Systems of linear equations

Learning objectives

What you should be able to do

Build two constraints from a context; solve with substitution, elimination, or intersections; choose efficient equation combinations; classify solution counts; determine unknown coefficients; and recognize when an expression is fixed even though individual variables are not.

A shared solution must satisfy both equations

A system is a group of conditions that must hold simultaneously. For two linear equations in two variables, a solution is one ordered pair that makes both equations true. A point lying on only one of the two lines is not a solution to the system.

When each equation represents a line, there are three geometric possibilities:

One solution, no solution, and infinitely many solutions. The sketches illustrate the relationships, not specific numerical systems.

Substitution replaces one variable with an equal expression from the other equation. It is convenient when a variable is already isolated or has coefficient 1 or −1. The result is an equation in just one variable; after solving it, substitute back to recover the other variable when needed.

Elimination adds or subtracts equations so one variable cancels. It is convenient when coefficients already match, are opposites, or can be made so with small multipliers. Multiply every term, including the constant, when scaling an equation.

Graphing locates the shared point visually or numerically. If two lines intersect at (r, s), then $x = r$ and $y = s.$ Read the requested coordinate or expression carefully. A decimal shown by a calculator may be rounded; verify a claimed exact value by substitution.

A method-selection shortcut

An isolated variable? Try substitution.

Matching or opposite coefficients? Try elimination.

Only a sum or other expression requested? Look for a useful equation combination first.

Numeric equations awkward to compute? Graph, then verify the intersection.

Unknown coefficient or number of solutions? Compare algebraic structure rather than relying on a picture.

Elimination is controlled equation combination

If $L 1 = R 1$ and $L 2 = R 2$ are both true, then $L 1 + L 2 = R 1 + R 2$ and $L 1 - L 2 = R 1 - R 2$ are true. Multiplying an equation by a nonzero constant also preserves it. These facts explain elimination: it is arithmetic on equal quantities, not a special rule that only works for systems.

Preserve enough information. A combined equation is a consequence of the system, but it generally does not replace both original equations by itself. After obtaining one variable, use an original equation to find the other. Keep the originals available for a final check.

The multiplier must reach the right side. If you triple $2x + 3y = 8,$ you get $6x + 9y = 24,$ not $6x + 9y = 8.$ When subtracting an equation, change the signs of every term being subtracted. Writing a whole row in parentheses can prevent sign errors.

Solve for the expression the question actually requests

Suppose a question asks for $x + y.$ Adding two equations may produce a multiple of $x + y$ directly. If it asks for $x + 2y,$ twice one equation minus the other may produce that expression. This can avoid fractions and unnecessary work.

A systematic way to look for a target is to compare coefficient pairs. The expressions $3x + 4y$ and $5x + 6y$ have coefficient pairs (3, 4) and (5, 6). Twice the first minus the second gives $(6, 8) - (5, 6) = (1, 2),$ the coefficients of $x + 2y.$ Apply the same combination to their right sides.

What “not enough information” really means

Two dependent equations may fail to determine x and y individually but still determine a particular combination. For example, if every solution satisfies $2x + 3y = 12,$ then every solution also satisfies $6x + 9y = 36.$ The question may ask for the invariant combination, not a unique point.

Creating a system from a context. Define the unknowns with units. One equation often counts a total number of objects; the other counts a total value, mass, volume, or cost. Coefficients must match the variable definitions. If a counts adult tickets, the adult ticket price multiplies a, not the child-ticket variable.

Mixture equations count the component, not just the mixture. If a blend has fruit-juice fraction 0.20 and volume u liters, it contains $0.20u$ liters of fruit juice. A total-volume equation and a total-fruit-juice equation express different conserved quantities. All percentages in a component equation must be converted consistently to fractions or decimals.

Context can reject a mathematical solution. A system may have one real solution but demand a negative count or a fractional number of indivisible items. That means there is no feasible solution in that situation. State the restrictions before interpreting the algebra.

Solution counts and parameter conditions

For two nonvertical lines $y = m 1 x + b 1$ and $y = m 2 x + b 2 :$

Comparison	Graph	Number of solutions
$m 1 \neq m 2$	Lines intersect	Exactly one.
$m 1 = m 2,$ $b 1 \neq b 2$	Distinct parallel lines	None.
$m 1 = m 2,$ $b 1 = b 2$	Same line	Infinitely many.

