Quadratic systems and the possible numbers of intersection points, solving a linear-quadratic system by graphing and by substitution, solving a system of two second-degree equations by elimination, and using two hyperbolas to locate a ship.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Solve Quadratic Systems
Objectives
Five outcomes. Chapter 3's three methods, now with conics in the system.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-663 — the lesson these objectives are drawn from
Warm-up
Chapter 3 solved linear systems by graphing, substitution and elimination.
Discussion prompt
Two lines meet in at most one point. How many times can a line cross a circle, and how did you know?
Hint: Try drawing a few lines across a circle.
Answer:
Zero, one or two. A line can miss the circle entirely, touch it once as a tangent, or cut straight through it.
\[ x^2+(mx+b)^2 = r^2: \; \text{a quadratic in } x \]
The algebra says the same thing: substituting gives a quadratic, and a quadratic has zero, one or two real roots. The methods do not change in this lesson; only the number of possible answers does.
Concept
A quadratic system contains at least one conic. Graphing, substitution and elimination all still work, but the resulting equation is now quadratic or higher, so a line and a conic can meet up to twice and two conics up to four times.
quadratic system — A system of equations in which at least one equation is second-degree, so that at least one graph is a conic section.
\[ x^2+y^2 = 10, \quad y = -3x+10 \]
The choice of method follows the shape of the system: substitution when one equation is linear, elimination when both are second-degree.
Figure (svg): Two columns comparing substitution with elimination for quadratic systems
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-660
Section
Section 1
Concept
A line and a conic can intersect in zero, one or two points. Two distinct conics can intersect in anywhere from zero to four points, so a quadratic system may have any of those numbers of solutions.
\[ \text{line and conic: } 0, 1, 2; \quad \text{two conics: } 0 \text{ to } 4 \]
The count is not chosen in advance. It is whatever the polynomial equation produced by substitution or elimination turns out to have as real roots.
Figure (svg): The possible numbers of intersection points for a line with a conic and for two conics
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-659 — Quadratic systems
Picture it
A line against a conic, and two conics against each other.
Figure (svg): The possible numbers of intersection points for a line with a conic and for two conics
One solution for a line means tangency. For two conics the odd counts happen when a solution sits on an axis of symmetry, where two would otherwise pair up.
Worked example
Reading the number of solutions off the resulting equation.
\[ \text{How many solutions do } x^2+y^2=10 \text{ with } y=-3x+10, \text{ and } x^2+4y^2=4 \text{ with } y=0.5x-3 \text{ have?} \]
First: substitute and simplify
Why: The quadratic becomes x squared minus 6x plus 9.
\[ (x - 3) ^{2} = 0 \]
First: count the real roots
Why: A repeated root.
Second: substitute and simplify
Why: Two x squared minus 12x plus 32, halved.
\[ x ^{2} - 6 x + 16 = 0 \]
Second: check the discriminant
Why: Thirty-six minus 64.
\[ -28,\text{ so no solution} \]
Figure (svg): The possible numbers of intersection points for a line with a conic and for two conics
\[ 1 \text{ solution}; \quad 0 \text{ solutions} \]
Verify: check the second against the picture
Why: The ellipse x squared over 4 plus y squared equal to 1 never rises above y equal to 1 or falls below negative 1, while the line y equals 0.5x minus 3 is below negative 2 for every x between negative 2 and 2. They cannot meet, which is exactly what the negative discriminant reported.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 659-659
Sorting
Read the resulting equation's roots.
Sort into buckets
Sort each result of substituting or eliminating.
For a line and a conic the count of real roots IS the count of solutions, since each x on the line gives exactly one y.
