Systems in three variables
Three unknowns tied together by three equations. How do you pull the numbers apart when every equation holds all three letters at once?
Where you are 7 in Systems of equations
When you reach for it
Some questions have three unknown numbers and three facts that tie them together. How many shots of each value did a player make? How was a sum of money split between three accounts? Which circle passes through three given points? Each fact becomes a linear equation, and every one of them may hold all three letters:
A solution is an ordered triple (x, y, z) that makes all three equations true at once. For the system above it is (4, -3, 2): put those numbers in and each line comes out true, which you can check in your head.
No single equation can give the answer, because each one has three unknowns. The move is the one from solving by elimination, used twice: remove one letter from two different pairs of equations. That leaves two equations in two letters, which you already know how to solve. Reach for it whenever the equations are linear (each letter to the first power, no letter multiplied by another) and there are as many equations as unknowns.
The procedure
Day 1: from three letters to two
Number the equations so you can talk about them.
Choose one letter to remove, and remove it twice. Any letter will do, but some are cheaper. Here x has coefficient 1 in (1) and -1 in (3), so adding those two removes it with no multiplying:
Call this (4). Adding is allowed for the reason it was allowed with two letters: a triple that makes both sides of (1) equal and both sides of (3) equal makes the sums of those sides equal too.
The second pair must bring in the equation not yet used. (4) came from (1) and (3), so the next elimination uses (2). Multiplying (1) by -2 gives -2x - 2y - 2z = -6, and adding (2) removes x again:
Call this (5). Multiplying an equation by a number other than 0 keeps exactly the same solutions, since the multiplying can be undone by dividing.
Both new equations are missing the same letter, x. That is the whole point: (4) and (5) now form a system in y and z alone.
Solve the two-letter system. The z terms are opposites, so add (4) and (5):
Put y = -3 into (4): -15 - z = -17, so z = 2.
Go back for the third letter. Any original equation that holds x will do; (1) is simplest: x + (-3) + 2 = 3, so x = 4. The solution is the ordered triple (4, -3, 2).
Check in all three originals. (1): 4 - 3 + 2 = 3. (2): 8 + 3 + 6 = 17. (3): -4 - 12 - 4 = -20. All three hold.
Why nothing was gained or lost along the way: the system (1), (4), (5) has exactly the same solutions as (1), (2), (3). Every triple that solves the old system solves the new one, by the adding argument above. And the old equations can be rebuilt from the new ones, since (3) is (4) minus (1) and (2) is (5) plus twice (1), so every triple that solves the new system solves the old one. Elimination swaps a system for an easier one with the same solution set.
Choosing the letter. Pick the letter whose coefficients make the multiplying light: a coefficient of 1 or -1, or two coefficients that are already opposites. If one equation is already missing a letter, remove that letter from the other two: one of the two eliminations has been done for you.
Day 2: when the letters all vanish
An equation ax + by + cz = d, with a, b and c not all 0, describes a flat plane in space, just as ax + by = c describes a line in the plane (see lines and planes in space). A solution of three such equations is a point on all three planes, and three planes need not share exactly one point.
Elimination tells you which picture you are in, without drawing anything.
- If you reach a single value for each letter, the planes share one point.
- If the letters all vanish and leave a false statement such as 0 = 5, no triple can make it true, so there is no solution. The system is called inconsistent. This happens when two of the planes are parallel, and also in the tent picture, where every pair of planes meets but no point lies on all three.
- If the letters all vanish and leave a true statement 0 = 0, one equation was a combination of the others and told you nothing new. There are infinitely many solutions. Usually they form a line: let one letter be any number t and write the other two in terms of t. That letter is called a free variable.
A false or true statement can appear at either stage: when you eliminate the first letter (two equations that are parallel or identical), or when you solve the two-letter system.
Worked examples
Day 1: is a triple a solution?
Is (3, 1, 2) a solution of the system below? Is (1, 5, 2)?
A triple is a solution only if it makes every equation true, so test each one in turn.
For (3, 1, 2): the first gives 6 + 1 - 2 = 5, the second 3 - 3 + 4 = 4, the third 9 + 2 + 2 = 13. All three hold, so (3, 1, 2) is a solution.
For (1, 5, 2): the first gives 2 + 5 - 2 = 5, which holds. The second gives 1 - 15 + 4 = -10, not 4. One false equation is enough: (1, 5, 2) is not a solution. It lies on the first plane but not on the second, and there is no need to test the third.
Day 1: a letter already missing
Solve the system.
