Solves systems in which at least one equation is not linear, predicting the number of solutions from the geometry and checking every candidate against the originals. Covers substitution and elimination for nonlinear systems, extraneous solutions, and the shaded regions that nonlinear inequalities describe.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 9 — Systems of Equations and Inequalities
§9.3 Systems of Nonlinear Equations and Inequalities: Two Variables, pp. 1090-1103
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1090-1103 — the pages these objectives are drawn from
Warm-up
The two-line answer of at most one no longer applies.
Discussion prompt
In how many points can a straight line cross a circle?
Hint: Try drawing a few positions.
Answer:
Two, if the line passes through the circle. One, if it just touches — a tangent. None, if it misses entirely.
So the count is zero, one, or two, where two lines allowed only zero, one, or infinitely many.
And a circle meeting a parabola can give up to four. The solution count is no longer bounded by a short list, which is why sketching first is worth the time.
Concept
When one equation is linear, solving it for a variable and substituting into the nonlinear one leaves a single-variable equation you already know how to solve.
\[ y=mx+b \;\to\; \text{substitute} \;\to\; \text{a quadratic in }x \]
The resulting equation's degree predicts the number of solutions: a quadratic gives at most two, a quartic at most four. That count is worth noting before solving, as a check on the answer.
Figure (svg): A worked ladder solving a line and circle system by substitution, reaching a quadratic in one variable
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1090-1094
Section
Section 1
Concept
Two curves can intersect in as many points as their shapes allow. Sketching them roughly, before solving, tells you how many answers to look for.
The degree of the equation that substitution produces gives the same information algebraically. A quadratic can have at most two roots, and a quartic at most four — which matches the geometric counts exactly.
| the two curves | possible intersections |
|---|---|
| line and line | 0, 1, or infinitely many |
| line and circle | 0, 1, or 2 |
| line and parabola | 0, 1, or 2 |
| circle and parabola | 0 through 4 |
| circle and circle | 0, 1, 2, or infinitely many |
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1090-1095
Picture it
Three positions, three different counts.
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
The middle case is a tangent line, which touches without crossing. Algebraically it corresponds to a repeated root, which is why the discriminant vanishes there.
Worked example
A sketch takes ten seconds.
\[ \text{How many solutions can } x^2+y^2=25 \text{ and } y=x+1 \text{ have?} \]
Identify the curves
Why: A circle and a line.
\[ \text{circle radius } 5,\text{ line} \]
Check the line's position
Why: Its intercept is small relative to the radius.
Read the count
Why: A line through a circle crosses twice.
Confirm algebraically
Why: Substitution gives a quadratic.
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
\[ \text{two solutions expected} \]
Verify: check the algebraic prediction agrees
Why: Substituting produces a quadratic, whose two roots correspond to the two crossings. Geometry and algebra give the same count, which is why either can serve as a check on the other.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1091-1093
Prediction
A circle and a parabola.
Predict first
How many times can they intersect?
Correct: Up to four.
Why: Substituting one into the other gives a quartic, which has at most four roots. Geometrically, a parabola can pass into and out of a circle twice on each side of its axis.
Worked example
One solution means a repeated root.
\[ \text{How many solutions does } x^2+y^2=4 \text{ and } y=2 \text{ have?} \]
Substitute
Why: The line into the circle.
\[ x ^{2} + 4 = 4 \]
Simplify
Why: Isolate the square.
\[ x ^{2} = 0 \]
Solve
Why: A repeated root.
\[ x = 0,\text{ twice} \]
Interpret
Why: One point of contact.
Figure (svg): The solution to Worked example a tangent case shown as a ladder of expressions, one row per legal move
\[ (0,2)\text{ only} \]
Verify: check the geometry
Why: The line at height 2 touches a circle of radius 2 exactly at its highest point and goes no further in. The repeated root is the algebraic signature of tangency, and it is what distinguishes touching from crossing.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1093-1095
Trap
\[ \text{substitution gives }x=3, \text{ so the solution is }(3,4) \]
Report the first root found
Why: The quadratic's second root is not pursued.
A second intersection point is missed entirely.
