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How do you solve a system of two linear equations, and when does a system have no solution or infinitely many?

Systems of two linear equations in two variables: solve by substitution and elimination, solve graphically, and determine when a system has one solution, no solution, or infinitely many.

A focused answer to the Digital SAT Algebra skill of systems of two linear equations: solving by substitution, elimination and graphing, and using slope and intercept to tell one solution from no solution or infinitely many.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this skill is asking
  2. The three solution methods
  3. A worked elimination
  4. Counting solutions
  5. Choosing a method under time pressure

What this skill is asking

A system of two linear equations asks for the point (or points) that satisfy both lines at once. Geometrically, that is where the two lines intersect. The Digital SAT (Algebra domain) tests solving systems by substitution, by elimination, and graphically, and asks you to identify when a system has exactly one solution, none, or infinitely many.

The three solution methods

Pick the method that fits the system's form.

A worked elimination

Elimination shines when the equations are in standard form.

Counting solutions

The SAT loves a question that gives a parameter and asks for the value that makes a system have no solution or infinitely many. Convert both equations to slope-intercept form and compare. Different slopes means the lines cross exactly once, so one solution, no matter the intercepts. Equal slopes means the lines are parallel; then look at the intercepts: different intercepts means the lines never meet (no solution), and equal intercepts means they are the same line (infinitely many solutions). Reducing the question to "compare slopes, then compare intercepts" makes these reliably quick.

Choosing a method under time pressure

On test day, let the form of the system choose your method. If an equation already reads y=…y = \ldots or x=…x = \ldots, substitute. If both are in Ax+By=CAx + By = C and a variable's coefficients are opposites or easy to match, eliminate. If the numbers are awkward or you just want a fast, safe answer, graph in Desmos and click the intersection. Many SAT systems are designed so one method is clearly fastest; recognising which saves real time. For "no solution / infinitely many" questions, skip solving and go straight to comparing slopes and intercepts.

Word problems are a common dressing for systems: two unknown quantities tied together by two conditions. The standard pattern is "the total number of items is NN" giving one equation, and "the total value (cost, points, weight) is VV" giving a second. For example, if adult tickets cost \12 and child tickets cost \8, and 30 tickets were sold for \300,then300, then a + c = 30and and 12a + 8c = 300$. Set up both equations from the two sentences, then solve by whichever method fits, usually elimination or substitution. The hardest part is the translation, not the algebra, so name your variables clearly and write one equation per condition before you start solving.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Digital SAT Math (style)1 marksSolve the system {2x+y=7xβˆ’y=2\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases} for xx. (A) 11 (B) 22 (C) 33 (D) 55
Show worked answer β†’

The correct answer is (C), x=3x = 3.

Add the two equations to eliminate yy: (2x+y)+(xβˆ’y)=7+2(2x + y) + (x - y) = 7 + 2, giving 3x=93x = 9, so x=3x = 3. (Then y=7βˆ’2(3)=1y = 7 - 2(3) = 1, and the solution is (3,1)(3, 1).)

Digital SAT Math (style)1 marksFor what value of kk does the system {y=3x+1y=kxβˆ’4\begin{cases} y = 3x + 1 \\ y = kx - 4 \end{cases} have no solution? (A) βˆ’3-3 (B) 00 (C) 11 (D) 33
Show worked answer β†’

The correct answer is (D), k=3k = 3.

Two lines have no solution when they are parallel but distinct: equal slopes, different intercepts. The first line has slope 33 and intercept 11; the second has slope kk and intercept βˆ’4-4. Setting the slopes equal, k=3k = 3, makes the lines parallel, and since the intercepts differ (1β‰ βˆ’41 \neq -4) the lines never meet, so there is no solution.

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