Solve systems of linear equations in two variables algebraically and graphically, and in three variables algebraically

Learning goal: Solve pairs of linear equations in two variables both graphically and algebraically by substitution and elimination, decide when a graph gives an exact solution and when it gives only an approximation, classify systems with one, no or infinitely many solutions, and extend the algebraic methods to solving systems of linear equations in three variables.

What a solution of a system is. A system of linear equations is a set of equations that must hold at the same time. A solution is an assignment of values to the variables that satisfies every equation in the system, not just one of them. For a pair of linear equations in two variables, each equation is a line in the plane, and the solutions are the points lying on both lines. The pairs can meet in exactly one point (one solution), never meet because they are parallel and distinct (no solution), or be the same line (infinitely many solutions). Algebraically, each case shows up as a specific outcome: a single value for each variable, a false statement such as 0 = 8, or a statement that is always true such as 0 = 0.

Solving graphically, and what a graph can and cannot tell you. Graph 3x − 2y = 0 and x + 2y = 8. The first line passes through (0, 0) and (2, 3), the second through (0, 4) and (8, 0). They cross at (2, 3). Check in both equations: 3(2) − 2(3) = 0 and 2 + 2(3) = 8. Because both equations are satisfied exactly, (2, 3) is the solution. A graph is exact only when the intersection lands on a point whose coordinates you can read with certainty and then confirm by substitution. If the lines cross at, say, x = 55/41, no ruler reading will give that number; the graph gives an approximation, roughly (1.3, −0.9), and the algebra gives the exact answer. So the graphical method is the way to see the situation and estimate the solution, and the algebraic methods are the way to know it.

Substitution. When one equation already expresses one variable in terms of the other, substitute that expression into the other equation. Solve y = 2x − 1 and 3x + 2y = 12. Replace y in the second equation: 3x + 2(2x − 1) = 12, so 3x + 4x − 2 = 12, 7x = 14, x = 2. Then y = 2(2) − 1 = 3. Check the second equation with both values: 3(2) + 2(3) = 12. The substitution step is valid because at a solution both equations are true, so y really does equal 2x − 1 there and may be replaced by it.

Elimination. When both equations are in the form Ax + By = C, multiply one or both by constants so that the coefficients of one variable are opposites, then add the equations. Solve 4x + 3y = 7 and 2x − 5y = −16. Multiply the second equation by −2: −4x + 10y = 32. Add to the first: 13y = 39, so y = 3. Substitute into the first: 4x + 9 = 7, so x = −1/2. Check the second equation: 2(−1/2) − 5(3) = −1 − 15 = −16. The solution is (−1/2, 3). Why is adding two equations legitimate? If A = B and C = D are both true at a solution, then A + C = B + D is also true there, so the new equation holds at every solution of the original system. Multiplying an equation by a nonzero constant is reversible, so no solutions are gained or lost.

A fully worked exact-versus-approximate model. Solve 5x + 3y = 4 and 2x − 7y = 9. To eliminate y, multiply the first equation by 7 and the second by 3: 35x + 21y = 28 and 6x − 21y = 27. Add: 41x = 55, so x = 55/41. Substitute into the first equation: 3y = 4 − 275/41 = 164/41 − 275/41 = −111/41, so y = −37/41. Check the second equation: 2(55/41) − 7(−37/41) = 110/41 + 259/41 = 369/41 = 9. The exact solution is (55/41, −37/41); as decimals that is about (1.3415, −0.9024). Someone reading a careful graph would report about (1.3, −0.9), which is a good estimate and not the solution. The estimate is still useful: if the algebra had produced (13, −9), the graph would immediately show a mistake.

Recognizing no solution and infinitely many solutions algebraically. Solve 2x − 4y = 6 and −x + 2y = 1. Multiply the second by 2: −2x + 4y = 2. Add to the first: 0 = 8. That statement is false for every x and y, so the system has no solution; both lines have slope 1/2 and different intercepts, so they are parallel and distinct. Now solve x − 3y = 2 and 3x − 9y = 6. Multiply the first by −3 and add: 0 = 0, which is true for every point on the line x − 3y = 2, so there are infinitely many solutions, such as (2, 0) and (5, 1), and the two equations describe the same line.

Three variables, algebraically. A linear equation in three variables, such as x + y + z = 6, has solutions that form a plane rather than a line, so a graph in the plane is no help; the system is solved algebraically. The strategy is to reduce three equations in three variables to two equations in two variables by eliminating one variable twice, then solve the smaller system, then substitute back. Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2. Add the first and third to eliminate z: 2x + 3y = 8. Add the second and third: 3x + y = 5. Now solve the pair: from 3x + y = 5, y = 5 − 3x; substitute, 2x + 3(5 − 3x) = 8, so 2x + 15 − 9x = 8, −7x = −7, x = 1, and y = 5 − 3 = 2. Then from the first equation z = 6 − 1 − 2 = 3. Check all three original equations: 1 + 2 + 3 = 6; 2 − 2 + 3 = 3; 1 + 4 − 3 = 2. The solution is (1, 2, 3). Every original equation must be checked, because the reduced pair was built from only some of them.