Worked example
Example 3, counted before it is solved.
\[ \text{Why can } 9x^2+y^2-90x+216=0 \text{ and } x^2-y^2-16=0 \text{ have an odd number of solutions?} \]
Eliminate and solve for x
Why: Adding gives a quadratic with roots 4 and 5.
\[ x = 4\text{ or } x = 5 \]
Find y at the first root
Why: Sixteen minus y squared minus 16 equals zero.
\[ y = 0,\text{ one point} \]
Find y at the second root
Why: Twenty-five minus y squared minus 16 equals zero.
\[ y = +- 3,\text{ two points} \]
Count
Why: One point from the first root, two from the second.
Figure (svg): An ellipse and a hyperbola meeting at three points, solved by elimination
\[ (4,0), \; (5,3), \; (5,-3) \]
Verify: notice where the odd one sits
Why: The solution 4 comma 0 lies on the horizontal axis, which is an axis of symmetry for both conics — so its mirror image is itself. Every other solution comes in a symmetric pair, which is why odd counts always involve a point on an axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660
Trap
\[ x^2+4y^2 = 4 \text{ and } y = 0.5x-3 \]
Expect at least one intersection
Why: Curves are assumed to cross somewhere.
\[ \text{find the two points} \quad \text{(there are none)} \]
The resulting quadratic has discriminant negative 28, so it has no real roots at all — and the graphs confirm it, since the line runs entirely below the ellipse.
\[ x^2-6x+16 = 0, \; b^2-4ac = -28 < 0 \]
Let the discriminant report the count
Why: No real roots means no intersection points.
\[ \text{the system has no solution} \]
No solution is a complete answer here, exactly as it was for the rational equation in Lesson 8.6. It is a fact about the system rather than a failure to find something.
Prediction
Commit before reasoning.
Predict first
Substituting a line into a circle gives x minus 3, all squared, equal to zero. What does that say geometrically?
Correct: The line is tangent to the circle, touching at exactly one point.
\[ (x-3)^2 = 0 \;\Longrightarrow\; \text{one point, } (3,1) \]
Why: A repeated root means the quadratic just touches zero rather than crossing it, and geometrically that is a line grazing the curve. A line through the centre would give two roots, the two ends of a diameter. This is the same reading of the discriminant used for quadratic functions in Chapter 4, applied to an intersection rather than to an x-intercept.
Matching
Line against conic.
Match the pairs
Why: The first three are the whole story for a line and a conic. The fourth needs a second-degree partner, since only two conics can produce an odd count above one.
Two truths and a lie
All three are about solution counts.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The ellipse x squared plus 4y squared equal to 4 and the line y equal to 0.5x minus 3 never meet, and the algebra says so with a discriminant of negative 28. No solution is an ordinary outcome for a system of any kind, linear or quadratic.
Section
Section 2
Concept
To graph a system, solve every equation for y. A conic that is not a function has to be entered as two separate halves, one for each sign of the square root, or the calculator will draw only part of it.
\[ y^2 = 7x-3 \;\Longrightarrow\; y = \pm\sqrt{7x-3} \]
Graphing gives approximate answers and shows the count at a glance, which makes it a good first move before an exact method.
Figure (svg): A sideways parabola and a line meeting at two points, solved by graphing
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-658 — Solve a linear-quadratic system by graphing
Picture it
Example 1: a sideways parabola crossed by a line.
Figure (svg): A sideways parabola and a line meeting at two points, solved by graphing
The lower half meets the line near 0.75 comma negative 1.5, and the upper half at 4 comma 5. Entering only one half would have found only one of them.
Worked example
Example 1.
\[ \text{Solve } y^2-7x+3=0 \text{ and } 2x-y=3 \text{ by graphing.} \]
Solve the first for y
Why: Isolate y squared, then take both roots.
\[ y = +- \sqrt{7 x - 3} \]
Solve the second for y
Why: Ordinary rearranging.
\[ y = 2 x - 3 \]
Enter all three curves
Why: Both halves of the parabola plus the line.
