Solve multi-step linear equations in one variable, including equations with the variable on both sides and with rational-number coefficients, and identify when an equation has one solution, no solution, or infinitely many solutions (MA.912.AR.2.1, MA.912.AR.2.2).

What this topic is asking

MA.912.AR.2 asks you to solve linear equations in one variable: multi-step, with fractions, and with the variable on both sides. You also classify an equation as having one solution, no solution, or infinitely many solutions. On the B.E.S.T. Algebra 1 EOC these are core Algebra and Modeling points, mostly equation-editor and multiple choice.

The balance method

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

Subtract first (undo the $+7$), then divide (undo the $\times 3$). 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 $\frac{x}{2} + 3 = \frac{x}{4} + 5$, multiply by $4$: $2x + 12 = x + 20$, then solve normally to $x = 8$.

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

One, none, or infinitely many

After simplifying, the variable sometimes cancels:

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

How the B.E.S.T. EOC examines this topic

• Equation editor. Solve for $x$ and type the value (including fractions).
• Multiple choice. Count solutions (one, none, infinitely many), or pick the solution.
• Editing task choice. Select the correct next step in a solution.

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

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), then they are equal for every $x$, 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.

Modeling with a linear equation

Many EOC items hand you a situation and expect you to build and solve the equation. The pattern is to name the unknown, translate each phrase into algebra, and solve. "A gym charges a \30 joining fee plus \15 per month; after how many months is the total \$135?" becomes$30 + 15m = 135$, so$15m = 105$and$m = 7$months. The fixed amount is a constant, the per-unit amount multiplies the variable, and the total is what the expression equals. The same structure handles "consecutive integers" (where the next integer is$x + 1$), perimeter ("length is 4 more than width"), and distance ($d = rt$). Defining the variable in words first, "let$m$ be the number of months," keeps the translation honest and earns the setup credit even before you solve.

Try this

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

• Cue. $6x - 2 = 4x + 10 \Rightarrow 2x = 12 \Rightarrow x = 6$.

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

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