5.2 Linear Equations & Systems of Equations

A linear equation in one variable has the form $ax + b = c$ with $a \ne 0$ and exactly one solution $x = (c - b)/a$. A linear equation in two variables $ax + by = c$ graphs as a line; a system of two such equations asks for the intersection of two lines. On the 21-question, 45-minute, no-calculator Quant section this is Official Guide Math Review 3.2 equalities: short arithmetic after a clean setup, not a calculator grind.

Two tactical choices dominate. Clear clutter first — fractions, decimals, and distributed parentheses — so the coefficients are small integers. Then pick substitution or elimination from the coefficients you actually have, not from a preferred method.

Clearing fractions and decimals

Fractions: multiply by the LCM of every denominator

Every term, including constants, is multiplied. A leftover unmultiplied term is the usual error.

Worked example. Solve $\dfrac{x}{2} + \dfrac{x}{3} = 5$.

The LCM of $2$ and $3$ is $6$. Multiply through:

$3x + 2x = 30 \implies 5x = 30 \implies x = 6$

Check: $6/2 + 6/3 = 3 + 2 = 5$.

Worked example (two variables). Solve

$\frac{x}{2} + \frac{y}{3} = 4, \qquad \frac{x}{4} - \frac{y}{6} = 1$

Multiply the first equation by $6$: $3x + 2y = 24$. Multiply the second by $12$: $3x - 2y = 12$. Add: $6x = 36$, so $x = 6$. Then $18 + 2y = 24$, so $y = 3$.

Decimals: multiply by $10$, $100$, or $1{,}000$

If the longest decimal has two places, multiply by $100$. Then reduce the integers by their GCF if it is obvious.

Worked example. Solve $0.05x + 0.2y = 1.5$ together with $x + y = 20$ for $x$.

Multiply the first equation by $100$: $5x + 20y = 150$. Divide by $5$: $x + 4y = 30$. Subtract $x + y = 20$: $3y = 10$, so $y = 10/3$ and $x = 20 - 10/3 = 50/3$. Leave the answer as a fraction. Quant almost never wants a repeating decimal.

One linear equation: isolate after clearing

Once denominators are gone:

1. Distribute parentheses.
2. Collect variable terms on one side and constants on the other.
3. Divide by the coefficient of $x$.

Worked example. Solve $3(2x - 4) - (x + 5) = 7$.

$6x - 12 - x - 5 = 7 \implies 5x - 17 = 7 \implies 5x = 24 \implies x = \frac{24}{5}$

If a later factor cancels, record the excluded value. Linear equations themselves do not create extra roots, but an equation that started as a rational equation can.

Two-variable systems: substitution versus elimination

Substitution, when a coefficient is $\pm 1$

Isolate the cheap variable and plug it into the other equation.

$\begin{cases} x - 2y = 4 \\ 3x + 5y = 23 \end{cases}$

From the first equation, $x = 2y + 4$. Substitute:

$3(2y + 4) + 5y = 23 \implies 6y + 12 + 5y = 23 \implies 11y = 11 \implies y = 1$

Then $x = 2(1) + 4 = 6$. Solution: $(6, 1)$.

Elimination (linear combination), when coefficients are easy multiples

Scale one or both equations so one variable has opposite coefficients, then add.

$\begin{cases} 3x + 4y = 18 \\ 2x - 3y = -5 \end{cases}$

Multiply the first equation by $3$ and the second by $4$:

$9x + 12y = 54, \qquad 8x - 12y = -20$

Add: $17x = 34$, so $x = 2$. Back-substitute: $3(2) + 4y = 18$, so $4y = 12$ and $y = 3$. Solution: $(2, 3)$.

If the question asks for $x - y$ or $2x + 3y$ rather than $x$ and $y$ separately, pause. A linear combination of the two given equations may already be the target. That is often faster than fully solving the system.

Unique, infinite, and no solution: coefficient tests

For

$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

compare cross-products rather than forming $a_1/a_2$ when a denominator could be zero.

