--- title: Systems of Linear Equations subtitle: Graphing, substitution, elimination, and mathematical modeling author: Mathematics Department organization: Your School recipient: Algebra I students date: 2026-09-13 version: "1.0" description: A proof-led Algebra I lesson on interpreting, classifying, solving, checking, and modeling with two-variable linear systems. keywords: [systems of linear equations, graphing, substitution, elimination, dependent systems, inconsistent systems, word problems, Algebra I] template: assignment theme: latex-article paper: us-letter titlePage: true toc: true tocDepth: 2 headingNumbers: false equationNumbers: false columns: 1 figurePlacement: none tableCaptionPosition: top header: Linear Systems | Three Exact Methods footer: Student lesson and practice pdfProfile: tagged language: en-US --- # Systems of Linear Equations A single linear equation in two variables describes many ordered pairs. A **system** asks which ordered pairs satisfy two conditions at the same time. Its solution is therefore not an answer obtained from one equation and then assumed to work in the other; it is a common solution that must survive both tests. For example, the equations $$ x+y=7,\quad 2x-y=5 $$ have the common solution $(4,3)$ because $$ 4+3=7,\quad 2(4)-3=5. $$ The ordered pair is the same point whether the system is viewed numerically, algebraically, or graphically. Graphing locates the common point, substitution expresses one coordinate in terms of the other, and elimination combines the equations without changing their shared solution set. ## Learning goals By the end of this lesson, you should be able to: - decide whether an ordered pair solves both equations; - interpret the solution as an intersection of two graphs; - distinguish one solution, no solution, and infinitely many solutions; - solve exactly by substitution and elimination and estimate by graphing; - justify why substitution and elimination preserve the solution set; - choose an efficient method and verify the result; - define variables, build a system from a context, and interpret its solution. ## 1. One solution set, three representations For an equation $E$, let $S(E)$ be the set of all ordered pairs satisfying it. The solution set of the system $$ E_1,\quad E_2 $$ is the intersection $$ S(E_1)cap S(E_2). $$ This notation says precisely what *simultaneous* means: retain only the pairs belonging to both sets. ### Why graph intersections are system solutions The graph of an equation is the set of all its ordered-pair solutions. A point lies at the intersection of two graphs if and only if it lies on both graphs. It therefore satisfies both equations if and only if it solves the system. This is a proof in both directions: - If $(p,q)$ solves the system, then it satisfies each equation, so it lies on both graphs and hence at their intersection. - If $(p,q)$ lies at an intersection, then it lies on each graph, so it satisfies each equation and hence solves the system. Graphing and algebra are not separate meanings of solution. They reveal the same common set in different forms. ## 2. The three possible outcomes Two distinct nonvertical lines either meet once or are parallel. Two equations can also name the same line. Consequently, a system of two linear equations in two variables has exactly one of three outcomes. :::figure{#fig:solution-types caption="Intersecting, parallel, and coincident lines give the three possible solution counts" alt="Three coordinate plots: y equals x plus 1 and y equals negative x plus 5 intersect at 2 comma 3; y equals one half x plus 1 and y equals one half x plus 3 are parallel; y equals negative one half x plus 3 and 2y equals negative x plus 6 coincide" width=98 placement=none} ![Three possible solution types for a pair of linear equations](solution-types.svg) ::: :::table{#tbl:classification caption="Geometric and algebraic classification of a two-line system"} | Solutions | Geometry | Solving produces | | --- | --- | --- | | One | One