Read the intersections
Why: Use the calculator's intersect feature twice.
\[ (0.75, -1.5)\text{ and } (4, 5) \]
Figure (svg): A sideways parabola and a line meeting at two points, solved by graphing
\[ (0.75,-1.5) \text{ and } (4,5) \]
Verify: substitute both into both equations
Why: At 4 comma 5: 25 minus 28 plus 3 is 0, and 8 minus 5 is 3. At 0.75 comma negative 1.5: 2.25 minus 5.25 plus 3 is 0, and 1.5 plus 1.5 is 3. Both check in both equations, which is what a solution of a SYSTEM has to do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-658
Fill the middle
Example 1.
Fill in the blanks
y^2 = 7x-3 \;\Longrightarrow\; y = +-\sqrt___
Why: Both signs are needed, since squaring loses the sign. Entering only the positive root would show only the upper half and miss any intersection below the axis.
Worked example
Guided Practice 1 and 2.
\[ \text{Solve } x^2+y^2=13 \text{ with } y=x-1, \text{ and } x^2+8y^2-4=0 \text{ with } y=-x+7. \]
First: solve for y
Why: The circle needs both halves.
\[ y = +- \sqrt{13 - x ^{2}}, y = x - 1 \]
First: read the intersections
Why: The line cuts the circle twice.
\[ (3, 2)\text{ and } (-2, -3) \]
Second: look at the ranges
Why: The ellipse never rises above about 0.71 in y.
\[ | y | \le 0.71 \]
Second: compare with the line
Why: The line is above 5 wherever the ellipse exists.
Figure (svg): The solution to Worked example two more by graphing shown as a ladder of expressions, one row per algebraic move
\[ (3,2), (-2,-3); \quad \text{no solution} \]
Verify: check the first pair algebraically
Why: At 3 comma 2: 9 plus 4 is 13, and 3 minus 1 is 2. At negative 2 comma negative 3: 4 plus 9 is 13, and negative 2 minus 1 is negative 3. Both check. For the second system, comparing the ranges was faster and more certain than squinting at a graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 659-659
Error analysis
A student graphs a sideways parabola on a calculator.
Annotate
On: \( \text{entered only } y = \sqrt{7x-3} \)
Any conic that fails the vertical line test needs two entries. Circles, ellipses, sideways parabolas and hyperbolas all do.
Sorting
Ask whether the curve is a function.
Sort into buckets
Sort each curve by how many calculator entries it needs.
Only a line or an upward or downward parabola escapes the split. Everything else in Chapter 9 needs both halves.
Ranking
Solving a system by graphing.
Put in order
Why: Step five matters because graphing gives approximations: a reading of 0.75 might really be 0.7499 or exactly three quarters, and only substitution settles it. It also catches a crossing that the chosen window hid.
Prediction
Commit before reasoning.
Predict first
What is the main risk of solving a quadratic system only by graphing?
Correct: A crossing may lie outside the chosen window, or two crossings may be too close to distinguish.
\[ \text{graph to see; substitute to be sure} \]
Why: The screen shows only a rectangle of the plane, and a hyperbola's branches or a wide parabola can carry an intersection far outside it. Nearly tangent curves are also hard to tell apart on a low-resolution display. Graphing is excellent for seeing how many solutions to expect and roughly where; substitution or elimination is what makes them exact and complete.
Section
Section 3
Concept
When one equation of the system is linear, solve it for a variable and substitute into the conic. The result is a single quadratic in one variable, and its roots give the solutions.
\[ x^2+(-3x+10)^2 = 10 \;\Longrightarrow\; (x-3)^2 = 0 \]
Substitute the resulting x back into the LINEAR equation rather than the conic. The conic would give two candidate y values, only one of which belongs to the solution.
Figure (svg): A circle and a line meeting at exactly one point, solved by substitution
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 659-659 — Solve a linear-quadratic system by substitution
Picture it
Example 2: a circle and a line meeting exactly once.
Figure (svg): A circle and a line meeting at exactly one point, solved by substitution
The quadratic collapsed to a perfect square, which is the algebra reporting tangency. The single point of contact is 3 comma 1.
