Systems of equations and intersections

The same pair must satisfy both constraints

A system groups equations that must hold together. In this lesson, x and y are real numbers unless a context narrows their domain. For the system x + y = 5 and x − y = 1, a solution is an ordered pair (x, y) satisfying both equations with the same x and y. The pair (3, 2) works: its sum is 5 and its difference is 1.

The pair (4, 1) satisfies the first equation but fails the second, since 4 − 1 = 3. Separate successes using different pairs do not solve the system. The order matters: (3, 2) assigns x = 3 and y = 2, while (2, 3) assigns different values. This is an AND condition on the two equations.

One linear equation describes a whole line

A linear equation in two unknowns has the form ax + by = c, with fixed real coefficients and a and b not both zero. Its real solutions form a straight line. For x + y = 5, choosing x determines y = 5 − x, so infinitely many pairs satisfy this single constraint.

When b is nonzero, rearrange to y = (c − ax)/b for plotting. When b is zero, use x = c/a, a vertical line; dividing by b would be invalid. A system’s solutions are precisely the points common to both graphs. A graph can suggest the answer, but an approximate picture is not an exact algebraic check.

Substitute an equal expression into the other equation

For x + y = 5, isolate y to obtain y = 5 − x. Substitute this expression for y into the other equation x − y = 1: x − (5 − x) = 1. Distributing the minus gives 2x − 5 = 1, so x = 3. Recover y = 5 − 3 = 2, then check both originals.

The system can be rewritten as y = 5 − x together with x − (5 − x) = 1. Keeping the first relation tells you how to recover y and why the substitution is reversible. Substituting into the same equation you just rearranged usually produces an identity instead of using the second constraint. Keep brackets around a substituted expression until its signs are handled.

Eliminate one unknown and retain another constraint

If two equalities hold, adding their left sides and their right sides gives another true equality. Adding x + y = 5 and x − y = 1 yields 2x = 6, hence x = 3. Substitute into either original to recover y = 2, and check the resulting pair in both.

The sum alone is not equivalent to the whole system: x = 3 permits every y until another constraint is retained. Replacing one equation by the sum while keeping the other is reversible, since subtraction recovers the replaced equation. More generally, you may add a multiple of one equation to the other, or scale a whole equation by a nonzero number, while preserving the system.

Create opposite coefficients across complete equations

For 2x + 3y = 12 and x − y = 1, multiply the complete second equation by 3 to get 3x − 3y = 3. Add it to the first: 5x = 15, so x = 3. The original second equation gives y = 2. Every term, including the right side, must receive the multiplier.

Substitution is often convenient when an unknown already has coefficient 1 or −1; elimination is convenient when coefficients already cancel or need small multipliers. Either method can work. Exact fractions are legitimate solutions: do not round a value before recovering the second unknown and checking both constraints.

Interpret the equation remaining after elimination

Two distinct nonparallel lines meet at one point, giving one ordered-pair solution. Distinct parallel lines have no common point. Coincident equations describe the same line, so the solutions are every pair on that line. Infinitely many solutions does not mean every pair in the plane.

Consider x + y = 5 together with 2x + 2y = 10. Subtract twice the first equation from the second to obtain 0 = 0; keep x + y = 5 as the remaining constraint. Changing the second right side to 11 gives 0 = 1, a contradiction and no solutions. These conclusions come from equivalent system steps, not from failing to find a candidate.

This three-case classification assumes each original equation defines a line. If both variable coefficients are zero, a statement such as 0 = 0 imposes no constraint, while 0 = 1 makes the whole system impossible. Such constant equations should be inspected directly rather than treated as a line.

Include vertical lines and be careful with graph readings

The system x = 2 and y = 3 has the unique solution (2, 3), even though a vertical line cannot be written as y = mx + b. Two distinct vertical lines are parallel; two copies of the same vertical line coincide. Use the equation’s actual geometry rather than a slope formula with division by zero.

A plotted crossing may be read only approximately, and a crossing can lie outside the visible window. Lines that look nearly parallel at one scale may still intersect. Use exact equations to establish the solution set, then use the graph to interpret it. A drawing’s window does not restrict the equations’ real domain.

