Systems of two linear equations

Official Digital SAT skill — Algebra domain.

What this question tests

Systems of two linear equations test your ability to represent a situation with two unknowns, write two equations that describe the same quantities, and then find values that satisfy both equations at once. On the Digital SAT, this can appear as a word problem with prices, mixtures, or totals, or as a more symbolic pair of equations where you solve for a specific variable. You may be asked for the value of one variable, the ordered pair solution, or to determine whether the system has one solution, no solution, or infinitely many solutions. The test emphasizes both the algebraic solving process and the modeling step, because a correct method on the wrong system still leads to a wrong answer. It is tested because systems connect algebra to real constraints: you must keep track of two relationships simultaneously and interpret what the solution means in context.

You should expect systems to be solvable by substitution, elimination, or by spotting structure (like matching coefficients) to reduce work. Many items are designed so arithmetic is manageable if you choose a smart method, such as eliminating a variable with minimal scaling. You also need to be comfortable identifying special cases: parallel lines give no solution, the same line written two ways gives infinitely many solutions, and intersecting lines give one solution. Finally, you must read the question carefully to know which variable is requested, because the “other variable” is a frequent trap and can look very plausible if you did most of the work correctly.

What to know

• A system of two linear equations is a pair of equations in the same variables (often $x$ and $y$), and a solution is a value pair that makes both equations true at the same time.
• Substitution means solving one equation for one variable (for example, $y=mx+b$ or $x=...$) and replacing that expression into the other equation to get a single-variable equation.
• Elimination means adding or subtracting the equations (sometimes after multiplying one or both by constants) to cancel one variable, using the property that equal quantities added to equal quantities remain equal.
• A system has one solution when the lines intersect once, no solution when the lines are parallel with different intercepts, and infinitely many solutions when the equations are equivalent (same line), which is often seen when both the slope and intercept match after simplification.
• When equations are written in standard form $Ax+By=C$, multiplying an entire equation by a nonzero constant gives an equivalent equation, but changing only one term’s sign without changing the others creates a different equation and leads to errors.
• In context problems, variables must be defined clearly (for example, $r$ for raspberries and $b$ for blackberries), and the coefficients represent per-unit contributions to totals (like cost), so mixing up variable meanings changes the model even if the algebra steps are correct.

How to approach it

1. First, define your variables in plain language before writing equations so each coefficient has a clear meaning and you avoid swapping what $x$ and $y$ represent.
2. Next, build two equations that represent two different constraints (for example, total cost and total quantity), and align like terms so the same variable appears in both equations with consistent meaning.
3. Then choose a solving method that minimizes arithmetic: use substitution if one equation is already solved for a variable, and use elimination if the coefficients are easy to match or cancel.
4. If using elimination, multiply entire equations as needed to create opposite coefficients for one variable, because this lets you add or subtract the equations to remove that variable and solve the remaining one cleanly.
5. After you find one variable, substitute it back into either original equation (not a partially altered one unless you are confident it is equivalent) to find the other variable, because partial errors often come from using a changed equation incorrectly.
6. Finally, check the solution in both original equations and in the context of the problem; this confirms both algebraic correctness and whether the values make sense (such as nonnegative counts for items).

Common traps

• Answering for the wrong variable is common because the other variable’s value can be computed along the way and feels like a “finished” result; avoid this by underlining the requested quantity and labeling your final answer with the variable name.
• Sign errors during elimination happen when students multiply or subtract equations and forget to change every term; prevent this by rewriting the entire scaled equation line by line before combining.
• Building the system incorrectly (like using the wrong price with a variable or mixing units) leads to confident but wrong arithmetic; avoid this by stating what each variable counts and matching coefficients to those units.
• Declaring no solution or infinite solutions without simplification can be wrong when equations look different but are actually equivalent or intersecting; reduce equations to a comparable form and compare both slope and intercept behavior.
• Using a modified equation for back-substitution without ensuring it is equivalent can introduce mistakes, especially after arithmetic slips; prefer substituting into one of the original equations to reduce risk.

Tips & shortcuts

• If one equation is already in the form $y=...$ or $x=...$, substitution is usually the fastest because it avoids extra scaling.
• For elimination, look for coefficients that are already equal or opposites (like $3x$ and $-3x$) to cancel a variable with minimal work.
• In word problems, write the variable definitions directly into the equation line (for example, $5.50r+3.00b=37$) so you can visually track which item each term represents.
• When checking special cases, compare the simplified equations: same left-hand side with different constants means no solution, while proportional left-hand sides with proportional constants means infinitely many solutions.

Worked example

A museum sold only adult and student tickets one day. The total number of tickets sold was $120$, and the total revenue was $1{,}470$ dollars. Adult tickets cost $15$ dollars each and student tickets cost $9$ dollars each. How many adult tickets were sold?

• A. $55$
• B. $65$ ✓ (correct answer)
• C. $70$
• D. $60$

Why: Let $a$ be the number of adult tickets and $s$ be the number of student tickets. The system is $a+s=120$ and $15a+9s=1{,}470$. From $a+s=120$, $s=120-a$. Substitute into the revenue equation: $15a+9(120-a)=1{,}470$, so $15a+1{,}080-9a=1{,}470$, which simplifies to $6a=390$. Therefore, $a=65$. The correct answer is $65$, which is choice C.

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Use the Practice Questions for this skill to drill it, then attempt a Timed Practice Test.