A quadratic has two roots, and each generally gives a solution to the system.
The sketch predicted two crossings, so finding only one is a signal that the work is incomplete.
Back-substitute every root and check each. The predicted count is what tells you when you are done.
Sorting
The pair of curves decides.
Sort into buckets
Sort each pair by the maximum count.
Faded example
Substitution produces a quadratic.
Fill in the blanks
\text22 \;\Longrightarrow\; \text______\text___
Why: A polynomial of degree n has at most n real roots, and each root generally gives one solution to the system. The degree is available before any solving and predicts the count.
Step zero
You are given a nonlinear system to solve.
Discussion prompt
What do you do before any algebra?
Hint: What would tell you when to stop?
Answer:
Sketch the two curves roughly and count the crossings. That takes ten seconds and tells you how many solutions to expect.
The count is a check on the answer: finding fewer means work is missing and finding more means something extraneous crept in.
It also identifies tangency in advance, where a repeated root will appear. Without the prediction there is no way to know when the solving is finished, which is what makes the sketch worth the time.
Section
Section 2
Concept
When one equation is linear, solve it for a variable and substitute into the other. The result is a single-variable polynomial equation of familiar degree.
Substituting the other way — the nonlinear into the linear — is possible but rarely helpful, since it usually leaves both variables present. Going from simple into complicated is the direction that reduces.
Figure (svg): A worked ladder solving a line and circle system by substitution, reaching a quadratic in one variable
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1095-1099
Picture it
Five lines from system to solutions.
Figure (svg): A worked ladder solving a line and circle system by substitution, reaching a quadratic in one variable
The quadratic in the middle is the whole point of the substitution: it converts a two-variable problem into one whose solution method is already familiar.
Worked example
Substitute and collect.
\[ \text{Solve } x^2+y^2=25 \text{ and } y=x+1. \]
Substitute the line
Why: Into the circle.
\[ x ^{2} + (x + 1) ^{2} = 25 \]
Expand and collect
Why: A quadratic.
\[ 2 x ^{2} + 2 x - 24 = 0 \]
Factor
Why: Divide by two first.
\[ (x - 3) (x + 4) = 0 \]
Back-substitute both roots
Why: Into the line.
\[ (3, 4)\text{ and } (-4, -3) \]
Figure (svg): A worked ladder solving a line and circle system by substitution, reaching a quadratic in one variable
\[ (3,4),\;(-4,-3) \]
Verify: check both in the circle
Why: Nine plus sixteen is 25 — correct. Sixteen plus nine is 25 — also correct. Both points are on the line too, since each was found from it. Two solutions matches the geometric prediction for a line crossing a circle.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1096-1098
Prediction
One equation is linear and one is a circle.
Predict first
Which way should the substitution go?
Correct: The linear into the circle.
Why: The linear equation gives one variable as a simple expression, which when substituted leaves a single-variable equation. Going the other way usually leaves both variables present and reduces nothing.
Worked example
Same route, different curve.
\[ \text{Solve } y=x^2-3 \text{ and } y=2x. \]
Set the expressions equal
Why: Both equal the vertical variable.
\[ x ^{2} - 3 = 2 x \]
Collect
Why: Bring to standard form.
\[ x ^{2} - 2 x - 3 = 0 \]
Factor
Why: Two roots.
\[ (x - 3) (x + 1) = 0 \]
Back-substitute
Why: Into either equation.
\[ (3, 6)\text{ and } (-1, -2) \]
Figure (svg): The solution to Worked example a line and a parabola shown as a ladder of expressions, one row per legal move
\[ (3,6),\;(-1,-2) \]
Verify: check in the parabola
Why: At x equal to 3, the square minus 3 is 6 — matching. At negative 1, one minus three is negative 2 — also matching. Setting the two expressions equal was a shortcut available because both equations were already solved for the same variable.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1098-1099
Error analysis
A student finds the horizontal values and recovers the vertical ones.
Annotate
On: \( x=3 \;\Longrightarrow\; \text{substitute into } x^2+y^2=25 \text{ to get } y=\pm 4 \)
Back-substituting into the simpler equation avoids inventing pairs that satisfy only one of the two. If the nonlinear equation must be used, every resulting pair has to be tested against the other.