Equation (1) has no z. So remove z from the other two as well, and the result pairs up with (1) directly. Multiply (3) by 3 to get 3x + 6y + 3z = 15 and add (2); the z terms cancel:
Now (1) and (4) are two equations in x and y. Equation (1) gives y = 4x - 1, so substitute into (4):
Then y = 4(1) - 1 = 3, and (2) gives 3 - 3z = 9, so z = -2.
The solution is (1, 3, -2). Check: 4 - 3 = 1, 3 + 6 = 9, and 1 + 6 - 2 = 5.
Day 1: multiplying both equations of each pair
Solve the system.
No coefficient is 1, so look for the letter that costs least. The y coefficients are 2, -3 and 4: the 2 and 4 match after doubling, and the 2 and -3 become opposites as 6 and -6. Remove y.
Pair (1) and (2): multiply (1) by 3 and (2) by 2, then add.
Call this (4). Pair (1) and (3), which brings in the unused equation: multiply (1) by -2 and add (3).
Call this (5). It gives x = 5z - 26, which goes into (4):
So x = 25 - 26 = -1. From (1), -3 + 2y - 20 = -19, so 2y = 4 and y = 2.
The solution is (-1, 2, 5). Check (2): -2 - 6 + 25 = 17. Check (3): -5 + 8 - 15 = -12.
Day 1: standard form first, and a fraction in the answer
Solve the system.
Elimination works column by column, so first put every equation in the form ax + by + cz = d, with the letters in the same order. Adding 2x to both sides of the first and subtracting y from both sides of the second:
The y coefficients 1, -1, 2 are the easiest. Adding (1) and (2) gives 6x + 3z = 12, and dividing by 3 gives
Twice (2) plus (3): 8x - 2y + 4z + 6x + 2y - z = 18 - 2, that is
From (4), z = 4 - 2x. Then (5) becomes 14x + 12 - 6x = 16, so 8x = 4 and \displaystyle x = \frac{1}{2}. Then z = 4 - 1 = 3, and (1) gives 1 + y + 3 = 3, so y = -1.
The solution is \displaystyle \left(\frac{1}{2}, -1, 3\right). Nothing in the method needs whole numbers; the arithmetic simply carries the fraction. Check in the original second equation: \displaystyle 4\left(\frac{1}{2}\right) + 2(3) = 8 and -1 + 9 = 8.
Day 2: no solution
Solve the system.
Remove x. Multiply (1) by -2 and add (2):
Multiply (1) by -3 and add (3):
Subtracting (5) from (4) removes both letters and leaves 0 = -2, which is false. A triple that solved the system would also solve (4) and (5), and then 0 = -2 would be true. So there is no solution: the system is inconsistent.
The reason is visible once you look for it. Adding (1) and (2) gives 3x + y + 2z = 5, while (3) says the same left side equals 7. No two of the planes are parallel (no equation’s left side is a multiple of another’s), so each pair meets in a line, but the three lines are parallel and no point lies on all three planes. This is the tent in the figure.
Day 2: infinitely many solutions
Solve the system.
Remove y. Adding (1) and (2) gives 3x + 3z = 6, so x + z = 2. Adding (2) and (3) gives 6x + 6z = 12, so again x + z = 2. The two new equations are the same, and subtracting one from the other leaves 0 = 0: true, whatever the letters are.
The true statement means one of the original equations repeated information already in the others. Here (3) is exactly twice (1) plus (2): 2x - 2y + 4z + 2x + y + z = 4x - y + 5z and 2 + 5 = 7. So there are really only two conditions on three unknowns, and the planes share a line, as in the open book of the figure.
Describe the line by letting z = t, where t is any real number. From x + z = 2, x = 2 - t. From (1), y = x + 2z - 1 = 2 - t + 2t - 1 = 1 + t. The solutions are
for every real t. For instance t = 0 gives (2, 1, 0) and t = 1 gives (1, 2, 1). Check the general triple in (2): 2(2 - t) + (1 + t) + t = 4 - 2t + 1 + 2t = 5. In (3): 4(2 - t) - (1 + t) + 5t = 8 - 4t - 1 - t + 5t = 7.
Day 2: points in a basketball game
In one game Zuzana made 16 successful shots worth 30 points in all. Free throws are worth 1 point, two-pointers 2 and three-pointers 3. She made as many two-pointers as free throws and three-pointers together. How many of each did she make?
