hand side of the second — they are literally the same equation written differently.

As the visual shows, two groups of \( (x+3) \) cover the exact same length as two \( x \)’s and one \( 6 \).

We can test \( x = 3 \) by substituting into both sides. Left side: \( 5(3) \; – \; 3 = 15 \; – \; 3 = 12 \). Right side: \( 2(3) + 9 = 6 + 9 = 15 \). Since \( 12 \neq 15 \), the two sides are not equal when \( x = 3 \), so \( x = 3 \) is not a solution. A solution must make both sides equal.

The actual solution is found by collecting \( x \) terms: \( 5x \; – \; 2x = 9 + 3 \), giving \( 3x = 12 \), so \( x = 4 \). Check: \( 5(4) \; – \; 3 = 17 \) and \( 2(4) + 9 = 17 \) ✓.

Notice that every term in \( x + 5 = 12 \) has been multiplied by 2 to produce \( 2x + 10 = 24 \): the \( x \) becomes \( 2x \), the 5 becomes 10, and the 12 becomes 24. Multiplying both sides of an equation by the same non-zero number always produces an equivalent equation, because if two things are equal, doubling both of them keeps them equal.

We can verify: \( x + 5 = 12 \) gives \( x = 7 \). And \( 2x + 10 = 24 \) gives \( 2x = 14 \), so \( x = 7 \). Both have the same solution ✓.

Example: \( 2x + 3 = 13 \) — check: \( 2(5) + 3 = 10 + 3 = 13 \) ✓

Another: \( x \; – \; 7 = \;-2 \) — check: \( 5 \; – \; 7 = \;-2 \) ✓

Creative: \( 3(x + 1) = 18 \) — uses brackets and still works. Check: \( 3(5 + 1) = 3(6) = 18 \) ✓. Students often don’t think of building equations with brackets.

Trap: \( x + 5 = 0 \) — this looks like it involves 5, and students may confuse “5 appears in the equation” with “\( x = 5 \) is the solution.” But actually \( x + 5 = 0 \) gives \( x = \;-5 \), not \( x = 5 \).

Example: \( 2x + 1 = 4 \) → \( 2x = 3 \) → \( x = \frac{3}{2} \). Check: \( 2 \times \frac{3}{2} + 1 = 3 + 1 = 4 \) ✓

Another: \( 5x = 3 \) → \( x = \frac{3}{5} \). Check: \( 5 \times \frac{3}{5} = 3 \) ✓

Creative: \( \frac{2x}{3} = 4 \) — here the equation contains a fraction, but the solution is an integer (\( 2x=12 \rightarrow x=6 \)). This is a great way to challenge the assumption that “fractions in the equation mean fractional answers”.

Trap: \( 6x + 3 = 9 \) — students might expect a fractional answer because 6 and 9 don’t obviously simplify, but actually \( 6x = 6 \), so \( x = 1 \) (a whole number). Looking “messy” doesn’t guarantee a fractional solution.

Example: \( x + 5 = 2 \) → \( x = \;-3 \). Check: \( -3 + 5 = 2 \) ✓

Another: \( 7x + 20 = 6 \) → \( 7x = \;-14 \) → \( x = \;-2 \). Check: \( 7(-2) + 20 = \;-14 + 20 = 6 \) ✓

Creative: \( 5 \; – \; 2x = 11 \) — the \( x \) term is subtracted, making it trickier to solve: \( -2x = 6 \), so \( x = \;-3 \). Check: \( 5 \; – \; 2(-3) = 5 + 6 = 11 \) ✓. Many students struggle when \( x \) has a negative coefficient.

Trap: \( -3x = \;-15 \) — students see the negative signs and assume the solution must be negative, but dividing two negatives gives a positive: \( x = (-15) \div (-3) = 5 \). The solution is positive!

Example: \( x + 3 = 7 \) and \( 2x \; – \; 1 = 7 \) — both have \( x = 4 \). Check: \( 4 + 3 = 7 \) ✓ and \( 2(4) \; – \; 1 = 7 \) ✓

Another: \( 5x = 30 \) and \( 3x \; – \; 2 = 16 \) — both have \( x = 6 \). Check: \( 5(6) = 30 \) ✓ and \( 3(6) \; – \; 2 = 16 \) ✓

Creative: \( \frac{x}{2} + 1 = 4 \) and \( 4(x \; – \; 1) = 20 \) — one uses a fraction, the other uses brackets, yet both give \( x = 6 \). Check: \( \frac{6}{2} + 1 = 3 + 1 = 4 \) ✓ and \( 4(6 \; – \; 1) = 4(5) = 20 \) ✓. Students rarely expect equations of such different forms to share a solution.