Faded example
Substituting a line into a circle.
Fill in the blanks
x^2+(x+1)^2}=25 \;\Longrightarrow\; 2x^2+2x-24=0
Why: Expanding the square gives x squared plus 2x plus 1, which combines with the other x squared. Collecting everything on one side produces the standard quadratic form.
Sorting
One choice avoids inventing pairs.
Sort into buckets
Sort each choice.
Explain it
The substitution goes one way and not the other.
Discussion prompt
Explain to a classmate why.
Hint: What is the goal of the step?
Answer:
The goal is a single-variable equation. Substituting the linear equation replaces one variable everywhere with an expression in the other, achieving exactly that.
Going the other way substitutes a two-variable relation into an equation that already has both, so nothing is eliminated.
The general principle is substitute the simpler into the more complicated, which is what actually reduces the problem. A good explanation adds that back-substitution should also use the simpler equation, for the same reason.
Section
Section 3
Concept
When both equations are quadratic, adding or subtracting can eliminate a squared term without any substitution, leaving something much simpler.
Two circles are the clearest case: subtracting their equations eliminates both squared terms at once, leaving a line. That line is the radical axis, and it passes through both intersection points.
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1099-1101
Picture it
Elimination locates the crossings without substituting.
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
For two circles the elimination leaves a straight line through both intersection points, which is a striking simplification of a system that looked entirely quadratic.
Worked example
Both squared terms vanish.
\[ \text{Solve } x^2+y^2=25 \text{ and } (x-3)^2+y^2=10. \]
Expand the second
Why: Ready to subtract.
\[ x ^{2} - 6 x + 9 + y ^{2} = 10 \]
Subtract from the first
Why: Both squares cancel.
\[ 6 x - 9 = 15 \]
Solve the linear result
Why: One value.
\[ x = 4 \]
Substitute back
Why: Into the first circle.
\[ y = \pm 3 \]
Figure (svg): The solution to Worked example subtract two circles shown as a ladder of expressions, one row per legal move
\[ (4,3),\;(4,-3) \]
Verify: check in both circles
Why: In the first, 16 plus 9 is 25 — correct. In the second, one squared plus 9 is 10 — also correct. Both squared terms cancelling turned a pair of quadratics into a single linear equation, which is what made this quick.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1100-1100
Prediction
Both equations have the same squared terms.
Predict first
What does subtracting give?
Correct: A linear equation.
Why: Both squared terms have the same coefficients in the two equations, so both cancel on subtraction, leaving only first-degree terms and constants. That line passes through both intersection points.
Worked example
Only one term matches.
\[ \text{Solve } x^2+y^2=13 \text{ and } x^2-y=7. \]
Note the matching term
Why: Both have a squared horizontal.
\[ \text{eliminate } x ^{2} \]
Subtract
Why: The squared terms cancel.
\[ y ^{2} + y = 6 \]
Solve the quadratic
Why: Factor.
\[ y = 2\text{ or } y = -3 \]
Back-substitute both
Why: Into the second equation.
Figure (svg): Three small coordinate planes showing a line meeting a circle in two points, one point and no points
\[ (\pm 3,2),\;(\pm 2,-3) \]
Verify: check the count
Why: A circle and a parabola can meet in up to four points, and four were found — consistent with the geometry. Each vertical value gave two horizontal ones because the surviving equation was quadratic in that variable.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1100-1101
Trap
\[ x^2=16 \;\Longrightarrow\; x=4 \]
Take the positive root only
Why: The square root is applied without the plus-or-minus.
Half the solutions are lost, and the count no longer matches the sketch.
A square equals a positive number in two ways, so both signs must be kept until one is ruled out.
Each may or may not satisfy the other equation, so both are candidates and both must be tested.
The predicted solution count is the check. Finding half as many as the sketch showed is the signal that a sign was dropped.
Faded example
Subtracting one equation from another.