SolutionLet f, t and h be the numbers of free throws, two-pointers and three-pointers. Each sentence of facts gives one equation. The count of shots, the count of points, and the comparison:
The third came from t = f + h by subtracting t from both sides. Subtracting (3) from (1) removes f and h at once: 2t = 16, so t = 8. Subtracting (1) from (2) gives t + 2h = 14, so 8 + 2h = 14 and h = 3. Then (1) gives f = 16 - 8 - 3 = 5.
Zuzana made 5 free throws, 8 two-pointers and 3 three-pointers. Check against the story: 5 + 8 + 3 = 16 shots, 5 + 16 + 9 = 30 points, and 8 = 5 + 3.
Day 2: the circle through three points
Every circle has an equation of the form x^2 + y^2 + Dx + Ey + F = 0: multiply out (x - h)^2 + (y - k)^2 = r^2 and move r^2 across, which gives D = -2h, E = -2k and F = h^2 + k^2 - r^2. Find the circle through (5, 3), (-1, 3) and (6, -4), and give its center and radius.
SolutionThe unknowns are the numbers D, E and F, not x and y. Once a point is put in, x^2 + y^2 is a known number, so each point gives an equation that is linear in D, E and F. The point (5, 3) gives 25 + 9 + 5D + 3E + F = 0. Doing the same for all three:
Every equation has F with coefficient 1, so subtraction removes it. (1) minus (2) gives 6D = -24: the first two points have the same y, so E went too, and D = -4. (1) minus (3) gives -D + 7E = 18, so 4 + 7E = 18 and E = 2. Then (1) gives -20 + 6 + F = -34, so F = -20.
The circle is x^2 + y^2 - 4x + 2y - 20 = 0. To read off its center, complete the square in x and in y:
This is the equation of a circle with center (2, -1) and radius 5. Check the third point: (6 - 2)^2 + (-4 + 1)^2 = 16 + 9 = 25.
Boundaries
Linear equations only. Elimination relies on adding and scaling terms like 3y column by column. A term such as xy, y^2 or \displaystyle \frac{1}{z} breaks it, and systems with a curve need other methods. The circle example is not an exception: there the unknowns were D, E and F, and each appeared to the first power.
Three equations do not promise one solution. Three equations in three unknowns have exactly one solution only when none of them is a combination of the others. Otherwise there is no solution or there are infinitely many, as on Day 2. Counting equations is not enough.
Fewer equations than unknowns never pin down one triple. Two equations in x, y and z are two planes, which are parallel, the same, or meet in a whole line. There is never exactly one solution.
The two eliminations must remove the same letter, and between them use all three equations. If the first removes x and the second removes y, the two results still have three letters between them. If both come from the same pair, the third equation is never used, and the second result is only a multiple of the first.
The multiplier must not be 0. Multiplying an equation by 0 turns it into 0 = 0 and throws its information away.
Exactly one, none, or infinitely many. A linear system never has exactly two solutions, or any other finite number above one. If two different triples solved it, every triple on the line through them would too (exercise 21).
Where this goes next
The next lesson in this unit, fitting a quadratic to data, uses this method to find the parabola through three points: each point gives one linear equation in the three coefficients, just as each point gave one equation for the circle here.
With more unknowns the same idea carries on: eliminate one letter from all but one equation, then repeat on the smaller system. Writing only the coefficients in a rectangle and working row by row is Gaussian elimination, the way computers solve systems with thousands of unknowns, and the rectangle itself is a matrix equation. Cramer’s rule gives each unknown as a quotient of determinants, and the determinant is 0 exactly in the Day 2 cases, when there is not one solution. The geometry of the Day 2 pictures is taken further in lines and planes in space.
Exercises
Day 1: is it a solution?
- Is (2, -1, 3) a solution of x + 4y - z = -5, 3x - y + 2z = 13, -x + 2y + 5z = 11?
- Is (0, 4, -1) a solution of 2x + y + 3z = 1, x - y - 2z = -2, 5x + 2y + z = 9?
Day 1: solve
Solve each system and check the answer in all three equations.
- x + y + z = 2, x - y + z = 6, x + y - z = -4
- x + 2y - z = 1, 2x - y + z = 8, x + 3y + 2z = 7
- 3x + 2y = 0, y - 4z = -1, 3x + z = -5
- 2x + 3y - z = 4, 3x - 2y + 2z = -5, 4x + y + 3z = -12
- x = 2y - z - 3, 3y + z = x - 1, 2x - y + 2z = 0
- 2x + 3y + z = 3, 4x - 3y + 2z = 3, 6x + 6y - z = 4
- \displaystyle \frac{x}{2} + \frac{y}{3} + z = 1, x + y + z = 4, 3x - 2y + z = -1
Day 2: none, or infinitely many
Solve each system. If it has no solution, say so; if it has infinitely many, describe them all.