Worked example
Example 2.
\[ \text{Solve } x^2+y^2=10 \text{ and } y=-3x+10 \text{ by substitution.} \]
Substitute for y
Why: The linear equation is already solved for y.
\[ x ^{2} + (-3 x + 10) ^{2} = 10 \]
Expand and combine
Why: Nine x squared minus 60x plus 100, plus x squared.
\[ 10 x ^{2} - 60 x + 90 = 0 \]
Divide and factor
Why: Every term over 10 gives a perfect square trinomial.
\[ (x - 3) ^{2} = 0 \]
Find y from the LINEAR equation
Why: Negative 3 times 3, plus 10.
\[ y = 1 \]
Figure (svg): A circle and a line meeting at exactly one point, solved by substitution
\[ (3, 1) \]
Verify: check in both equations
Why: Nine plus 1 is 10, and negative 9 plus 10 is 1. Both hold. Note the book's caution: substituting x equal to 3 into the CIRCLE would give y equal to plus or minus 1, and negative 1 fails the line — an extraneous candidate created by using the wrong equation for the back-substitution.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 659-659
Fill the middle
Example 2.
Fill in the blanks
x^2+(-3x+10)^2 = x^2+9x^2-60x+100
Why: Ten squared is 100, and the middle term is twice negative 3x times 10, which is negative 60x. Expanding the square carefully is where most substitution errors happen.
Worked example
Guided Practice 5 and 6.
\[ \text{Solve } y^2-2x-10=0 \text{ with } y=-x-1, \text{ and } 9x^2-y^2-36=0 \text{ with } y=4x-8. \]
First: substitute and expand
Why: X plus 1, squared, equals 2x plus 10.
\[ x ^{2} + 2 x + 1 = 2 x + 10 \]
First: solve
Why: The linear terms cancel, leaving x squared equal to 9.
\[ x = 3\text{ or } x = -3 \]
Second: substitute and simplify
Why: Nine x squared minus 16x squared plus 64x minus 64 equals 36.
\[ 7 x ^{2} - 64 x + 100 = 0 \]
Second: factor and solve
Why: Seven x minus 50 times x minus 2.
\[ x = 2\text{ or } x = \frac{50}{7} \]
Figure (svg): The solution to Worked example two more by substitution shown as a ladder of expressions, one row per algebraic move
\[ (3,-4),(-3,2); \quad (2,0), \left(\tfrac{50}{7},\tfrac{144}{7}\right) \]
Verify: check one from each
Why: For the first, at 3 comma negative 4: 16 minus 6 minus 10 is 0, and negative 3 minus 1 is negative 4. For the second, at 2 comma 0: 36 minus 0 minus 36 is 0, and 8 minus 8 is 0. Both check in both of their equations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 659-659
Trap
\[ x = 3 \text{ in } x^2+y^2 = 10 \]
Find y from the circle
Why: Either original equation is assumed to serve.
\[ y = \pm 1: \; (3,1) \text{ and } (3,-1) \quad \text{(one is wrong)} \]
The point 3 comma negative 1 satisfies the circle but not the line, since negative 3 times 3 plus 10 is 1, not negative 1.
\[ x = 3 \text{ in } y = -3x+10 \;\Longrightarrow\; y = 1 \]
Back-substitute into the LINEAR equation
Why: It gives one y per x, so no false candidate can appear.
\[ \text{the only solution is } (3,1) \]
If you do use the conic, every candidate must be tested in the other equation — which is more work than simply choosing the linear one.
Ranking
Solving a linear-quadratic system.
Put in order
Why: Step five specifies the linear equation deliberately: using the conic there would produce two candidate values of y for each x, only one of which satisfies the line. Choosing the right equation is easier than testing extra candidates.
Matching
Substitute, then solve.
Match the pairs
Why: All four counts appear here: one, two, two and none. The first is tangency and the last is a line that misses its ellipse entirely, which the negative discriminant reports without any graphing.
Prediction
Commit before reasoning.