The misconception and why it is wrong. A common error is to eliminate a different variable in each of the two combinations when reducing a three-variable system, ending with one equation in x and y and another in y and z. That is not a two-variable system and cannot be solved as one. The same variable must be eliminated in both combinations. A second error is to check a solution in only one equation; a pair such as (4, 7) can satisfy y = 2x − 1 and still fail 3x + 2y = 12, so the check has to cover every equation in the system.

About the numbers. Every price, fee, amount and business in this lesson's examples, tasks and questions is made up for practice; none describes a real company, product or offer.

Worked substitution: Solve y = 3x − 4 and 2x + y = 11. Substitute: 2x + 3x − 4 = 11, so 5x = 15 and x = 3; then y = 3(3) − 4 = 5. Check both equations: 5 = 9 − 4 and 6 + 5 = 11. Solution (3, 5). Graphically, the line y = 3x − 4 passes through (0, −4) and (3, 5), the line y = 11 − 2x passes through (0, 11) and (3, 5), and the crossing at (3, 5) can be read exactly because it lands on integer coordinates.

Worked elimination with an exact fractional solution: Solve 4x + 6y = 9 and 2x − 3y = 1. Multiply the second by 2: 4x − 6y = 2. Add: 8x = 11, so x = 11/8. Substitute into the second: 11/4 − 3y = 1, so 3y = 7/4 and y = 7/12. Check the first: 4(11/8) + 6(7/12) = 11/2 + 7/2 = 9. Solution (11/8, 7/12) ≈ (1.375, 0.583). A graph would show the crossing a little to the left of x = 1.4 and a little above y = 0.5, an estimate that agrees with the exact values; on an ordinary sketch the crossing can only be estimated, and the algebra and the substitution check are what establish the exact coordinates.

Worked three-variable system: Solve 2x + y − z = −6, x − 3y + 2z = 1, 3x + 2y + z = −1. Eliminate z twice. First and third added: 5x + 3y = −7. Twice the first added to the second: 4x + 2y − 2z + x − 3y + 2z = −12 + 1, so 5x − y = −11. From 5x − y = −11, y = 5x + 11; substitute into 5x + 3y = −7: 5x + 15x + 33 = −7, 20x = −40, x = −2, and y = −10 + 11 = 1. From the first equation, z = 2x + y + 6 = −4 + 1 + 6 = 3. Check all three: −4 + 1 − 3 = −6; −2 − 3 + 6 = 1; −6 + 2 + 3 = −1. Solution (−2, 1, 3).

Worked classification without solving completely: For 6x − 9y = 12 and 2x − 3y = 5, multiply the second by 3 to get 6x − 9y = 15. The left sides are identical and the right sides differ, so subtracting gives 0 = −3, a contradiction: no solution. In slope-intercept form both lines have slope 2/3 with intercepts −4/3 and −5/3, so they are parallel and distinct, which is the graphical version of the same fact.

Solution-type record: For each of the systems (a) 3x − 2y = 0 and x + 2y = 8, (b) 2x − 4y = 6 and −x + 2y = 1, (c) x − 3y = 2 and 3x − 9y = 6, draw the two lines on one set of axes, state from the drawing whether they meet once, never, or everywhere, then solve algebraically by elimination and record the final algebraic statement (a value, a false statement, or a true statement) beside the drawing. Explain in one sentence per system why the drawing and the algebra agree.

Independent performance: Design three systems of your own: one pair of linear equations in two variables whose solution has integer coordinates, one pair whose exact solution involves fractions with a denominator larger than 10, and one system of three linear equations in three variables built by choosing the solution first and then writing three equations it satisfies. Equations that share a chosen solution can still be dependent (one may be a combination of the others), so if your elimination ends in a true statement such as 0 = 0 instead of a value, the system has infinitely many solutions: change one equation and solve again. Solve the first pair both graphically and algebraically and explain why the graph gives the exact solution in this case. Solve the second pair algebraically, give the exact solution and a two-decimal approximation, and explain what a graph could and could not tell you about it. Solve the three-variable system algebraically, eliminating the same variable twice, and check the solution in all three original equations. Success criteria: each two-variable solution is checked in both equations; the three-variable solution is checked in all three; the reduction of the three-variable system eliminates one variable consistently; and the distinction between an exact solution and an approximation read from a graph is stated in your own words.