Open Practice Tests

Nonlinear systems

Learning objectives

Find the points shared by every relationship, and solve only as far as the requested target requires.

What a system’s solution represents

A solution of a two-variable system is an ordered pair (x, y) that makes both equations true. Graphically, it is a point of intersection. An x-value found by elimination is only part of the answer when the question requests the full ordered pair; substitute back to find its corresponding y-value.

A nonlinear system can have no solution, one solution, several solutions, or infinitely many. Do not transfer the “at most one point unless the lines coincide” rule from two linear equations to a nonlinear system. A line can meet a parabola twice; a line can meet a higher-degree polynomial graph more than twice.

Choose substitution, elimination, or structure

Substitution is natural when an equation already gives y = f(x) or x = g(y). If both equations give y, set their right sides equal. Solve the resulting one-variable equation and substitute each candidate into a convenient original equation.

Elimination is often shorter when the equations contain the same squared terms. Subtracting two circle equations can cancel x2 and y2, leaving a line. When one squared variable remains after substitution, take both signs if both are allowed.

Structure can identify a requested expression without finding either coordinate. For example, a known sum x + y and a known sum of squares x2 + y2 reveal xy, and hence (x − y)2. If the target is symmetric in x and y, solving separately for the two coordinates may be unnecessary.

Given structure

Useful first move

y = f(x) and y = g(x)

Set f(x) = g(x).

x + y = s and xy = p

Use y = s − x or the quadratic z2 − sz + p = 0.

Matching squared terms

Subtract equations before substituting.

A line and a rational curve

Substitute and record denominator restrictions.

A line and a radical curve

Substitute and preserve the principal-root sign.

A parameter and one intersection

Reduce to one variable; check the degree, then use the discriminant if quadratic.

A system requires a simultaneous check

A pair that satisfies only one equation is not a partial solution of the system. Also, do not mix the x-coordinate from one intersection with the y-coordinate from another. Keep each pair together during substitution and verification.

Count intersections with a discriminant when appropriate

For a parabola y = ax2 + bx + c, a ≠ 0, and a nonvertical line y = mx + n, intersections satisfy

ax2 + (b − m)x + (c − n) = 0.

The discriminant D = (b − m)2 − 4a(c − n)

therefore determines whether there are two intersections, one, or none. For this particular pairing, one intersection means the line is tangent to the parabola. A vertical line x = d is a different case: it meets the graph of a quadratic function once but is not classified by this substituted quadratic. For a line and the circle x2 + y2 = r2, substitution also usually produces a quadratic. The coordinates must satisfy the circle equation as well as the line. If a graph is not a function of x, a single allowed x can correspond to two different y values, so back-substitution must be complete.

Check parameter-induced cancellation. In a system of two quadratics, their difference may become linear for a special parameter. It may even become an identity or contradiction. Inspect that special case before using a discriminant. A vanished leading coefficient is a change of equation type, not a reason to insert zero into a formula with a forbidden denominator.

Interpret systems in context and on a calculator

Two models sharing an input can represent different quantities until the question asks when their outputs are equal. A revenue and a cost model meet at break-even quantities. A length-and-area model supplies simultaneous constraints. Label each equation with its meaning, use consistent units, and reject coordinates outside the stated domain.

Graphing method. Enter the two original equations and inspect their intersections. Adjust the window to the relevant context and check whether more intersections may lie outside it. Use algebra to verify exact coordinates or a parameter value. For radical or rational equations, graph the original relation rather than an unrestricted equation created by squaring or cancellation.

The five‐step system check

1. State the target and original restrictions. 2. Choose substitution, elimination, or a direct identity. 3. Solve the reduced equation without losing zero or negative possibilities. 4. Recover the needed coordinates or target. 5. Verify each surviving result in both original equations and the context.

A useful connection. An equation f(x) = g(x) and the two-graph system y = f(x), y = g(x) contain the same input condition. The system adds the shared output. This is why methods from quadratic equations, rational equations, radicals, and function interpretation converge in this final section.

15 worked examples

7.01. Set two expressions for the same output equal

Find all solutions of the system y = x2 and y = 2x + 3.

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Recognize the structure. Both expressions equal y, so they must equal each other at an intersection.

Work it through. Set x2 = 2x + 3, then factor:

x2 − 2x − 3 = (x − 3)(x + 1) = 0.

Thus x = 3 or x = −1. Use y = x2 to get the paired outputs 9 and 1.

Answer: (3, 9) and (−1, 1).

Check. 2(3) + 3 = 9 and 2(−1) + 3 = 1, so both pairs also satisfy the line.

Avoid the trap. The values 3 and −1 are only the x-coordinates. A question asking for solutions of a two-variable system requires the paired coordinates.

7.02. Read intersections and answer the requested coordinate difference

The graphs of y = (x − 1)2 and y = 4 intersect at two points. What is the difference between their larger and smaller x-coordinates?

The parabola y = (x − 1) squared and horizontal line y = 4 intersect at x = −1 and x = 3.−1134xy
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Recognize the structure. The line fixes the output at 4, so solve (x − 1)2 = 4.

