Solve multi-step linear equations and inequalities in one variable, rearrange literal equations for a chosen variable, and represent inequality solutions on a number line.

What this topic is asking

The Algebra category requires confident solving of linear equations and inequalities in one variable (the A-REI standards) and rearranging literal equations (A-CED). On the Grade 10 MCAS these appear as quick selected-response items, short-answer questions with a number-line graph, and as the algebraic engine inside larger modeling problems. The skills are routine, but two points (the sign-flip rule and distributing correctly) are where students most often lose marks.

Solving multi-step linear equations

The strategy is to simplify each side, then isolate the variable using inverse operations.

1. Distribute to clear parentheses.
2. Combine like terms on each side.
3. Gather the variable on one side and constants on the other.
4. Divide by the coefficient.

For $7x - 4 = 2x + 11$: subtract $2x$ to get $5x - 4 = 11$, add 4 for $5x = 15$, divide for $x = 3$. Always check by substituting back into the original, which catches sign errors before they cost a multiple-choice point: $7(3) - 4 = 17$ and $2(3) + 11 = 17$.

Watch the distribution of a negative: $-2(x - 3) = -2x + 6$, not $-2x - 6$. The $-2$ multiplies both the $x$ and the $-3$, and a negative times a negative is positive.

Inequalities and the sign-flip rule

An inequality is solved with the same algebra as an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the inequality. Adding or subtracting never flips it.

Why the flip happens: $-2 < 3$ is true, but multiplying both sides by $-1$ gives $2$ and $-3$, and $2 > -3$, so the direction must reverse to stay true.

Literal equations

A literal equation has several letters, and you solve for one in terms of the others using the same inverse operations, treating the rest as constants.

To solve $A = \frac{1}{2}bh$ for $h$: multiply both sides by 2 to get $2A = bh$, then divide by $b$ for $h = \frac{2A}{b}$. To solve $P = 2l + 2w$ for $w$: subtract $2l$ to get $P - 2l = 2w$, then divide by 2 for $w = \frac{P - 2l}{2}$. The MCAS often phrases these as "solve for" or "express in terms of" a named variable.

Equations with no solution or infinitely many

Not every linear equation has a single solution. When the variable terms cancel, the equation collapses to a statement about constants:

• A false statement like $3 = 7$ means no solution: no value of $x$ works. For example $2x + 4 = 2x - 1$ simplifies to $4 = -1$, which is impossible.
• A true statement like $5 = 5$ means infinitely many solutions (an identity): every value of $x$ works. For example $3(x + 2) = 3x + 6$ simplifies to $6 = 6$.

The MCAS sometimes asks how many solutions an equation has, so recognizing these collapses, rather than panicking when the $x$ disappears, is the skill being tested.

Clearing fractions first

When an equation contains fractions, multiplying every term by the least common denominator clears them and makes the algebra cleaner. For $\dfrac{x}{3} + \dfrac{1}{2} = \dfrac{5}{6}$, the LCD is 6: multiply through to get $2x + 3 = 5$, so $2x = 2$ and $x = 1$. Multiply every term, including those without a fraction, or the equation becomes unbalanced.

Graphing solutions on a number line

The boundary point uses an open circle for $<$ or $>$ (the point is not included) and a closed circle for $\leq$ or $\geq$ (the point is included). Shade toward the values that satisfy the inequality: to the right for "greater than", to the left for "less than". After flipping a sign, re-read which direction the final inequality points, not the original. A compound inequality like $-3 < x \leq 4$ is graphed as a single shaded segment, open at $-3$ and closed at $4$.

Try this

Q1. Solve $3(x + 2) = 2x + 9$.

• Cue. $3x + 6 = 2x + 9$, so $x = 3$.

Q2. Solve $P = a + b + c$ for $b$.

• Cue. $b = P - a - c$.