Use full-equation multiples in standard form. If all coefficients and the constant in one equation are the same multiple of the corresponding entries in the other, the two equations represent the same line. If the variable coefficients share a multiplier but the constants do not, the lines are distinct and parallel.

Elimination is safer than a ratio formula. Ratios such as $A 1 /A 2$ can be undefined when a coefficient is zero. Instead, scale and subtract rows. A resulting contradiction such as $0 = 6$ means no solution. A resulting identity such as $0 = 0$ means that row adds no new restriction; keep track of any remaining equation.

Do not assume every parameter choice still produces a line

An equation $Ax + By = C$ is a line only when A and B are not both zero. A parameter can sometimes make an equation become $0 = 0$ (no restriction) or $0 = C$ with $C \neq 0$ (impossible). Simplify those cases directly before applying a two-line comparison.

A compact uniqueness test, with its conditions. For

$A 1 x + B 1 y = C 1,$ $A 2 x + B 2 y = C 2,$

eliminating y produces $(A 1 B 2 - A 2 B 1)x = C 1 B 2 - C 2 B 1 .$ If $A 1 B 2 - A 2 B 1 \neq 0,$ the system has a unique solution. If it is zero, check consistency; it does not tell you whether there are no solutions or infinitely many. This test is a derived shortcut, not a formula you must memorize.

A parameter is not automatically another graphing variable. If the unknown is k in $kx + 6y = 10,$ the question may be asking which line to choose. A slider can suggest a value but does not prove it is exact or show that all other cases were considered. Coefficient comparison is decisive.

Final check. Substitute a claimed pair into both original equations. For a no-solution answer, exhibit the contradiction or unequal parallel intercepts. For infinitely many solutions, exhibit the common line or a valid full-equation multiple. For an invariant expression, show why it has the same value for every allowed pair.

15 worked examples

4.01. Substitute an expression that is already isolated

The system y = 2x + 1 and x + y = 10 has solution (x, y). What is y?

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Recognize the structure. Replace y in the second equation with the equal expression $2x + 1.$

Work it through. Substitute and solve:

$x + (2x + 1) = 10 3x = 9 x = 3.$

Then $y = 2(3) + 1 = 7.$

Answer: 7.

Check. The pair (3, 7) gives $7 = 2(3) + 1$ and $3 + 7 = 10.$

Avoid the trap. The first variable found is x, but the question requests y. Always return to the target after the algebra is complete.

4.02. Opposite coefficients invite addition

Solve the system x + y = 13 and x − y = 5.

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Recognize the structure. Adding eliminates y immediately.

Work it through. Add the left and right sides:

$(x + y) + (x - y) = 13 + 5 2x = 18.$

Thus $x = 9.$ Substitute into $x + y = 13$ to get $y = 4.$

Answer: $(x, y) = (9, 4).$

Check. $9 + 4 = 13$ and $9 - 4 = 5.$ Both equations hold.

Another route. Subtracting the second equation from the first gives $2y = 8$ directly. Either order is efficient.

Avoid the trap. An ordered-pair answer must preserve the variable order: (9, 4), not (4, 9).

4.03. Create opposite coefficients with small multipliers

If 3x + 2y = 22 and 5x − 3y = 5, what is y?

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Recognize the structure. Multiplying the first equation by 3 and the second by 2 creates $+6y$ and $-6y.$

Work it through. The scaled equations are

$9x + 6y = 66,$ $10x - 6y = 10.$

Add to obtain $19x = 76,$ so $x = 4.$ Substitute into the first original equation: $12 + 2y = 22,$ giving $y = 5.$

Answer: 5.

Check. The second original equation gives $5(4) - 3(5) = 20 - 15 = 5.$

Avoid the trap. Both right sides must be scaled. Checking an original equation helps catch an error made while creating the multiplied rows.

4.04. Read the common point, then answer the target

The graph shows y = −x + 6 and y = 2x − 3. If (a, b) is their intersection, what is a + b?

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Recognize the structure. The same point lies on both lines: (3, 3).

Work it through. The coordinates are $a = 3$ and $b = 3,$ so $a + b = 6.$ Algebra confirms $-x + 6 = 2x - 3,$ giving $x = 3$ and $y = 3.$

Answer: 6.