Predict first
A system pairs a conic with the line 2x minus y equal to 3. Which variable is easier to isolate?
Correct: y, since its coefficient is negative one and no fractions appear.
\[ 2x-y=3 \;\Longrightarrow\; y = 2x-3, \text{ no fractions} \]
Why: Solving for y gives y equal to 2x minus 3 with integer coefficients, while solving for x gives x equal to y plus 3, all over 2 — and squaring a fraction in the substitution invites arithmetic slips. Choosing the variable with coefficient 1 or negative 1 is a habit worth keeping from Chapter 3, and it matters more here because the expression is about to be squared.
Section
Section 4
Concept
When both equations are second-degree, add or subtract multiples of them so that one squared term cancels. What remains is often a quadratic in one variable, or sometimes a line.
\[ 9x^2+y^2-90x+216=0 \text{ plus } x^2-y^2-16=0 \]
Each surviving value of x may give two values of y, so the count of solutions can exceed the count of roots. Substituting back into a conic is unavoidable here.
Figure (svg): An ellipse and a hyperbola meeting at three points, solved by elimination
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660 — Solve a quadratic system by elimination
Picture it
Example 3: three intersection points.
Figure (svg): An ellipse and a hyperbola meeting at three points, solved by elimination
Adding removed the y squared terms and left a quadratic in x. One root gave a single point on the axis and the other gave a symmetric pair.
Worked example
Example 3.
\[ \text{Solve } 9x^2+y^2-90x+216=0 \text{ and } x^2-y^2-16=0. \]
Add the equations
Why: The y squared terms have opposite signs and cancel.
\[ 10 x ^{2} - 90 x + 200 = 0 \]
Divide and factor
Why: Every term over 10, then factor.
\[ (x - 4) (x - 5) = 0 \]
Back-substitute the first root
Why: Sixteen minus y squared minus 16 is 0.
\[ y = 0 \]
Back-substitute the second root
Why: Twenty-five minus y squared minus 16 is 0.
\[ y = +- 3 \]
Figure (svg): An ellipse and a hyperbola meeting at three points, solved by elimination
\[ (4,0), \; (5,3), \; (5,-3) \]
Verify: check one solution in both equations
Why: At 5 comma 3: the first gives 225 plus 9 minus 450 plus 216, which is 0; the second gives 25 minus 9 minus 16, also 0. Both hold. Note that here the back-substitution HAD to use a conic, since neither equation is linear — so both signs of y must be kept and both tested.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660
Fill the middle
Example 3.
Fill in the blanks
(9x^2+y^2-90x+216)+(x^2-y^2-16) = 10x^2-90x+200
Why: Nine x squared plus x squared is 10x squared, and the y squared terms cancel because their coefficients are opposite. Choosing which term to eliminate is the whole art here.
Worked example
Example 4, Step 1.
\[ \text{Add } x^2-y^2-16x+32=0 \text{ and } -x^2+y^2-8y+8=0. \]
Add the equations
Why: Both x squared and y squared cancel.
\[ -16 x - 8 y + 40 = 0 \]
Divide by negative 8
Why: Two x plus y minus 5 equals zero.
\[ 2 x + y = 5 \]
Solve for y
Why: A linear equation has appeared.
\[ y = -2 x + 5 \]
Recognise what happened
Why: The system has become linear-quadratic.
Figure (svg): The solution to Worked example elimination leaving a line shown as a ladder of expressions, one row per algebraic move
\[ y = -2x+5 \]
Verify: explain why a line appeared
Why: Both conics had x squared and y squared with equal and opposite coefficients, so both cancelled at once and only first-degree terms survived. That line is not one of the original curves — it is the line through their intersection points, which is exactly what makes it useful for finding them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660
Error analysis
A student back-substitutes into a conic after eliminating.
Annotate
On: \( x = 5: \; 25-y^2-16 = 0 \;\Longrightarrow\; y = 3 \)
When the back-substitution uses a conic rather than a line, both signs must be kept and each candidate tested in the other equation.