Trap: \( 2x + 4 = 10 \) and \( 2x + 6 = 14 \) — these look very similar (both start with \( 2x \), both have small constants), so students might assume they share a solution. But \( 2x + 4 = 10 \) gives \( x = 3 \), and \( 2x + 6 = 14 \) gives \( x = 4 \). Similar-looking equations can have different solutions.

This is only sometimes true — it depends on the equation. Some equations have positive solutions, some have negative solutions, and some have \( x = 0 \).

True case: \( 2x + 1 = 7 \) gives \( 2x = 6 \), so \( x = 3 \) (positive). False case: \( 3x + 10 = 1 \) gives \( 3x = \;-9 \), so \( x = \;-3 \) (negative). Also: \( 4x + 5 = 5 \) gives \( 4x = 0 \), so \( x = 0 \) (neither positive nor negative).

This is always true — it is a fundamental property of equality. If two things are equal, adding the same amount to both keeps them equal: if \( a = b \), then \( a + c = b + c \) for any value of \( c \). This works regardless of whether \( c \) is positive, negative, zero, a fraction, or even an algebraic expression.

Students sometimes doubt this when the number being added is negative (but adding a negative is the same as subtracting, which also preserves equality) or when it’s a fraction (but fractions are just numbers, so the same principle applies).

This is only sometimes true. Many equations with \( x \) on both sides do have a solution — and in fact most that students encounter in the classroom will have exactly one solution.

True case (no solution): \( x + 1 = x + 3 \) simplifies to \( 1 = 3 \), which is a contradiction, so there is no value of \( x \) that works. False case (has a solution): \( 2x + 1 = x + 4 \) gives \( 2x \; – \; x = 4 \; – \; 1 \), so \( x = 3 \). There’s even a third possibility: \( 3x + 2 = 3x + 2 \) simplifies to \( 0 = 0 \), which is true for all values of \( x \) (infinitely many solutions).

This is never true. Finding \( x = 0 \) means the equation does have a solution — that solution is zero. The value zero is a perfectly valid number that \( x \) can take. “No solution” means there is no value of \( x \) that makes the equation true; “\( x = 0 \)” means the value zero is the one that works.

Analogy: Saying “the solution is 0” is like saying the temperature is 0°C. That is a specific, valid temperature. Saying “no solution” is like saying the thermometer doesn’t exist.

The student has made the “partial division” error — they divided the \( 2x \) term by 2 (getting \( x \)) and divided 22 by 2 (getting 11), but forgot to also divide the 8 by 2. When dividing both sides of an equation by a number, every term on each side must be divided.

The correct step would be: \( (2x + 8) \div 2 = 22 \div 2 \), giving \( x + 4 = 11 \), so \( x = 7 \).

The student has the right answer but uses the “Difference Method” misconception. They are just finding the gap between the numbers (\(4\) and \(1\)) without respecting the negative sign. The actual distance between \(4\) and \(-1\) is indeed 5, but the student’s logic of “just add them” will fail if the signs were different.

Consider \( 2x – 4 = 3x + 1 \): the student’s method would give \( 4+1=5 \), but the correct answer is \( x = -5 \). The signs matter!

The student made the “multiply only the first term” error when expanding the bracket. They correctly multiplied \( x \) by 5 to get \( 5x \), but left the \( +2 \) unchanged instead of also multiplying it by 5. The correct expansion of \( 5(x + 2) \) is \( 5x + 10 \), not \( 5x + 2 \). Both terms inside the bracket must be multiplied.

The correct solution: \( 5(x + 2) = 30 \) → \( 5x + 10 = 30 \) → \( 5x = 20 \) → \( x = 4 \). Check: \( 5(4 + 2) = 5(6) = 30 \) ✓. Alternatively, dividing both sides by 5 first: \( x + 2 = 6 \), so \( x = 4 \).

The student made the “ignoring the negative coefficient” error. The coefficient of \( x \) is \( -3 \), so we must divide both sides by \( -3 \). The correct solution is \( x = 12 \div (-3) = \;-4 \).

Sanity Check: We know \( 3 \times 4 = 12 \). If we change 3 to -3, the answer cannot still be 4. It must change sign too.