Solve linear equations with rational coefficients by expanding and collecting like terms

Learning goal: Solve linear equations in one variable whose coefficients are rational numbers (fractions, decimals and negatives), including equations whose solutions require expanding expressions using the distributive property and collecting like terms, and check every solution by substitution.

Rational number coefficients. A coefficient is the number multiplying a variable. Rational coefficients include fractions such as 2/3, decimals such as 0.4, and negatives such as −3. The solving moves are the same as with whole numbers, because adding, subtracting, multiplying and dividing both sides by the same nonzero number preserves equivalence whatever kind of number it is. What changes is the arithmetic: adding fractions needs a common denominator, dividing by a fraction means multiplying by its reciprocal, and a negative sign in front of parentheses reverses every sign inside.

Expanding with the distributive property. The distributive property says a(b + c) = ab + ac and a(b − c) = ab − ac. Expanding an expression means using it to remove parentheses. Every term inside the parentheses is multiplied, not just the first one: 4(x − 3) is 4x − 12, not 4x − 3. With a negative multiplier, −2(3x − 4) = −6x + 8, because −2 times −4 is +8. A subtraction sign in front of parentheses acts like multiplying by −1: 5 − (2x + 7) = 5 − 2x − 7.

Collecting like terms. Like terms have the same variable part: 3x and −(1/2)x are like terms, and 7 and −4 are like terms, but 3x and 7 are not. Collect them by adding their coefficients: (2/3)x − (1/2)x = (4/6 − 3/6)x = (1/6)x. Collect on each side separately before moving terms across the equals sign, so that each side is as simple as possible.

A fully worked model. Solve (2/3)(x + 6) − (1/2)x = 10. Expand the product: (2/3)x + (2/3)(6) = (2/3)x + 4, so the equation is (2/3)x + 4 − (1/2)x = 10. Collect like terms on the left: (2/3 − 1/2)x = (4/6 − 3/6)x = (1/6)x, giving (1/6)x + 4 = 10. Subtract 4 from both sides: (1/6)x = 6. Multiply both sides by 6, the reciprocal of 1/6: x = 36. Check in the original equation: (2/3)(36 + 6) = (2/3)(42) = 28, and (1/2)(36) = 18, so the left side is 28 − 18 = 10, which matches the right side.

The same equation by clearing denominators. An alternative first step multiplies both sides by the least common denominator of all the fractions, here 6. Multiplying by 6 is legitimate because 6 is nonzero. The equation becomes 6 × (2/3)(x + 6) − 6 × (1/2)x = 60, which is 4(x + 6) − 3x = 60. Expand: 4x + 24 − 3x = 60. Collect: x + 24 = 60. Subtract 24: x = 36, the same answer. Clearing denominators trades fraction arithmetic for larger whole numbers; either route is correct, and doing both is a strong check.

Decimals. Solve 0.4(x − 5) + 1.2 = 0.1x + 3.4. Expand: 0.4x − 2 + 1.2 = 0.1x + 3.4, since 0.4 × 5 = 2. Collect on the left: 0.4x − 0.8 = 0.1x + 3.4. Subtract 0.1x: 0.3x − 0.8 = 3.4. Add 0.8: 0.3x = 4.2. Divide by 0.3: x = 14. Check: 0.4(14 − 5) + 1.2 = 0.4(9) + 1.2 = 3.6 + 1.2 = 4.8, and 0.1(14) + 3.4 = 1.4 + 3.4 = 4.8. Multiplying both sides by 10 at the start, to get 4(x − 5) + 12 = x + 34, is an equally valid route and gives 4x − 8 = x + 34, 3x = 42, x = 14.

Negatives in front of parentheses. Solve 5 − 2(3x − 4) = −3(x + 1) + 25. Expand carefully: 5 − 6x + 8 = −3x − 3 + 25. Collect each side: 13 − 6x = 22 − 3x. Add 6x to both sides: 13 = 22 + 3x. Subtract 22: −9 = 3x. Divide by 3: x = −3. Check the left side: 5 − 2(3(−3) − 4) = 5 − 2(−9 − 4) = 5 − 2(−13) = 5 + 26 = 31. Right side: −3(−3 + 1) + 25 = −3(−2) + 25 = 6 + 25 = 31. Both sides equal 31.

Misconceptions to avoid. Distributing to only the first term inside the parentheses is the most frequent error; the second most frequent is dropping the sign change when the multiplier is negative. Students also combine unlike terms, writing 3x + 4 as 7x. And when the coefficient is a fraction, some divide by its numerator only; dividing by 1/6 means multiplying by 6. The check by substitution catches every one of these errors, which is why it is part of the solution and not an optional extra. The three possibilities from the previous lesson still apply here: after expanding and collecting, an equation with rational coefficients can end at x = a, at a = a, or at a false statement.

Worked fraction coefficients on both sides: (3/4)x − 5 = (1/4)x + 7. Subtract (1/4)x from both sides: (3/4 − 1/4)x − 5 = 7, which is (1/2)x − 5 = 7. Add 5: (1/2)x = 12. Multiply by 2: x = 24. Check: (3/4)(24) − 5 = 18 − 5 = 13 and (1/4)(24) + 7 = 6 + 7 = 13.

Worked equation with fractions as whole sides: (x + 2)/3 = (2x − 1)/4. Multiply both sides by 12, the least common denominator: 4(x + 2) = 3(2x − 1). Expand: 4x + 8 = 6x − 3. Subtract 4x: 8 = 2x − 3. Add 3: 11 = 2x. Divide by 2: x = 11/2 = 5.5. Check: (5.5 + 2)/3 = 7.5/3 = 2.5 and (2(5.5) − 1)/4 = 10/4 = 2.5.

Worked negative fraction multiplier: −(1/2)(4x − 10) + 3x = 2(x + 4) − 1. Expand the left: −(1/2)(4x) = −2x and −(1/2)(−10) = +5, giving −2x + 5 + 3x. Collect: x + 5. Expand the right: 2x + 8 − 1 = 2x + 7. Now x + 5 = 2x + 7. Subtract x: 5 = x + 7. Subtract 7: x = −2. Check the left: −(1/2)(4(−2) − 10) + 3(−2) = −(1/2)(−18) − 6 = 9 − 6 = 3. Right: 2(−2 + 4) − 1 = 2(2) − 1 = 3.

Worked decimal identity and impossibility: 1.5(2x + 4) = 3x + 6 expands to 3x + 6 = 3x + 6, so subtracting 3x gives 6 = 6 and every number is a solution. By contrast 2.5x − 1 = 2.5x + 0.5 gives −1 = 0.5 after subtracting 2.5x, so it has no solutions. Rational coefficients do not change the three possibilities; they only change the arithmetic on the way there.

Representation task: Solve (2/3)(x + 6) − (1/2)x = 10 twice: once by working with the fractions directly and once by first multiplying both sides by 6. Write the two solutions side by side, line by line, and draw an arrow between the corresponding steps. State which step in each route uses the distributive property and which collects like terms. Finish with a substitution check written out in full.

Independent performance: Write three original equations, each containing at least one fractional or decimal coefficient and at least one set of parentheses with a negative multiplier. Exactly one of the three should have no solutions and exactly one should have infinitely many. Solve all three showing every transformation, and for the one-solution equation check by substituting into the original equation. Exchange with a partner, who solves your equations by a different route (clearing denominators versus not) and confirms the results. Success criteria: each expansion multiplies every term inside the parentheses; sign changes with negative multipliers are correct; like terms are collected on each side before terms are moved; the check uses the original equation and shows the arithmetic; the solution set of each equation is stated as one number, every number, or no number.