Sorting
Look at the degrees.
Sort into buckets
Sort each system.
Two circles are the friendliest elimination of all: subtracting them cancels both squared terms and leaves the line through their intersections.
Comparison
Fill the blanks. Different systems, different tools.
Comparison matrix
| Question | Substitution | Elimination |
|---|---|---|
| Best when | one equation is linear | both are second-degree |
| The move | solve for a variable and replace it | add or subtract to cancel a squared term |
| Back-substitute into | the linear equation | a conic, keeping both signs |
| Risk | expanding a square wrongly | losing one sign of y |
Each method's risk sits at its final step, which is where the checking effort belongs.
Prediction
Commit before reasoning.
Predict first
Adding the two hyperbolas of Example 4 gives y equal to negative 2x plus 5. What is that line?
Correct: The line through both intersection points of the two conics.
\[ (-1,7) \text{ and } \left(\tfrac{7}{3},\tfrac{1}{3}\right) \text{ both satisfy } y = -2x+5 \]
Why: Any point satisfying both original equations satisfies their sum, so both intersection points lie on that line. It is not part of either conic — most of its points satisfy neither — but it passes through exactly the points where they meet, which is why substituting it back finds them. The same trick works for two circles, where subtracting gives the line through their two crossings.
Section
Section 5
Concept
Radio signals from two pairs of stations put a ship on two hyperbolas. Solving the system gives the candidate positions, and a single extra fact about the situation selects which one is real.
\[ y = -2x+5, \; 3x^2-4x-7=0 \]
Mathematics supplies the candidates and context supplies the choice, exactly as it did for extraneous roots and for the direction of travel in earlier lessons.
Figure (svg): Two hyperbolas from radio stations, meeting at two possible ship positions
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660 — Solve a real-life quadratic system
Picture it
Example 4: two crossings, one of them east of the vertical axis.
Figure (svg): Two hyperbolas from radio stations, meeting at two possible ship positions
The candidates are negative 1 comma 7 and seven thirds comma one third. Being east of the axis makes the second the ship's position.
Worked example
Example 4.
\[ \text{Solve } x^2-y^2-16x+32=0 \text{ and } -x^2+y^2-8y+8=0, \text{ then choose the point east of the vertical axis.} \]
Add to eliminate both squares
Why: The x squared and y squared terms both cancel.
\[ y = -2 x + 5 \]
Substitute into the first equation
Why: X squared minus the square of negative 2x plus 5, minus 16x plus 32.
\[ 3 x ^{2} - 4 x - 7 = 0 \]
Factor and solve
Why: X plus 1 times 3x minus 7.
\[ x = -1\text{ or } x = \frac{7}{3} \]
Find both y values and choose
Why: Seven and one third; east means positive x.
\[ (\frac{7}{3}, \frac{1}{3}) \]
Figure (svg): Two hyperbolas from radio stations, meeting at two possible ship positions
\[ \left(\tfrac{7}{3}, \tfrac{1}{3}\right) \]
Verify: check the chosen point in both equations
Why: At seven thirds comma one third: the first gives 49 over 9 minus 1 over 9 minus 112 over 3 plus 32, which is 48 over 9 minus 112 over 3 plus 32, or 0. The second gives negative 49 over 9 plus 1 over 9 minus 8 over 3 plus 8, also 0. Both hold, and the point has positive x as required.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660
Fill the middle
Example 4, Step 2.
Fill in the blanks
x^2-(-2x+5)^2-16x+32 = -3x^2-4x-7
Why: One x squared minus 4x squared is negative 3x squared, and multiplying through by negative 1 gives the tidier 3x squared minus 4x minus 7. Either form factors to the same roots.
