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How do you solve multi-step linear equations in one variable, including those with variables on both sides, and identify equations with one, no, or infinitely many solutions?

Solve multi-step linear equations in one variable, including equations with the variable on both sides and with rational-number coefficients, and classify an equation as having one solution, no solution, or infinitely many solutions (A.EI.1).

A Virginia SOL Algebra I answer on A.EI.1: the balance method, clearing fractions, variables on both sides, modeling with linear equations, and identifying one, no, or infinitely many solutions.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. The balance method
  3. Fractions and variables on both sides
  4. One, no, or infinitely many solutions
  5. Modeling with a linear equation
  6. Why a canceled variable signals the special cases
  7. How the SOL examines this topic
  8. Try this

What this topic is asking

A.EI.1 asks you to solve linear equations in one variable: multi-step, with rational-number coefficients, and with the variable on both sides. You also classify an equation as having one solution, no solution, or infinitely many solutions. On the Virginia Algebra I SOL these are core Equations and Inequalities items, mostly fill-in-the-blank (type the value) and multiple choice.

The balance method

An equation stays true if you do the same thing to both sides. To isolate xx, undo operations in reverse order:

3x+7=22    3x=15    x=5.3x + 7 = 22 \implies 3x = 15 \implies x = 5.

Subtract first (undo the +7+7), then divide (undo the ×3\times 3). Always reverse the order of operations when solving.

Fractions and variables on both sides

Clear fractions by multiplying every term by the least common denominator. For x2+3=x4+5\frac{x}{2} + 3 = \frac{x}{4} + 5, multiply by 44: 2x+12=x+202x + 12 = x + 20, then solve normally to x=8x = 8.

When the variable is on both sides, gather it on one side first.

One, no, or infinitely many solutions

After simplifying, the variable sometimes cancels:

  • If you reach a true statement with no variable (such as 6=66 = 6), every value of xx works: infinitely many solutions.
  • If you reach a false statement (such as 2=92 = 9), no value works: no solution.
  • If you reach x=x = a number, there is exactly one solution.

Modeling with a linear equation

Many SOL items hand you a situation and expect you to build and solve the equation. Name the unknown, translate each phrase into algebra, and solve. "A gym charges a \30joiningfeeplus joining fee plus \1515 per month; after how many months is the total \135?"becomes?" becomes 30 + 15m = 135,so, so 15m = 105and and m = 7months.Thefixedamountisaconstant,theperunitamountmultipliesthevariable,andthetotaliswhattheexpressionequals.Definingthevariableinwordsfirst,"let months. The fixed amount is a constant, the per-unit amount multiplies the variable, and the total is what the expression equals. Defining the variable in words first, "let m$ be the number of months," keeps the translation honest.

Why a canceled variable signals the special cases

The no-solution and infinite-solution cases are not exceptions to the method, they are what the method reports when the two sides are secretly the same kind of expression. If both sides simplify to the same line (same slope and same intercept), they are equal for every xx, so the variable cancels and a true statement remains: infinitely many solutions. If both sides have the same slope but different intercepts (parallel, never equal), the variable cancels but the constants disagree, a false statement: no solution. Geometrically, you are asking where two lines meet: identical lines meet everywhere, parallel lines meet nowhere, and lines with different slopes meet once. Seeing the algebra as a question about line intersections explains all three outcomes at once, and it connects this topic directly to solving systems of equations.

How the SOL examines this topic

  • Fill-in-the-blank. Solve for xx and type the value, including fractions.
  • Multiple choice. Count the solutions (one, none, infinitely many), or pick the solution.
  • Drag-and-drop. Order the steps of a solution, or drag the value that makes the equation true.

A clarifying idea: solving is just undoing, in reverse order, whatever was done to build the expression around xx. If xx was multiplied then had a number added, you subtract first and divide last, peeling the layers off in the opposite order they were applied.

Try this

Q1. Solve 2(3x1)=4x+102(3x - 1) = 4x + 10. [2 points]

  • Cue. 6x2=4x+102x=12x=66x - 2 = 4x + 10 \Rightarrow 2x = 12 \Rightarrow x = 6.

Q2. How many solutions does 3x+5=3x+53x + 5 = 3x + 5 have? [1 point]

  • Cue. The sides are identical, so infinitely many.

Exam-style practice questions

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

SOL (style)2 marksFill in the blank. Solve 5x3=2x+125x - 3 = 2x + 12 for xx and type the value.
Show worked answer →

The solution is x=5x = 5.

Collect the variable on one side: subtract 2x2x from both sides to get 3x3=123x - 3 = 12. Add 33: 3x=153x = 15. Divide by 33: x=5x = 5. Check by substituting: 5(5)3=225(5) - 3 = 22 and 2(5)+12=222(5) + 12 = 22, so both sides match. Moving the 2x2x without changing its sign is the common slip on a typed answer.

SOL (style)1 marksMultiple choice. How many solutions does 4(x+2)=4x+54(x + 2) = 4x + 5 have? (A) no solution (B) exactly one (C) two (D) infinitely many
Show worked answer →

The correct answer is (A).

Distribute the left side: 4x+8=4x+54x + 8 = 4x + 5. Subtract 4x4x from both sides: 8=58 = 5, which is false. A false numeric statement means no value of xx works, so there is no solution. If both sides had reduced to the same true statement (like 5=55 = 5), it would be infinitely many solutions.

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