Work it through. x − 1 = ±2 gives x = −1 and x = 3. Their requested difference is 3 − (−1) = 4.

Answer: 4.

Avoid the trap. The midpoint input is 1, and the distance from that midpoint to either root is 2. Neither is the full difference between the two inputs.

7.03. Recognize a repeated root as one intersection

Find the solution of y = x2 − 4x + 1 and y = 2x − 8.

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Recognize the structure. Equating the outputs may produce a perfect-square equation.

Work it through. x2 − 4x + 1 = 2x − 8 x2 − 6x + 9 = (x − 3)2 = 0.

So x = 3, and the line gives y = 2(3) − 8 = −2.

Answer: (3, −2).

Check. The quadratic gives 9 − 12 + 1 = −2 as well. Its difference from the line is (x − 3)2, so equality occurs only at x = 3.

Avoid the trap. A squared factor repeats one root; it does not create two distinct intersection points.

7.04. Prove nonintersection from a range bound

How many real solutions does the system y = x2 + 2 and y = −1 have?

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Recognize the structure. The quadratic’s smallest output is 2, but the line requires output −1.

Work it through. For every real x, x2 + 2 ≥ 2. Therefore it cannot equal −1. Equivalently, substitution would require x2 = −3, which has no real solution.

Answer: 0.

Check. The horizontal line lies strictly below the parabola’s vertex and every other point on the graph.

Avoid the trap. A finite graph window is not needed to prove this claim. The nonnegative-square argument covers every real input.

7.05. Use a known sum and product

Find all real ordered pairs satisfying x + y = 7 and xy = 12.

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Recognize the structure. Use the linear equation to replace one variable in the product equation.

Work it through. From y = 7 − x,

x(7 − x) = 12 x2 − 7x + 12 = (x − 3)(x − 4) = 0.

Thus x = 3 gives y = 4, and x = 4 gives y = 3.

Answer: (3, 4) and (4, 3).

Check. Both pairs have sum 7 and product 12.

Avoid the trap. Interchanging coordinates creates a different ordered pair when the coordinates are unequal. Both must be included unless the context distinguishes or restricts them.

7.06. Intersect a line with a circle

Find all real solutions of x2 + y2 = 25 and y = x + 1.

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Recognize the structure. Substitute the line into the circle, preserving the square of the whole expression.

Work it through. x2 + (x + 1)2 = 25 2x2 + 2x − 24 = 0.

Divide by 2 and factor: (x + 4)(x − 3) = 0. Thus x = −4 or 3. The line gives y = −3 or 4, respectively.

Answer: (−4, −3) and (3, 4).

Check. The squared coordinates sum to 25 for both pairs, and each has y = x + 1.

Avoid the trap. (x + 1)2 includes 2x. Omitting that cross term would change the intersection coordinates.

7.07. Compare two quadratic functions

Find all intersections of y = x2 + 2x + 1 and y = 2x2 − x − 3.

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Recognize the structure. Set the outputs equal and combine all terms before factoring.

Work it through. x2 + 2x + 1 = 2x2 − x − 3 x2 − 3x − 4 = 0.

Factor (x − 4)(x + 1) = 0, giving x = 4 and x = −1. Substitute into y = (x + 1)2 to find y = 25 and y = 0.

Answer: (4, 25) and (−1, 0).

Check. The second rule gives 32 − 4 − 3 = 25 at 4 and 2 + 1 − 3 = 0 at −1.

Avoid the trap. An intersection at y = 25 can lie outside a default graph window. Not seeing it immediately is not evidence that it does not exist.

7.08. Eliminate matching squared terms before substituting

Solve the system x2 + y2 = 25 and x2 + (y − 4)2 = 9.

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Recognize the structure. Subtracting the first equation from the second cancels x2 and then y2.

Work it through. (y − 4)2 − y2 = 9 − 25 −8y + 16 = −16.

Thus y = 4. Substitute into the first equation: x2 + 16 = 25, so x2 = 9 and x = ±3.

Answer: (3, 4) and (−3, 4).

Check. In the second equation, (y − 4)2 = 0 and x2 = 9 for both pairs.

Avoid the trap. Once y is known, x2 = 9 produces both signs. A single y-value does not imply a single ordered pair.

7.09. Keep the rational function’s original restriction

Find all solutions of y = 6/x and y = x − 1.

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Recognize the structure. The first equation requires x ≠ 0. Substitute the line, then clear the denominator on that domain.

Work it through. 6x = x − 1 6 = x2 − x (x − 3)(x + 2) = 0.

The inputs 3 and −2 are both allowed. The line gives outputs 2 and −3.

Answer: (3, 2) and (−2, −3).

Check. 6/3 = 2 and 6/(−2) = −3. Each output also equals its input minus 1.

Avoid the trap. A rational graph can have intersections on both branches. Restricting attention to positive inputs would miss a valid solution here.

7.10. Do not add a false intersection by squaring

Find all real solutions of y = x + 2 and y = x.