Avoid the trap. An intersection is a pair. The question asks for the sum of its coordinates, not either coordinate alone.

4.05. Use a count equation and a revenue equation

A theater sells 40 tickets. Adult tickets cost $12 and child tickets cost $8. The total revenue is $408. How many adult tickets were sold?
A) 18 B) 22 C) 34 D) 40

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Recognize the structure. One equation counts tickets; a second counts dollars.

Work it through. Let a and c count adult and child tickets. Then $a + c = 40$ and $12a + 8c = 408.$ Multiply the count equation by 8 and subtract:

$(12a + 8c) - (8a + 8c) = 408 - 320 4a = 88.$

Thus $a = 22,$ leaving $c = 18.$

Answer: B, 22.

Another route. Forty child tickets would bring in $320. Each replacement by an adult ticket adds $4, so the extra $88 requires 22 replacements.

Avoid the trap. The number 18 counts children, not adults. Keep the variable definitions attached to the results.

4.06. Conserve both total volume and a component

A drink maker mixes a blend containing 20% fruit juice with a blend containing 50% fruit juice. The result is 10 liters containing 32% fruit juice. Assuming volumes add, how many liters of the 50% blend are used?

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Recognize the structure. Total liters and liters of actual fruit juice give different equations.

Work it through. Let u and v be liters of the 20% and 50% blends. Then

$u + v = 10,$ $0.20u + 0.50v = 0.32(10) = 3.2.$

Substitute $u = 10 - v:$ $2 - 0.20v + 0.50v = 3.2.$ Thus $0.30v = 1.2,$ so $v = 4.$

Answer: 4 liters.

Check. The other blend contributes 6 liters. Actual juice totals $0.20(6) + 0.50(4) = 1.2 + 2 = 3.2$ liters.

Avoid the trap. A simple average of 20% and 50% would assume equal volumes. The unknown amounts must weight the percentages.

4.07. Interpret an intersection as equal total cost

Rental plan A costs $30 plus $8 per hour. Plan B costs $66 plus $5 per hour. For how many hours do the plans have the same total cost?

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Recognize the structure. At an intersection, both cost formulas give the same output for the same number of hours.

Work it through. Set the totals equal:

$30 + 8h = 66 + 5h 3h = 36 h = 12.$

Answer: 12 hours.

Check. At 12 hours, plan A costs $30 + 96 = 126$ dollars, and plan B costs $66 + 60 = 126$ dollars. The intersection is (12, 126).

Key takeaway. Plan A begins $36 cheaper but gains $3 in relative cost each hour. The difference closes after $36/3 = 12$ hours.

4.08. Equal coefficient ratios do not guarantee the same line

How many solutions does the system 2x + 3y = 7 and 4x + 6y = 18 have?
A) Zero B) One C) Two D) Infinitely many

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Recognize the structure. The second left side is twice the first, but its right side is not.

Work it through. Doubling the first equation gives $4x + 6y = 14.$ The second requires the same expression to equal 18. Subtraction produces $0 = 4,$ which is impossible.

Answer: A, zero solutions.

Check. In slope-intercept form the slopes match, but the intercepts are $7/3$ and 3. These are distinct parallel lines.

Avoid the trap. Proportional variable coefficients alone imply parallel or coincident lines. The constants decide which.

4.09. Recognize two versions of the same constraint

How many real ordered-pair solutions satisfy both 6x − 9y = 12 and 2x − 3y = 4?

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Recognize the structure. The first equation is exactly three times the second, including the constant.

Work it through. Multiplying $2x - 3y = 4$ by 3 produces $6x - 9y = 12.$ Therefore every point on the line $2x - 3y = 4$ satisfies both equations.

Answer: Infinitely many real ordered-pair solutions.

Check. Both (2, 0) and (5, 2) work, and the full line contains infinitely many points. The multiplier argument proves the complete solution set, not just these two examples.

Avoid the trap. Infinitely many solutions does not mean every pair in the plane works. The pairs must still lie on the common line.

4.10. Find a sum without solving for each variable

If 5x + 2y = 17 and 2x + 5y = 11, what is x + y?

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Recognize the structure. Adding gives the same coefficient on x and y.

Work it through. Add the equations:

$7x + 7y = 28 7(x + y) = 28.$

Divide by 7 to get $x + y = 4.$

Answer: 4.