intersection | An ordered pair | | None | Distinct parallel lines | A contradiction, such as $0=5$ | | $\infty$ | The same line | An identity, such as $0=0$ | ::: A **consistent** system has at least one solution. A **dependent** system contains equations that express the same constraint; neither equation contributes a new line. An inconsistent system contains incompatible constraints. ### A coefficient test and its reason Write a general pair in standard form: $$ \begin{aligned} a_1x+b_1y&=c_1,\\ a_2x+b_2y&=c_2. \end{aligned} $$ Multiply the first equation by $b_2$, the second by $b_1$, and subtract. Every common solution must satisfy $$ (a_1b_2-a_2b_1)x=c_1b_2-c_2b_1. $$ Let $D=a_1b_2-a_2b_1$. - If $D\ne0$, division determines exactly one $x$, and either original equation then determines exactly one $y$. The system has one solution. - If $D=0$, the $x$ and $y$ coefficients are proportional. If the constants share the same proportion, the equations describe one line; otherwise they describe distinct parallel lines. The exceptional case $D=0$ must be classified rather than divided away. This coefficient structure will later become the determinant condition in Cramer's rule. ### Worked example 1: verify before claiming Does $(3,2)$ solve $$ x+y=5,\quad 2x-y=4? $$ Substitution into both equations gives $3+2=5$ and $2(3)-2=4$. Both statements are true, so $(3,2)$ is a solution. Testing only the first equation would be insufficient. ### Worked example 2: classify proportional equations Compare $$ 2x-3y=6,\quad 4x-6y=12. $$ The second equation is exactly twice the first. They have the same graph and infinitely many solutions. In contrast, $$ 2x-3y=6,\quad 4x-6y=15 $$ has proportional left sides but incompatible constants. Doubling the first equation would require $4x-6y=12$, not $15$, so the second system is inconsistent. ## 3. Solving by graphing To solve by graphing, draw both complete lines on one coordinate plane and read their common point. Two well-separated points determine each line; the intersection should then be checked in the original equations. Graphing is especially useful when the equations are already in slope-intercept form, when a visual estimate is acceptable, or when the solution type matters more than an exact coordinate. A graph drawn by hand may not distinguish $2.33$ from $2.37$, so an algebraic method is preferable when exact values are required. ### Worked example 3: an exact grid intersection Solve $$ y=x+1,\quad y=-x+5. $$ The first line passes through $(0,1)$ and $(2,3)$. The second passes through $(0,5)$ and $(2,3)$. Their only intersection is $(2,3)$. Check: $$ 3=2+1,\quad 3=-2+5. $$ ### Worked example 4: graphing reveals dependence The equations $$ y=-\frac12x+3,\quad 2y=-x+6 $$ look different, but dividing the second by $2$ reproduces the first. Every point on one graph is on the other, so the system has infinitely many solutions. A graph shows one coincident line; algebra explains why. ### Worked example 5: approximate from a graph, exact by algebra Consider $$ y=0.5x+0.7,\quad y=-0.25x+3.1. $$ A careful graph suggests an intersection near $(3.2,2.3)$. Equating the two expressions confirms it: $$ 0.5x+0.7=-0.25x+3.1 \quad\Longrightarrow\quad 0.75x=2.4 \quad\Longrightarrow\quad x=3.2, y=2.3. $$ ## 4. Solving by substitution Suppose one equation states $y=f(x)$. Any common solution must have exactly that value of $y$, so replacing $y$ by $f(x)$ in the other equation cannot discard a common solution. The converse is also essential. If $x=p$ solves the substituted equation and $y=f(p)$, then the pair satisfies the equation $y=f(x)$ by construction and satisfies the other equation because the substituted statement is true. Thus substitution preserves the entire common solution set. This method is efficient when a variable is isolated or has coefficient $1$ or $-1$. ### Worked example 6: an isolated variable Solve $$ y=3x-4,\quad x+y=8. $$ Replace $y$ in the second equation: $$ x+(3x-4)=8 \quad\Longrightarrow\quad 4x=12 \quad\Longrightarrow\quad x=3. $$ Then $y=3(3)-4=5$. Check: $5=9-4$ and $3+5=8$, so the solution is $(3,5)$. ### Worked example 7: isolate before substituting Solve $$ 2x+y=1,\quad y=5x-13. $$ Substitution gives $$ 2x+(5x-13)=1,\quad 7x=14,\quad x=2. $$ Then $y=5(2)-13=-3$. The pair $(2,-3)$ satisfies both original equations. ### Worked example 8: substitution classifies the system For $$ y=-x+2,\quad 2x+2y=8, $$ substitution produces $$ 2x+2(-x+2)=8 \quad\Longrightarrow\quad 4=8. $$ No value of $x$ makes this false statement true, so the system has no solution. If the result had been an identity such as $4=4$, every point satisfying the first equation would also satisfy the second. ## 5. Solving by elimination Elimination replaces one equation by a linear combination chosen to remove a variable. The method is not a trick; it is an equivalence theorem. ### Equivalent-system theorem Let the original equations be $E_1$ and $E_2$. Replace $E_2$ by $$ E_2+kE_1 $$ for any real number $k$, while keeping $E_1$. Every solution of the original system satisfies both equations, so it also satisfies their stated linear combination. Conversely, a pair satisfying $E_1$ and $E_2+kE_1$ also satisfies $$ (E_2+kE_1)-kE_1=E_2. $$ The transformation is reversible, so the old and new systems have exactly the same solutions. This proves why adding a multiple of one equation to the other is valid. Multiplying an equation by a nonzero constant is also reversible. Multiplication by zero is not: it destroys all the information in that equation. ### Worked example 9: coefficients already oppose Solve $$ 3x+2y=16,\quad 5x-2y=16. $$ Adding eliminates $y$: $$ 8x=32,\quad x=4. $$ Then $3(4)+2y=16$, so $y=2$. The solution $(4,2)$ satisfies both original equations. ### Worked example 10: scale one equation Solve $$ x+2y=8,\quad 3x-y=3. $$ Multiply the second equation by $2$ and add: $$ \begin{aligned} x+2y&=8,\\ 6x-2y&=6,\\ 7x&=14. \end{aligned} $$ Thus $x=2$ and $y=3$. Scaling was selected to create opposite $y$-coefficients with little arithmetic. ### Worked example 11: fractions need not be removed first Solve $$ \frac12x+y=5,\quad \frac32x-y=3. $$ Adding the equations immediately gives $2x=8$, hence $x=4$ and $y=3$. Clearing denominators would also work, but it adds a step without simplifying this system. ### Worked example 12: elimination exposes dependence For $$ 4x-6y=10,\quad 6x-9y=15, $$ multiplying the first equation by $3$ and the second by $-2$ gives $$ 12x-18y=30,\quad -12x+18y=-30. $$ Adding produces $0=0$. The equations are proportional, so the system is dependent and has infinitely many solutions. The identity does not mean every ordered pair works; it means every point on the common line works. ## 6. Choosing a method No method is universally best. Choose from the structure visible before doing arithmetic. :::table{#tbl:method-choice caption="A practical method-selection guide"} | Visible structure | Method | | --- | --- | | Easy lines; estimate or classification wanted | Graphing | | One variable already isolated | Substitution | | Equal, opposite, or easy-multiple coefficients | Elimination | | Fractions cancel when equations are added | Elimination | | Two formulas describe the same quantity | Substitution | ::: Graphing preserves the geometry of the problem. It displays the two complete solution sets, makes parallel and coincident lines visible, and gives an immediate estimate of an intersection. Its weakness is numerical precision: a point read from a hand-drawn graph may be approximate even when the underlying solution is exact. Substitution and elimination preserve the same common solution set through reversible equality operations. Substitution is often shortest when a variable is isolated; elimination is often shortest when coefficients can be made opposite. Neither method may turn an identity into one ordered pair or a contradiction into an answer. Those outcomes classify the original constraints. A strong choice is therefore structural rather than