Fill in the blanks
(x^2+y^2)-(x^2-y)=13-7 \;\Longrightarrow\; y^2}+y=6
Why: The squared horizontal terms cancel, leaving a quadratic in the vertical variable alone. Subtracting is legitimate for the same reason it was in §9.1: any solution satisfies both equations, so it satisfies their difference.
Sorting
Look for a linear equation or matching terms.
Sort into buckets
Sort each system.
Explain it to yourself
Two quadratic equations yield a linear one.
Discussion prompt
Explain why, and what that line means.
Hint: What cancels?
Answer:
Both circle equations contain the same squared terms with the same coefficients, so subtracting removes both at once. Only first-degree terms and constants survive.
That linear equation is satisfied by every point on both circles, so it passes through both intersection points.
Which means the line is the chord joining them. A quadratic system has collapsed to a linear one, and a good explanation notes that the line exists even when the circles do not intersect — it is just that no point of it lies on either circle then.
Section
Section 4
Concept
Squaring, clearing radicals, or multiplying by a variable expression can produce candidates that satisfy the rewritten equation but not the original. Every candidate must be tested.
Testing in a rewritten equation catches nothing, because extraneous solutions satisfy the rewritten form by construction — that is where they came from. Only the originals distinguish them.
Figure (svg): A contrast between the operations that can introduce extraneous solutions and those that cannot
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1097-1101
Picture it
The left column requires a check afterwards.
Figure (svg): A contrast between the operations that can introduce extraneous solutions and those that cannot
The caption is the rule. Testing in a rewritten equation is a check that always passes and therefore tells you nothing.
Worked example
Clearing the radical introduces one.
\[ \text{Solve } y=\sqrt{x+3} \text{ and } y=x-3. \]
Set them equal
Why: Both give the vertical variable.
\[ \sqrt{x + 3} = x - 3 \]
Square both sides
Why: Clears the radical.
\[ x + 3 = x ^{2} - 6 x + 9 \]
Collect and factor
Why: A quadratic.
\[ (x - 6) (x - 1) = 0 \]
Test both candidates
Why: In the original.
Figure (svg): A contrast between the operations that can introduce extraneous solutions and those that cannot
\[ (6,3)\text{ only} \]
Verify: check the rejected candidate
Why: At x equal to 1, the radical gives 2 and the line gives negative 2. They are negatives of each other, which squaring cannot distinguish — so that candidate satisfies the squared equation and fails the original. Testing against the original is what caught it.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1098-1100
Prediction
Squaring was used during the solution.
Predict first
Where must each candidate be tested?
Correct: In both original equations.
Why: Extraneous candidates satisfy the squared equation by construction, so testing there detects nothing. Only the originals distinguish a genuine solution from one squaring invented.
Worked example
A candidate can satisfy one and fail the other.
\[ \text{Verify } (-4,-3) \text{ solves } x^2+y^2=25 \text{ and } y=x+1. \]
Test in the circle
Why: Sum of squares.
\[ 16 + 9 = 25 \]
Test in the line
Why: Substitute.
\[ -4 + 1 = -3 \]
Compare both
Why: Both hold.
Conclude
Why: It is a solution.
Figure (svg): The solution to Worked example check both equations shown as a ladder of expressions, one row per legal move
\[ \text{a genuine solution} \]
Verify: consider a point that would fail
Why: The point (4, -3) also satisfies the circle, but the line would require the vertical value to be 5. Satisfying one equation is not enough, which is why both must be checked on every candidate.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1100-1101
Error analysis
A student verifies candidates after clearing a radical.
Annotate
On: \( \text{test }x=1\text{ in } x+3=x^2-6x+9 \)
An extraneous solution is defined by satisfying the derived equation and failing the original. Testing in the derived form can never distinguish them, which makes writing the original down separately worth doing before starting.
Sorting
Only irreversible steps do.
Sort into buckets
Sort each operation.
Faded example
Testing a candidate in a radical equation.
Fill in the blanks
\sqrt2=-2, \quad 1-3=___ \;\Longrightarrow\; \text___
Why: The two sides come out as negatives of each other, which squaring could not distinguish. Testing in the original exposes the disagreement immediately.
Explain it
Some candidates have to be thrown away.