Worked example (infinite). $2x - 4y = 6$ and $x - 2y = 3$. The second equation is exactly half the first, including the constant, so they are the same line. Infinitely many solutions: all $(x, y)$ with $x - 2y = 3$.

Worked example (none). $2x - 4y = 6$ and $x - 2y = 4$. Variable coefficients are proportional ($2:1$), but $6/4 \ne 2/1$. Parallel distinct lines; no solution.

Worked example (unique). $2x - 4y = 6$ and $x + 2y = 4$. The $y$-coefficient ratio is $-4/2 = -2$, while the $x$-ratio is $2/1 = 2$. Slopes differ, so there is one intersection.

If a parameter $k$ appears, impose the matching condition and solve for $k$. For infinitely many solutions, all three ratios must match, including the constants. Matching only the $x$ and $y$ coefficients can describe a parallel family (no solution) instead.

Symmetric three-variable pair sums

Cyclic systems of the form

$\begin{cases} x + y = A \\ y + z = B \\ z + x = C \end{cases}$

should not be chased one variable at a time. Add all three:

$2x + 2y + 2z = A + B + C \implies x + y + z = \frac{A + B + C}{2}$

Then each singleton is the triple sum minus one pair:

$z = (x + y + z) - A, \quad x = (x + y + z) - B, \quad y = (x + y + z) - C$

Worked example. $x + y = 14$, $y + z = 19$, $x + z = 17$.

$x + y + z = (14 + 19 + 17)/2 = 50/2 = 25$

So $z = 25 - 14 = 11$, $x = 25 - 19 = 6$, $y = 25 - 17 = 8$.

If the question asks for $2x + 2y + 2z$, you already have it: $A + B + C$. Do not solve for $x$, $y$, and $z$ individually.

Word-problem translation into linear systems

Name the unknowns, write one equation per independent sentence, then apply the methods above. Count: $n$ unknowns need $n$ independent linear equations for a unique solution.

Tickets. Adult tickets cost $A$ dollars and child tickets cost $C$ dollars. Four adults and two children cost $$52$. Two adults and five children cost $$47$.

$\begin{cases} 4A + 2C = 52 \\ 2A + 5C = 47 \end{cases}$

Divide the first equation by $2$: $2A + C = 26$, so $C = 26 - 2A$. Substitute into the second: $2A + 5(26 - 2A) = 47$, hence $2A + 130 - 10A = 47$, $-8A = -83$, $A = 83/8$. Then $C = 26 - 83/4 = (104 - 83)/4 = 21/4$. Exact fractions, not $10.375$.

Mixture. A $20%$ acid solution is mixed with a $50%$ solution to make $12$ liters of $30%$ solution. Let $x$ be liters of $20%$ solution. Then $12 - x$ is liters of $50%$ solution, and the acid balance is

$0.20x + 0.50(12 - x) = 0.30(12)$

Multiply by $100$: $20x + 50(12 - x) = 360$, so $20x + 600 - 50x = 360$, $-30x = -240$, $x = 8$. Eight liters of $20%$ and four liters of $50%$.

Two quantities with a total and a difference. The sum of two numbers is $40$ and their difference is $12$. Then $x + y = 40$ and $x - y = 12$. Add: $2x = 52$, $x = 26$, $y = 14$. This is elimination on a pair that is already aligned.

Traps

• Clearing fractions but forgetting to multiply the constant term.
• Declaring infinitely many solutions because $a_1/a_2 = b_1/b_2$ without checking $c_1/c_2$. That case is no solution when the constants fail to match.
• Dividing one equation by $x$ (or by $y$) and silently dropping the $x = 0$ line that might still satisfy the original system.
• Solving a three-variable pair-sum system for each variable when the stem asked for $x + y + z$ or $2(x + y + z)$.
• Treating two word-problem sentences that restate the same constraint as two independent equations; the system is then underdetermined.