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How do you solve a linear equation in one variable, including ones with variables on both sides, and what does it mean when an equation has no solution or infinitely many?

Solve linear equations in one variable, including those with variables on both sides and with the distributive property, and recognize no-solution and identity cases (Ohio A-REI.3, A-REI.1).

An Ohio Algebra I answer on solving linear equations (A-REI.3): clearing parentheses and fractions, collecting variables on one side, and recognizing equations with no solution or infinitely many solutions.

Generated by Claude Opus 4.811 min answer

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

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Jump to a section
  1. What this topic is asking
  2. The solving routine
  3. Clearing fractions
  4. No solution and infinitely many
  5. How Ohio examines this topic
  6. Why every step preserves the solution
  7. Checking, and what a check catches
  8. Try this

What this topic is asking

Ohio standard A-REI.3 asks you to solve linear equations in one variable, and A-REI.1 asks you to justify the steps. That includes equations with parentheses, fractions, and variables on both sides, plus the special cases of no solution and infinitely many solutions. This is a bedrock skill in the Expressions and Equations reporting category, and it appears on Part 1 where no calculator helps.

The solving routine

The reliable order of operations for solving is the reverse of building an expression.

Clearing fractions

When an equation has fractions, multiply every term by the least common denominator first; this avoids fraction arithmetic later.

For x2+13=56\dfrac{x}{2} + \dfrac{1}{3} = \dfrac{5}{6}, the LCD is 66. Multiplying every term by 66 gives 3x+2=53x + 2 = 5, so 3x=33x = 3 and x=1x = 1.

No solution and infinitely many

Most linear equations have exactly one solution, but two special cases arise when the variable cancels.

The key is to finish the algebra: if you ever reach a statement with no variable, judge whether it is true or false to decide between infinitely many and none.

How Ohio examines this topic

  • Equation response. Type the single solution value.
  • Multiple choice. Decide how many solutions an equation has, or pick the correct value.
  • Drag and drop. Order the solving steps, or match an equation to "one / none / infinitely many."

Because Part 1 bans the calculator, the arithmetic, especially distributing a negative and combining signed terms, must be clean.

Why every step preserves the solution

A-REI.1 is really asking "why are these steps allowed?" The answer is that each move, adding the same amount to both sides, multiplying both sides by the same nonzero number, distributing, applies an operation equally to both sides, so any value that made the original equation true still makes the new one true, and vice versa. That is why solving never changes the solution set: you are rewriting the same balance in a simpler form. The one move to watch is multiplying or dividing by a variable expression that could be zero, which is why for linear equations we only ever multiply by known nonzero numbers (like a common denominator). Understanding this justification is what lets you trust a chain of steps and explain it on a reasoning item.

Checking, and what a check catches

Substituting your answer back into the original equation is the fastest insurance on a no-calculator part. A check catches a distribution slip or a sign error immediately: if the two sides do not match, you know to retrace. It is especially worth doing when you distributed a negative, like βˆ’(xβˆ’3)=βˆ’x+3-(x - 3) = -x + 3, since dropping that sign flip is the single most common mistake in this topic. Build the habit of a ten-second check on every equation-response item.

Try this

Q1. Solve 4xβˆ’7=2x+94x - 7 = 2x + 9. [2 points]

  • Cue. 2x=162x = 16, so x=8x = 8.

Q2. How many solutions does 3(x+1)=3x+33(x + 1) = 3x + 3 have? [1 point]

  • Cue. 3x+3=3x+33x + 3 = 3x + 3 gives 3=33 = 3 (true), so infinitely many.

Exam-style practice questions

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

Ohio Algebra I EOC (style)2 marksEquation response. Solve for xx: 3(xβˆ’4)=2x+53(x - 4) = 2x + 5.
Show worked answer β†’

The solution is x=17x = 17.

Distribute the 33: 3xβˆ’12=2x+53x - 12 = 2x + 5. Subtract 2x2x from both sides: xβˆ’12=5x - 12 = 5. Add 1212: x=17x = 17. Check: 3(17βˆ’4)=3(13)=393(17 - 4) = 3(13) = 39 and 2(17)+5=392(17) + 5 = 39, so it works. Clearing the parentheses first, then collecting the variable on one side, is the standard route, and equation-response items expect the single value.

Ohio Algebra I EOC (style)1 marksMultiple choice. How many solutions does 2(x+3)=2x+72(x + 3) = 2x + 7 have? (A) none (B) one (C) two (D) infinitely many
Show worked answer β†’

The correct answer is (A).

Distribute: 2x+6=2x+72x + 6 = 2x + 7. Subtract 2x2x from both sides: 6=76 = 7, which is false. A false numeric statement means no solution, because no value of xx can make 66 equal 77. If both sides had reduced to a true statement like 6=66 = 6, there would be infinitely many solutions instead.

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