Discussion prompt
Explain to a classmate where they come from.
Hint: What does squaring lose?
Answer:
Squaring turns a statement that two things are equal into a statement that they are equal in size. That is weaker, so more values satisfy it.
Any value where the two sides are negatives of each other now qualifies, even though it failed the original. That is exactly an extraneous solution.
So the only cure is to test every candidate in the original equations. A good explanation stresses that this is part of the method after any squaring, not an optional precaution — the answer is not correct until the test is done.
Section
Section 5
Concept
A nonlinear inequality's solution is a region. Graph the boundary curve, then test one point in each region to decide which satisfies the inequality.
One test point per region is enough because the inequality's truth cannot change without crossing the boundary. That is why the boundary matters: it is precisely where the two sides are equal.
Figure (svg): A shaded region showing the solution set of a nonlinear inequality, bounded by a parabola with the interior shaded
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1101-1103
Picture it
The parabola is the boundary and the interior is shaded.
Figure (svg): A shaded region showing the solution set of a nonlinear inequality, bounded by a parabola with the interior shaded
One test point settles which side to shade. Choosing a point well away from the boundary makes the arithmetic easy and the conclusion unambiguous.
Worked example
Boundary, test, shade.
\[ \text{Graph } y>x^2. \]
Graph the boundary
Why: The parabola.
\[ y = x ^{2} \]
Choose the line style
Why: Strict inequality.
Test a point inside
Why: The point zero and one.
\[ 1 > 0,\text{ true} \]
Shade
Why: The region containing it.
Figure (svg): A shaded region showing the solution set of a nonlinear inequality, bounded by a parabola with the interior shaded
\[ \text{interior, boundary excluded} \]
Verify: test a point outside
Why: At the point (0, -1) the inequality would require negative 1 to exceed 0, which is false — so the outside is correctly unshaded. Two test points, one per region, settle the whole picture.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1101-1102
Prediction
The inequality is strict.
Predict first
How should the boundary be drawn?
Correct: Dashed, since its points are excluded.
Why: A strict inequality is not satisfied where the two sides are equal, which is exactly the boundary. Dashing records that the curve itself is not part of the solution set.
Worked example
Shade the overlap.
\[ \text{Graph } y>x^2 \text{ and } x^2+y^2<9. \]
Graph both boundaries
Why: A parabola and a circle.
Shade for the first
Why: Inside the parabola.
Shade for the second
Why: Inside the circle.
Take the overlap
Why: Both conditions.
Figure (svg): The solution to Worked example a system of inequalities shown as a ladder of expressions, one row per legal move
\[ \text{the overlap of the two regions} \]
Verify: test a point in the overlap
Why: The point (0, 1) has 1 greater than 0 and 1 less than 9, so it satisfies both — and it lies in the shaded overlap. A point in only one region, such as (0, 5), fails the circle condition and is correctly excluded.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1102-1103
Trap
\[ y>x^2 \;\Longrightarrow\; \text{shade above the curve, since it says greater} \]
Read the inequality symbol as a direction
Why: Greater is taken to mean the region above.
For a boundary that is not a function graph, above has no clear meaning and the guess fails.
Test an actual point. Substituting one point settles the region without any interpretation of the symbol.
For a circle or a sideways parabola, above and below are not well defined, and only a test point works.
The test costs one substitution and is reliable for every boundary shape, so there is no reason to guess.
Faded example
Testing the point (0, 1) in an inequality.
Fill in the blanks
1>0^2} \;\Longrightarrow\; 1>0, \text___
Why: Substituting the test point's coordinates gives a numerical statement that is either true or false. True means that whole region satisfies the inequality, since the truth cannot change without crossing the boundary.
Sorting
A test point decides.
Sort into buckets
Sort each situation for the inequality that the vertical exceeds the square of the horizontal.
Explain it to yourself
A single substitution decides a whole region.
Discussion prompt
Explain why testing one point settles it.
Hint: Where can the truth value change?
Answer:
The inequality's truth can only change where the two sides become equal, which is exactly the boundary curve.
So within a region bounded by that curve, the truth value is constant — every point in it agrees with every other.