Another route. Solving separately gives $x = 3$ and $y = 1,$ whose sum is 4. That route is valid but unnecessary for this target.

Avoid the trap. The right-side sum, 28, is seven times the requested expression. Do not stop before dividing by the common coefficient.

4.11. Build the target with a weighted combination

If 3x + 4y = 18 and 5x + 6y = 28, what is x + 2y?

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Recognize the structure. Twice the first left side minus the second equals $x + 2y.$

Work it through. Apply that combination to both sides:

$2(3x + 4y) - (5x + 6y) = 2(18) - 28.$

Simplifying gives $x + 2y = 36 - 28 = 8.$

Answer: 8.

Check. The system’s solution is (2, 3), which gives $x + 2y = 2 + 6 = 8.$ Finding it is a check, not a necessary step.

Key takeaway. Useful combinations are not limited to canceling a variable. They can also create the exact expression being requested.

4.12. Choose the coefficient that makes the lines inconsistent

For what value of k does the system 3x + 2y = 8 and kx + 6y = 10 have no solution?

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Recognize the structure. To cancel y, triple the first equation. Match the resulting x coefficient, then inspect the constants.

Work it through. Three times the first equation is $9x + 6y = 24.$ To have the same left side as the second, k must be 9. For $k = 9,$ the system demands $9x + 6y = 24$ and $9x + 6y = 10,$ a contradiction.

Answer: 9.

Key takeaway. For $k \neq 9,$ subtraction gives $(k - 9)x = -14,$ with a unique x and then a unique y. This shows that 9 is the only no-solution value.

Avoid the trap. The constants do not need to match for no solution; they need to disagree after the variable coefficients match.

4.13. Match an entire equation with two unknown constants

The system (a + 1)x + 6y = 15 and 2x + 3y = b has infinitely many solutions. What is a + 2b?

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Recognize the structure. The coefficient 6 on y fixes the multiplier from the second equation to the first as 2.

Work it through. Twice the second equation is $4x + 6y = 2b.$ For this to be the first equation,

$a + 1 = 4,$ $2b = 15.$

Thus $a = 3$ and $b = 15/2.$ The target is $a + 2b = 3 + 15 = 18.$

Answer: 18.

Check. With those constants, the first equation is exactly twice the second.

Avoid the trap. Match the full coefficient $a + 1$ with 4, not a alone. The requested expression also saves you from needing a decimal for b.

4.14. Use an extra condition on the intersection point

The lines 2x + y = 11 and px − 3y = 1 intersect at a point on the line y = x − 1. What is p?

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Recognize the structure. Use the condition on the intersection with the equation that has no unknown parameter.

Work it through. Substitute $y = x - 1$ into $2x + y = 11:$

$2x + x - 1 = 11 x = 4,$ $y = 3.$

Now require this point to satisfy $px - 3y = 1:$ $4p - 9 = 1,$ so $p = 10/4 = 5/2.$

Answer: $\frac{5}{2}$

Check. The point (4, 3) lies on all three stated lines. For this parameter the first two lines have different slopes, so their intersection is unique.

Avoid the trap. The relation $y = x - 1$ is information about the shared point. Use it to find that point before solving for the coefficient.

4.15. A constant expression can survive infinitely many solutions

Every solution of the system 2x + 3y = 12 and 4x + 6y = 24 gives the same value for kx + 9y. What is k?

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Recognize the structure. The equations are dependent. The requested expression must be a fixed multiple of their common constraint.

Work it through. Multiplying $2x + 3y = 12$ by 3 gives $6x + 9y = 36.$ Thus $k = 6$ makes the expression constant for every solution. To see why no other value works, compare two allowed points: (6, 0) gives $6k$ and (0, 4) gives 36. Equality requires $6k = 36,$ hence $k = 6.$

Answer: 6.

Key takeaway. The variables are not individually determined, but the combination $6x + 9y$ is. “Infinitely many solutions” does not automatically mean the question has no numeric answer.

Mastery check and error prevention

Attempt the exit check below, then reveal each answer and explanation.

Topic practice

Before leaving this section

Check that you can select an efficient method, scale every term correctly, test a pair in both originals, and distinguish a shared line from the whole plane. Before solving for both variables, inspect whether a sum or multiple already gives the target.