Worked example
Understanding why two answers appear at all.
\[ \text{Check that } (-1, 7) \text{ also satisfies both equations, and say why it is rejected.} \]
Test in the first equation
Why: One minus 49 plus 16 plus 32.
\[ 0,\text{ it satisfies} \]
Test in the second equation
Why: Negative 1 plus 49 minus 56 plus 8.
\[ 0,\text{ it satisfies} \]
Note it is a genuine solution
Why: Both hyperbolas really do pass through it.
Apply the context
Why: The ship is east of the vertical axis, so x must be positive.
Figure (svg): The solution to Worked example the rejected candidate shown as a ladder of expressions, one row per algebraic move
\[ (-1,7) \text{ satisfies both, but } x < 0 \]
Verify: distinguish rejection from extraneousness
Why: An extraneous root fails the original equations, as in Lesson 8.6. This candidate passes both, and is discarded only because the ship is known to be east. That distinction matters: the mathematics is complete either way, and the extra fact is doing modelling work rather than error-checking.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 660-660
Trap
\[ (-1,7) \text{ and } \left(\tfrac{7}{3},\tfrac{1}{3}\right) \]
Call the first extraneous
Why: Any rejected answer is assumed to have failed the equations.
\[ \text{claim } (-1,7) \text{ does not satisfy the system} \quad \text{(wrong)} \]
It satisfies both equations exactly. Nothing algebraic rules it out.
\[ \text{both are solutions of the SYSTEM} \]
Reject on the stated physical condition instead
Why: The ship is east of the vertical axis, so its x-coordinate is positive.
\[ -1 < 0 \;\Longrightarrow\; \text{not this one} \]
Two hyperbolas genuinely cross twice, which is why real navigation systems need a rough position to start from, or a third pair of stations.
Sorting
Two different reasons look similar.
Sort into buckets
Sort each rejection.
Both kinds get discarded, but only the first indicates that a step in the algebra was not reversible. Naming which is which keeps the reasoning honest.
Prediction
Commit before reasoning.
Predict first
Two pairs of stations put a ship on two hyperbolas that cross twice. What does that imply?
Correct: A further fact is needed to choose between the two candidates, such as a rough known position.
\[ \text{two hyperbolas} \;\Longrightarrow\; \text{up to four crossings} \]
Why: Two conics can meet more than once, so two measurements narrow the position to a short list rather than a point. In practice a navigator knows roughly where the ship is and picks the nearby candidate, or a third station adds another curve. Satellite navigation faces the same issue in three dimensions, which is why four satellites are used rather than three.
Ranking
Locating a position from two conics.
Put in order
Why: Step five is the only one that is not algebra, and it is the one that turns a pair of answers into a position. Reporting both candidates without applying it would leave the question unanswered.
Comparison
Fill the blanks. Chapter 3's tools, applied to conics.
Comparison matrix
| Method | Best for | What to watch |
|---|---|---|
| Graphing | seeing how many solutions there are | split any non-function into two entries |
| Substitution | a system with one linear equation | back-substitute into the linear equation |
| Elimination | two second-degree equations | keep both signs of y |
| All three | give the same solutions | check every answer in both originals |
The final row is the one that never changes: a solution of a system has to satisfy every equation in it, not just the one it came from.
Pattern
Choose the method from the shape of the system.
Graphing is worth doing first to see how many solutions to expect, but it gives approximations rather than exact values.
Check
Substitution. A repeated root means something.
Check your understanding
Solve x^2 + y^2 = 10 and y = -3x + 10.
Answer: A
Why: The quadratic becomes a perfect square, so the line is tangent at one point.
Check
Elimination. Both signs of y.
Check your understanding
How many solutions does 9x^2 + y^2 - 90x + 216 = 0 with x^2 - y^2 - 16 = 0 have?
Answer: A
Why: The root x = 4 gives one point on the axis and x = 5 gives a symmetric pair.
Check
Choosing among candidates.
Check your understanding
A system gives (-1, 7) and (7/3, 1/3), and the ship is east of the vertical axis. Which is it?
Answer: A
Why: East of the vertical axis means positive x, which only the second candidate has.