Testing one point therefore settles the whole region. The boundary is where the change happens, and a good explanation notes that this is why graphing the boundary first is not optional: it is what defines the regions to be tested.
Comparison
Fill the blanks from memory. Two things change and one does not.
Comparison matrix
| linear | nonlinear | |
|---|---|---|
| solution count | 0, 1, or infinitely many | any number up to the degree |
| main method | substitution or elimination | the same two |
| extraneous solutions | cannot occur | possible after squaring |
| inequality solution | a half-plane | a curved region |
The methods carry over unchanged. What is new is that the answer count must be predicted and every candidate must be checked.
Pattern
Five steps, and the first and last bracket the work.
Steps 1 and 5 work together: the predicted count tells you whether the answer is complete, and the test tells you whether it is correct.
Check
Solution counts.
Check your understanding
In how many points can a line meet a circle?
Answer: A
Why: The line can miss, touch as a tangent, or pass through. Substitution gives a quadratic, whose zero, one or two real roots match those three cases exactly.
Check
Extraneous solutions.
Check your understanding
After squaring, where must each candidate be tested?
Answer: A
Why: Extraneous candidates satisfy the squared equation by construction, so testing there always passes. Only the originals, which still carry the sign information squaring destroyed, can reject them.
Check
Inequalities.
Check your understanding
How do you decide which region to shade?
Answer: A
Why: The inequality's truth value is constant within each region, since it can only change at the boundary. One substitution therefore settles the whole region, and it works for any boundary shape.
Real world
Trilateration finds a position from distances to known points, and it is a nonlinear system.
Discussion prompt
Two distance measurements from known points give two circles. Why is a third measurement needed?
Hint: How many points do two circles share?
Answer:
Each measured distance says the receiver lies on a circle around that known point. Two circles generally intersect in two points, so the position is narrowed to two candidates.
Subtracting the two circle equations gives a line through both — exactly the elimination step from this section — but the line does not say which of the two points is right.
A third circle passes through only one of them, resolving the ambiguity. This is why positioning systems need one more measurement than the number of coordinates, and the extra one exists solely to discard the second solution the nonlinear system produced.
Commit first
State your confidence along with your answer.
Predict first
Why can nonlinear systems have more solutions than linear ones?
Correct: Curves can cross each other more than once.
Why: Two distinct lines cross at most once, but a line can enter and leave a circle, and two curves can weave in and out repeatedly. The degree of the equation substitution produces gives the same bound algebraically.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate found two solutions and is not sure whether to keep both. What do you tell them?
Hint: Two checks decide it.
Answer:
Test each in both original equations. A candidate satisfying only one is not a solution to the system, and squaring can produce candidates that satisfy neither.
Then compare the count to a sketch. If the curves clearly cross twice and both candidates check out, both are genuine.
If the sketch shows only one crossing but two candidates survived the check, something is wrong with the sketch or the check — and a good explanation notes that the two methods disagreeing is itself useful information, since it says exactly where to look.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where correct algebra still produces wrong answers, which makes it the costliest. The first is what tells you when the work is finished.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch a line meeting a circle in two points, one point and none, labelling each with its solution count. Beside them, solve one system by substitution all the way to both solutions. Underneath, list the operations that can invent extraneous solutions and write where candidates must be tested. Finally, sketch one shaded region with its test point marked.
If your extraneous list names squaring and multiplying by a variable, and your test point is marked on the shaded region, the section's two habits are on the page.
Recap
Five things, and the first and last are the ones that make the middle reliable.
| if you remember one thing | it should be this |
|---|---|
| about counts | sketch first; it tells you when to stop |
| about substitution | put the simpler equation into the harder one |
| about squaring | test every candidate in the originals, never the derived form |
| about inequalities | one test point per region, and never guess the side |
Section 9.4 stays with inequalities but returns to linear ones, where the shaded regions become polygons — the setting for linear programming.
OpenStax, Precalculus, §9.3 Systems of Nonlinear Equations and Inequalities: Two Variables §9.3, pp. 1090-1103 — everything on these slides traces back here
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