habitual. Inspect the equations, predict what should happen, choose an efficient transformation, and then verify the result in both original equations. The three methods should agree because they describe the same intersection in different forms. A complete solution should still contain four ideas: state the system, transform it validly, give the ordered pair or classification, and check or interpret the result. ## 7. Modeling with systems A word problem becomes a system only after two quantities and two independent constraints are identified. Define variables with units, translate each constraint, solve, and then ask whether the values are meaningful in the context. ### Worked example 13: ticket sales A school sold $120$ tickets to a concert. Adult tickets cost $\$12$ and student tickets cost $\$7$. Revenue was $\$1{,}040$. Let $a$ and $s$ be the numbers of adult and student tickets. $$ a+s=120,\quad 12a+7s=1040. $$ Substitute $s=120-a$: $$ 12a+7(120-a)=1040,\quad 5a=200. $$ Hence $a=40$ and $s=80$. The counts total $120$, and the revenue is $40(12)+80(7)=1040$ dollars. ### Worked example 14: mixture concentration How many liters of a $20\%$ solution and a $50\%$ solution make $12$ liters of a $35\%$ solution? Let $x$ and $y$ be the respective volumes in liters. $$ x+y=12,\quad 0.20x+0.50y=0.35(12). $$ The second equation tracks pure solute, not total liquid. Since $0.35(12)=4.2$, elimination or substitution gives $x=6$ and $y=6$. The final mixture contains $0.20(6)+0.50(6)=4.2$ liters of solute. ### Rate model A student spends $2.5$ hours traveling, walking at $3$ miles per hour for $w$ hours and biking at $12$ miles per hour for $b$ hours. The total distance is $21$ miles. $$ w+b=2.5,\quad 3w+12b=21. $$ Subtracting three times the first equation from the second gives $9b=13.5$, so $b=1.5$ and $w=1$. Both times are nonnegative, sum to $2.5$ hours, and produce $3(1)+12(1.5)=21$ miles. ### Break-even as an intersection A club pays a fixed setup cost of $\$120$ and $\$3$ per item, so $C=120+3x$. It sells each item for $\$9$, so $R=9x$. Break-even occurs when cost equals revenue. :::figure{#fig:break-even caption="The cost and revenue lines intersect at 20 items and 180 dollars" alt="A coordinate graph shows C equals 120 plus 3x and R equals 9x intersecting at the labeled break-even point 20 comma 180" width=94 placement=none} ![Break-even model with cost and revenue lines](break-even-model.svg) ::: Solving $120+3x=9x$ gives $x=20$ and $C=R=180$. The coordinates mean that selling $20$ items exactly covers the $\$180$ total cost. Below $20$ items cost exceeds revenue; above $20$ items revenue exceeds cost. ## 8. Common errors worth diagnosing - **Checking one equation only:** a system solution must satisfy every original equation. - **Reading a graph too precisely:** hand-drawn graphs support estimates unless the intersection is exact and clearly marked. - **Substituting into the same equation:** after isolating a variable in one equation, substitute into the other independent equation. - **Changing only one side:** any addition or multiplication applied to an equation must preserve equality. - **Multiplying by zero:** this turns an equation into $0=0$ and loses its constraint. - **Misreading $0=0$:** it signals dependent equations after valid transformations, not that every point in the plane is a solution. - **Misreading $0=c$ for $c\ne0$:** it is a contradiction and proves the system is inconsistent. - **Ignoring units or constraints:** negative ticket counts or fractional people expose a flawed model or unsuitable data. # Practice: 40 exercises Show enough reasoning to reveal the method. For every unique solution, check the ordered pair in both original equations. For a dependent system, describe the common line or give a parameter form. For an inconsistent system, state the contradiction or parallel-line relationship. ## A. Meaning and classification 1. Determine whether $(3,2)$ solves $x+y=5$ and $2x-y=4$. Justify using both equations. 