Real world
A satellite receiver at the origin needs to know where a ground station sits. It measures that the station is 5 kilometres away, and a second receiver at 8 comma 0 measures the same station at 5 kilometres from itself.
Discussion prompt
Set up and solve the system, and explain why a third measurement is normally taken.
Hint: Each measurement is a circle.
Answer:
\[ x^2+y^2 = 25; \qquad (x-8)^2+y^2 = 25 \]
\[ \text{subtracting: } -16x+64 = 0 \;\Longrightarrow\; x = 4 \]
\[ 16+y^2 = 25 \;\Longrightarrow\; y = \pm 3 \]
The station is at 4 comma 3 or 4 comma negative 3 — two candidates, and nothing in the measurements distinguishes them. That is the same two-crossing problem the ship faced.
Subtracting the two circles eliminated both squared terms and left the vertical line x equal to 4, exactly as Example 4's addition left a line. That line is the perpendicular bisector of the segment joining the two receivers, which is Lesson 9.1's construction arriving from a completely different direction. A third receiver, not on the line through the first two, gives a third circle that passes through only one of the two candidates — which is why positioning systems always use one more station than the dimension they are working in.
Commit first
Answer, then rate your confidence honestly.
Predict first
After substituting and finding x equal to 3, may you use either original equation to find y?
Correct: Prefer the linear one — the conic can produce a candidate that fails the other equation.
\[ x=3 \text{ in the circle: } y = \pm 1; \quad \text{in the line: } y = 1 \]
Why: In Example 2, substituting x equal to 3 into the circle gives y equal to plus or minus 1, but only positive 1 satisfies the line. The extra candidate is not wrong arithmetic — it is a genuine point of the circle that simply is not on the line, and so is not a solution of the SYSTEM. Using the linear equation avoids it entirely, since a line gives exactly one y for each x. When there is no linear equation, as in an elimination problem, both signs must be kept and each candidate tested.
Explain it
They can solve two linear equations and have just met conics.
Discussion prompt
In four sentences or fewer, explain how solving a system with a conic differs from solving two lines.
Hint: Think about how many answers there can be.
Answer:
The methods are the same: substitute one equation into the other, or add them to cancel a term. What changes is the equation you end up with — it has a square in it, so it can have two answers rather than one.
That means the curves can cross twice, touch once, or miss entirely. So always check every answer in both original equations, and expect that sometimes there is no answer at all.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For expanding, write the square out in full before combining anything. For eliminating, look for squared terms whose coefficients are already opposite or easily made so. For signs, write plus or minus explicitly the moment you take a square root. For choosing, underline the condition in the question before you start solving.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a systems page. Top: sketch the three possible line-and-conic pictures and beside each write what the resulting quadratic's discriminant does. Middle left: work Example 1 by graphing, showing both halves of the parabola written out separately, and then confirm both answers algebraically. Middle right: work Example 2 by substitution in full, and write beside it what happens if you back-substitute into the circle instead of the line. Bottom left: work Example 3 by elimination, marking clearly where one root gives one point and the other gives two. Bottom right: work Example 4, showing the line that appears after adding, and write one sentence on the difference between a rejected candidate and an extraneous root.
If your Example 3 answer has only two points, check the root x equal to 5 again: it gives y equal to plus or minus 3, which is two solutions rather than one.
Recap
Five things, and Chapter 9 is complete.
| If you see | Then |
|---|---|
| One linear equation | Substitute it into the conic |
| Two second-degree equations | Eliminate a squared term |
| Both squared terms cancelling | A line appears; substitute it back |
| A perfect square after substituting | The line is tangent; one solution |
| A negative discriminant | No solution at all |
| Two valid candidates | Use the stated condition to choose |
That closes Chapter 9. Chapter 10 turns to counting methods and probability, where permutations, combinations and the binomial theorem take over.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.7 Solve Quadratic Systems §9.7, pp. 658-663 — everything on these slides traces back here
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