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Recognize the structure. The radical gives y ≥ 0, and the line then requires x ≥ 0.

Work it through. Set x + 2 = x and square:

x + 2 = x2 (x − 2)(x + 1) = 0.

The candidate −1 violates x ≥ 0. Only x = 2 remains, and y = 2.

Answer: (2, 2).

Check. At (−1, −1), the line is satisfied but the radical gives 1 = 1, not −1. At (2, 2), both equations hold.

Avoid the trap. The graph of y2 = x + 2 has a lower branch that is absent from y = x + 2. Squaring can accidentally add that branch.

7.11. Check each branch of an absolute‐value graph

Find all real solutions of y = |x − 1| and y = x + 1.

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Recognize the structure. The absolute-value formula changes at x = 1.

Work it through. For x ≥ 1, equality requires x − 1 = x + 1, which is impossible. For x < 1, |x − 1| = 1 − x, so

1 − x = x + 1 x = 0.

This input belongs to the second branch, and y = 1.

Answer: (0, 1).

Check. |0 − 1| = 1 and 0 + 1 = 1. The line is parallel to the right branch of the V and intersects only the left branch.

Avoid the trap. An absolute-value graph does not always create two intersections with a line. The slopes and allowed branches determine the count.

7.12. Find both line slopes that create tangency

For which real values of m does the system y = x2 and y = mx − 4 have exactly one real solution?

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Recognize the structure. Substitution produces a genuine quadratic whose discriminant must be zero.

Work it through. x2 − mx + 4 = 0, D = (−m)2 − 16 = m2 − 16.

Set D = 0, giving m2 = 16 and m = ±4.

Answer: m = −4 or m = 4.

Check. For m = 4, the difference is (x − 2)2 and the point is (2, 4). For m = −4, it is (x + 2)2 and the point is (−2, 4).

Avoid the trap. The equation m2 = 16 has two real solutions. A negative slope can also produce a tangent line.

7.13. Count intersections when two quadratic terms can cancel

For which real values of k does the system y = x2 +2x and y = kx2 +6x−4 have exactly one real solution?

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Recognize the structure. The difference has leading coefficient 1 − k, which can vanish.

Work it through. Equating the outputs gives

(1 − k)x2 − 4x + 4 = 0.

If k = 1, this becomes −4x + 4 = 0, giving one point with x = 1. If k ≠ 1, the equation is quadratic and needs

D = 16 − 16(1 − k) = 16k = 0,

so k = 0.

Answer: k = 0 or k = 1.

Check. At k = 0, the intersection is (2, 8); at k = 1, it is (1, 3). Each pair satisfies its corresponding original system.

Avoid the trap. A discriminant-only method misses k = 1, where two quadratics have a linear difference and still meet once.

7.14. Interpret equality between revenue and cost

For an integer quantity n with 0 ≤ n ≤ 30, revenue and cost are modeled by R(n) = 60n − 2n2 and C(n) = 12n + 160, in dollars. What is the larger break-even quantity?

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Recognize the structure. Break-even means the two outputs are equal, not that revenue alone is zero.

Work it through. Set R(n) = C(n):

60n − 2n2 = 12n + 160 n2 − 24n + 80 = 0.

Factor (n − 4)(n − 20) = 0. Both 4 and 20 are allowed integer quantities, and 20 is larger.

Answer: 20 items.

Check. R(20) = 1200 − 800 = 400 and C(20) = 240 + 160 = 400.

Avoid the trap. Solving R(n) = 0 would find quantities with zero revenue, not quantities with zero profit. Translate the contextual target before choosing an equation.

7.15. Find a symmetric target without finding either coordinate

Real numbers x and y satisfy x + y = 13 and x2 + y2 = 89. What is (x − y)2?

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Recognize the structure. The target can be written entirely in terms of the two known expressions.

Work it through. Use the identity

(x − y)2 = 2(x2 + y2) − (x + y)2.

It follows by expanding both sides, or by eliminating xy from the sum-square and difference-square formulas. Substitute the given values: (x − y)2 = 2(89) − 132 = 178 − 169 = 9.

Answer: 9.

Check. The pairs (8, 5) and (5, 8) satisfy the system. Both have squared difference 9, even though their unsquared differences have opposite signs.

Avoid the trap. The target is a squared difference. There is no need to choose which variable is larger or to assign a sign to x − y.

What this section should change in your approach

A system is a common-condition problem, not a command to solve for everything. Choose the elimination or substitution that simplifies the structure, preserve every original restriction, and recover only the quantity the question asks for. One-variable solution counts are useful only after you confirm the reduced equation’s degree and domain.

Section exit check

Attempt these five questions without the lesson or worked solutions. Give exact answers unless a decimal is requested. Open each model answer after attempting the question. These are learning checks, not an official score scale.

Topic practice

Mastery check

Every pair must satisfy both equations. Explain why Question 4 permits a discriminant test, and why squaring in Question 5 creates a candidate that is not an actual intersection.