Linear Equations and Systems

Linear Equations and Systems

Solving a linear equation means isolating the variable: apply the same operation to both sides of the equation until the variable stands alone. This chapter applies that method to equations, inequalities, and systems of two equations. Linear equations are the most heavily tested Algebra skill on the ACT, and the questions are designed so that a careless step produces a tempting wrong answer, so the habits below matter as much as the algebra.

Equations: gather, then divide

Move all the variable terms to one side and all the constants to the other, then divide by the coefficient. Work one operation at a time and do the same thing to both sides. The most common slip is a sign error while moving a constant across the equals sign, for example writing $9 - 7$ when the step actually calls for $9 + 7$.

Inequalities: the one new rule

Inequalities follow the same steps as equations, with one exception. Multiplying or dividing both sides by a negative number reverses the inequality symbol. Skipping this step produces an incorrect answer choice even when every other step is correct. Dividing by a positive number never changes the direction.

A compound inequality such as $-1 \le 3x - 4 < 8$ is solved by applying each step to every part. Add $4$ to all three parts to get $3 \le 3x < 12$, then divide all three parts by $3$ to get $1 \le x < 4$. Keep each boundary's strictness ($<$ or $\le$) exactly where it started; the answer choices often swap them.

Solving for a letter in a formula

A "solve for" question treats every other letter as if it were a number. To solve $P = 2l + 2w$ for $w$, first subtract $2l$ from both sides to get $P - 2l = 2w$, then divide the entire right side by $2$: $w = \frac{P - 2l}{2}$. A common wrong choice divides only one term.

Systems of two linear equations

A system of two linear equations identifies the point where two lines intersect. Two methods solve a system. Substitution solves one equation for a variable and substitutes that expression into the other equation. Elimination adds or subtracts the two equations to remove a variable. The more efficient method depends on the form of the given equations.

When one equation already isolates a variable, as in $y = 2x - 1$, substitution requires a single step: replace that variable in the other equation and solve. When both equations appear in standard form, $ax + by = c$, elimination is typically faster, because adding or subtracting the equations can remove a variable directly.

Worked Example. Solve the system $y = 2x - 1$ and $3x + y = 14$. Find the value of $x$.

$y$ is already isolated, so substitute directly into the second equation: $3x + (2x - 1) = 14$, so $5x - 1 = 14$, $5x = 15$, and $x = 3$.

Solve for the requested quantity

A question often asks for an expression built from a variable, such as $2x + 3$ or the sum of two variables, instead of the variable alone. Solving the equation correctly but reporting the value of the wrong quantity produces an incorrect answer even when every calculation is correct. Before you pick a choice, reread the last sentence of the question.

Worked Example. Solve $5x - 3 = 2x + 12$. Then find the value of $3x - 1$.

Gather the $x$-terms on one side and the constants on the other: $3x = 15$, so $x = 5$. Then $3x - 1 = 3(5) - 1 = 14$. Notice that $5$ is on the answer list as a trap.

Systems with no solution or infinitely many

Two lines that never meet are parallel: same slope, different intercepts. Two lines that coincide have the same slope and the same intercept. To find an unknown coefficient $k$, scale one equation so a pair of coefficients match, then compare. For $2x + ky = 6$ and $4x + 8y = 12$, doubling the first equation gives $4x + 2ky = 12$. Matching it to $4x + 8y = 12$ requires $2k = 8$, so $k = 4$, and the constants already agree, so there are infinitely many solutions. If the coefficients matched but the constants did not, there would be no solution.

Word problems and models

Two competing costs described in words are a classic linear model. A repair shop charges a $15 diagnostic fee plus $2 per part, while a second shop charges $10 plus $3 per part. The two costs are equal when $15 + 2x = 10 + 3x$, which happens at $x = 5$ parts. A more advanced version presents the system as a word problem: two numbers, a total, and a ratio between them. Translating the given information into two equations is the primary skill tested; solving is a direct application of the methods above.

Strategy checklist

• Underline what the question asks for ($x$, $x+y$, $3x-1$, a coefficient) before solving.
• Flip an inequality only when you multiply or divide by a negative.
• Use substitution when a variable is already isolated, elimination when both equations are in standard form.
• Plug your answer back in when time allows; a ten-second check catches sign errors.