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How do you solve a linear equation in one variable, and how does each step follow from the properties of equality?

Solve linear equations in one variable, including those with variables on both sides and with coefficients represented by letters, justifying each step (TN A1.A.REI.A.1, A1.A.REI.B.3).

A TNReady Algebra I answer on solving linear equations (TN A1.A.REI.A.1, B.3), the properties of equality, clearing fractions, variables on both sides, and recognizing no-solution and identity cases.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The properties of equality
  3. A solving routine
  4. No solution and infinitely many solutions
  5. How TNReady examines this topic
  6. Why every step is reversible
  7. A worked equation with variables on both sides
  8. Try this

What this topic is asking

Two standards combine here. A1.A.REI.B.3 is the procedure, solve a linear equation in one variable, including with the variable on both sides and with letter coefficients. A1.A.REI.A.1 is the reasoning, explain each step as following from the properties of equality. On TNReady, you both solve and (on some items) identify which property justifies a step.

The properties of equality

Every solving step is one of these, applied to both sides at once.

A1.A.REI.A.1 may show a worked solution and ask which property justifies a particular line, so name the move, not just the number.

A solving routine

No solution and infinitely many solutions

When you simplify and the variable disappears, read the leftover statement:

  • A false numeric statement (6=βˆ’16 = -1) means the equation is never true: no solution.
  • A true numeric statement (6=66 = 6) means the equation is always true: infinitely many solutions (an identity).

These appear as quick multiple-choice items, and the key is to keep simplifying until the variable cancels, then judge the remaining statement.

How TNReady examines this topic

  • Numeric response. Solve for the variable and enter the exact value, including fractions.
  • Multiple choice. Identify the number of solutions, or which property justifies a step.
  • Subpart 1 (no calculator). Linear solving is a core no-calculator fluency skill.

A clarifying idea is that letter coefficients are handled exactly like numbers: to solve ax+b=cax + b = c for xx, subtract bb and divide by aa, giving x=cβˆ’bax = \frac{c - b}{a}. This bridges directly to the literal-equations standard, where the whole equation is in letters.

Why every step is reversible

The properties of equality work because each is reversible: if you add 33 to both sides, you can subtract 33 to get back, so the solution set never changes. This is the deep reason a check should always succeed, the steps only rewrite the same equation in simpler form. The one operation that can break this is multiplying or dividing by an expression that might be zero, which is why the multiplication property specifies a nonzero quantity. In pure linear equations you divide only by the numeric coefficient, so this rarely bites, but it becomes important with rational equations where a denominator could vanish.

A worked equation with variables on both sides

Variables on both sides are the most common Subpart 1 format, and the move is always to gather them on one side.

It usually saves arithmetic to move the smaller variable term, so the variable you keep stays positive. Subtracting 2x2x rather than 7x7x above kept the coefficient positive, avoiding a sign flip.

Try this

Q1. Solve 3(2x+1)=4x+113(2x + 1) = 4x + 11. [2 points]

  • Cue. 6x+3=4x+11β‡’2x=8β‡’x=46x + 3 = 4x + 11 \Rightarrow 2x = 8 \Rightarrow x = 4.

Q2. How many solutions does 2(x+4)=2x+82(x + 4) = 2x + 8 have? [1 point]

  • Cue. 2x+8=2x+8β‡’8=82x + 8 = 2x + 8 \Rightarrow 8 = 8, true: infinitely many solutions.

Exam-style practice questions

Practice questions written in the style of TDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TNReady (style)2 marksNumeric response. Solve for xx: 5(xβˆ’2)=3x+85(x - 2) = 3x + 8.
Show worked answer β†’

The solution is x=9x = 9.

Distribute first: 5xβˆ’10=3x+85x - 10 = 3x + 8. Subtract 3x3x from both sides to gather variables: 2xβˆ’10=82x - 10 = 8. Add 1010: 2x=182x = 18. Divide by 22: x=9x = 9. Each step uses a property of equality, doing the same operation to both sides, which is what A1.A.REI.A.1 asks you to justify. Checking: 5(9βˆ’2)=355(9 - 2) = 35 and 3(9)+8=353(9) + 8 = 35, so the solution is correct.

TNReady (style)1 marksMultiple choice. How many solutions does 4x+6=4xβˆ’14x + 6 = 4x - 1 have? (A) no solution (B) one solution (C) two solutions (D) infinitely many solutions
Show worked answer β†’

The correct answer is (A).

Subtract 4x4x from both sides: 6=βˆ’16 = -1, a false statement. When the variable cancels and a false statement remains, the equation has no solution (the two sides never meet). If a true statement like 6=66 = 6 remained, there would be infinitely many solutions (an identity). Recognizing these special cases is a frequent quick item.

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