2. Determine whether $(-1,4)$ solves $3x+y=1$ and $x-2y=-9$. Justify using both equations. 3. Without graphing, predict the number of solutions to $y=2x-1$ and $y=-x+5$. Explain using slopes. 4. Classify $y=-3x+2$ and $6x+2y=7$ as one solution, no solution, or infinitely many solutions. 5. Classify $2x-y=6$ and $4x-2y=12$. State whether the system is consistent and whether it is dependent. 6. A cost graph and a revenue graph meet at $(10,70)$, where $x$ is items sold and $y$ is dollars. Interpret both coordinates. 7. Classify $3x+2y=8$ and $6x+4y=17$ by comparing coefficients and constants. 8. Write a different-looking equation that forms a dependent system with $2x-5y=7$. Explain why it works. ## B. Solve by graphing Graph each pair on one coordinate plane. Give the intersection or classify the system. 9. $y=x+1$ and $y=-2x+7$ 10. $x+y=1$ and $y=x-3$ 11. $2x-y=4$ and $x+y=5$ 12. $y=\frac12x+2$ and $y=-x-1$ 13. $y=-3x+4$ and $6x+2y=8$ 14. $3x-y=6$ and $6x-2y=10$ 15. Estimate the intersection of $y=0.5x+0.7$ and $y=-0.25x+3.1$ to the nearest tenth. Then verify algebraically. 16. Explain why a graph can establish that two lines are parallel or coincident but may not provide a reliable exact coordinate for a non-grid intersection. ## C. Solve by substitution 17. $y=3x-4$ and $x+y=8$ 18. $x=2y+1$ and $3x+y=17$ 19. $2x+y=1$ and $y=5x-13$ 20. $4x-3y=18$ and $x=y+2$ 21. $x+y=7$ and $2x-3y=-1$ 22. $0.5x+y=4$ and $y=0.25x+1$ 23. $y=2x+5$ and $4x-2y=-10$ 24. $y=-x+2$ and $2x+2y=8$ ## D. Solve by elimination 25. $x+y=9$ and $x-y=1$ 26. $2x+3y=12$ and $2x-y=4$ 27. $3x+2y=16$ and $5x-2y=16$ 28. $2x+5y=1$ and $3x-5y=14$ 29. $x+2y=8$ and $3x-y=3$ 30. $2x-3y=-4$ and $5x+2y=19$ 31. $\frac12x+y=5$ and $\frac32x-y=3$ 32. $4x-6y=10$ and $6x-9y=15$ ## E. Build and solve a model Define variables, write a system, solve it, and interpret the answer with units. 33. A school concert sold $120$ tickets. Adult tickets cost $\$12$, student tickets cost $\$7$, and total revenue was $\$1{,}040$. How many of each ticket were sold? 34. A jar contains $38$ dimes and quarters worth $\$6.65$. How many coins of each type are present? 35. How many liters of a $20\%$ solution and a $50\%$ solution should be mixed to make $12$ liters of a $35\%$ solution? 36. A student walks at $3$ miles per hour and bikes at $12$ miles per hour. The student travels for $2.5$ hours and covers $21$ miles. How long is spent in each mode? 37. A club's cost is $C=180+6x$ dollars for $x$ items, while revenue is $R=15x$ dollars. Find and interpret the break-even point. 38. A rectangle has perimeter $66$ centimeters. Its length is $3$ centimeters more than twice its width. Find its dimensions. 39. The digits of a two-digit number sum to $11$. The original number is $27$ greater than the number formed by reversing its digits. Find the original number. 40. A museum sold $210$ tickets. Adult admission was $\$14$, child admission was $\$8$, and total receipts were $\$2{,}400$. How many adult and child tickets were sold? # Bridge to Cramer's rule Elimination of $$ a_1x+b_1y=c_1,\quad a_2x+b_2y=c_2 $$ produced the coefficient $D=a_1b_2-a_2b_1$. When $D\ne0$, the system has exactly one solution; when $D=0$, classification is required. Cramer's rule packages this same elimination structure into determinants. It is a useful next lesson only after the meaning of a system, the exceptional cases, and the logic of equivalent equations are secure. # References - Common Core State Standards for Mathematics, HSA-REI.C.5, HSA-REI.C.6, HSA-REI.D.10, and HSA-CED.A.3. - OpenStax, *Elementary Algebra 2e*, Sections 5.1-5.4, for a modern open treatment of graphing, substitution, elimination, applications, and method choice. - Ankit Kumar Chauhan, "System of Linear Equations," Matherama, for the author-owned consistency framework and transition to determinant methods. The explanations, problems, and diagrams in this packet are newly written for this editable resource. Independent mathematics-teacher review and classroom trial remain outstanding.