Translate two independent facts before solving

Suppose a stipulated model counts x adult tickets and y child tickets. There are 8 tickets altogether, and the exact prices are 5 and 3 currency units, with a total of 32. The constraints are x + y = 8 and 5x + 3y = 32. Subtract three times the first from the second: 2x = 8, so x = 4 and y = 4. Both the count and total-cost checks work.

Names and units determine the equations: a count is not a price or a total cost. Here x and y must be nonnegative integers. A fractional or negative algebraic pair would fail that context even if it satisfied both real equations. Another equation that is only a rescaled copy of the first adds no independent information.

Test a pair against two exact constraints

The model uses six stated systems, including a fractional intersection, parallel and coincident lines, and a vertical/horizontal pair. Choose a candidate ordered pair independently of the system. The table evaluates both original left sides exactly and compares them with their right sides; success on only one is not success on the system.

Both axes use the same fixed scalar unit spacing. A coloured line segment represents only the visible part of an unbounded line. Fraction labels and checks are exact; plotted positions are limited by screen precision. A marked intersection represents the algebraically derived unique solution, while the independently chosen candidate can lie elsewhere.

Check both equations and explain why the method is complete

Write the result as an ordered pair with each unknown named, substitute into both originals, and retain exact values and units. Then inspect any context restrictions. Checking only an elimination equation or only the equation used to recover y can miss a copying error in the other original constraint.

A correct candidate check proves that the candidate works. Equivalent transformations and the remaining constraints explain whether it is the only solution, whether an entire common line remains, or why no pair can work. Larger systems, matrices and determinants will develop these ideas later; they are not needed for the two-equation methods here.

Worked examples

Try it yourself

1. Does (2, 3) solve x + y = 5 and 2x − y = 1? Does (1, 4)?

   Show the answer and reasoning

   (2, 3) does; (1, 4) does not.

   Both pairs have sum 5. The second left side is 1 for (2, 3) but −2 for (1, 4). Both equations must hold for the same pair.

2. Solve y = 2x + 1 and x + y = 7 by substitution.

   Show the answer and reasoning

   (x, y) = (2, 5).

   Substitute into the second to get x + 2x + 1 = 7, so x = 2 and y = 5. The original rule gives 5 = 2 × 2 + 1, and the sum is 7.

3. Solve 2x + y = 7 and x − y = 2 by adding, then recover the other unknown.

   Show the answer and reasoning

   (3, 1).

   Adding eliminates y and gives 3x = 9. With x = 3, the retained second equation gives y = 1. The original left sides are 7 and 2.

4. Solve x + y = 1 and x − y = 2 without rounding.

   Show the answer and reasoning

   (3/2, −1/2).

   Adding gives 2x = 3. Recover y = 1 − 3/2 = −1/2. Exact substitution gives sum 1 and difference 2; a negative y is allowed in the stated real domain.

5. Compare x − 2y = 3 with 2x − 4y = 6, and with 2x − 4y = 7.

   Show the answer and reasoning

   The first pair defines the same line; the second pair has no solution.

   Subtract twice the first equation from the second. The remaining equality is respectively 0 = 0 or 0 = 1. For the coincident case, every solution must still satisfy x − 2y = 3; not every pair in the plane qualifies.

6. Solve x = −2 and y = x + 5. Must both lines have finite slopes?

   Show the answer and reasoning

   (−2, 3); no.

   Substitution gives y = 3. The first graph is vertical and has no finite slope, but its equation supplies the x-coordinate directly.

7. Adding x + y = 5 and x − y = 1 gives x = 3. Why is (3, 100) not a solution?

   Show the answer and reasoning

   The added relation is only one consequence; the original constraints must still hold.

   The pair has x = 3, but its original sum is 103 and difference is −97. Keep one original equation when replacing the other by their sum, then recover y = 2.

What to carry forward

• A solution is one ordered pair that satisfies both original equations together.
• Substitution replaces an unknown with an equal expression; elimination combines constraints while retaining enough information to recover the original system.
• Distinct nonparallel lines give one point, distinct parallel lines give none, and coincident lines give every pair on their common line.
• Vertical lines and constant statements require their actual equations, rather than invalid slope or division formulas.
• Check exact values in both originals, then check units and the problem’s permitted domain.