- x + y - 2z = 3, 2x - y + z = 1, 4x + y - 3z = 9
- x + 2y + z = 3, 2x - y + 3z = 1, 3x + y + 4z = 4 (let y = t)
- x - 2y + 3z = 5, 2x - 4y + 6z = 7, x + y + z = 2
- x - y + 2z = 3, 2x - 2y + 4z = 6, -3x + 3y - 6z = -9
- Find the value of k for which the system x + y + z = 1, x - y + 2z = 3, 2x + 4y + kz = 5 does not have exactly one solution, and say whether it then has none or infinitely many.
Day 2: problems
- A farm has 30 animals, all chickens, goats and cows, with 84 legs in all. There are twice as many goats as cows. How many of each animal are there?
- In a triangle, the largest angle is 10^\circ less than the sum of the other two, and the middle angle is 15^\circ more than the smallest. Find the three angles.
- A sum of 10\,000 dollars is split between three accounts paying 2\%, 3\% and 5\% a year. The total interest after one year is 290 dollars, and the amount at 2\% equals the other two amounts together. How much is in each account?
- Find the circle x^2 + y^2 + Dx + Ey + F = 0 through (2, 7), (3, 0) and (-6, 3), and give its center and radius.
Reasoning
- Anouk removes z from equations (1) and (2) of a system, then removes z again from (1) and (2) using different multipliers. Explain why her two new equations cannot pin down x and y.
- Explain why, when one equation of a system is already missing a letter, it pays to remove that same letter from the other two equations.
- Show that if two different triples P and Q both solve a system of linear equations, then so does every triple (1 - s)P + sQ, for any real s. Why does this mean a linear system can never have exactly two solutions?
- Explain why a system of two linear equations in x, y and z can never have exactly one solution.
Answers
- Yes. 2 - 4 - 3 = -5, 6 + 1 + 6 = 13 and -2 - 2 + 15 = 11: all three hold.
- No. The first two hold (0 + 4 - 3 = 1 and 0 - 4 + 2 = -2), but the third gives 0 + 8 - 1 = 7, not 9.
- (1, -2, 3). (1) minus (2) gives 2y = -4, so y = -2; (1) minus (3) gives 2z = 6, so z = 3; then x = 2 + 2 - 3 = 1.
- (3, 0, 2). Remove z: (1) plus (2) gives 3x + y = 9, and twice (1) plus (3) gives 3x + 7y = 9. Subtracting, 6y = 0, so y = 0 and x = 3; then (1) gives z = 2.
- (-2, 3, 1). (1) minus (3) removes x: 2y - z = 5. With y = 4z - 1 from (2), 8z - 2 - z = 5, so z = 1, y = 3, and (1) gives 3x = -6, x = -2.
- (1, -1, -5). Remove z: twice (1) plus (2) gives 7x + 4y = 3; three times (1) plus (3) gives 10x + 10y = 0, so y = -x. Then 3x = 3, x = 1, y = -1, and z = 2 + (-3) - 4 = -5.
- (4, 2, -3). Standard form: x - 2y + z = -3, -x + 3y + z = -1, 2x - y + 2z = 0. Minus twice the first plus the third gives 3y = 6, so y = 2. The first two added give y + 2z = -4, so z = -3, and then x = 4 + 3 - 3 = 4.
- \displaystyle \left(\frac{1}{2}, \frac{1}{3}, 1\right). (1) plus (2) gives 6x + 3z = 6, so 2x + z = 2; twice (1) minus (3) gives -2x + 3z = 2. Adding, 4z = 4, so z = 1 and \displaystyle x = \frac{1}{2}; then 1 + 3y + 1 = 3, so \displaystyle y = \frac{1}{3}.
- (2, 3, -1). Multiply the first by 6: 3x + 2y + 6z = 6. Adding the third gives 6x + 7z = 5; twice the second plus the third gives 5x + 3z = 7. Then 17x = 34, so x = 2, z = -1 and y = 4 - 2 + 1 = 3.
- No solution. (1) plus (2) gives 3x - z = 4; (2) plus (3) gives 6x - 2z = 10, that is 3x - z = 5. Subtracting leaves 0 = 1, which is false.
- Infinitely many: (8 - 7t,\ t,\ 5t - 5) for every real t. (1) plus twice (2) and (2) plus (3) both give 5x + 7z = 5, so the letters vanish to 0 = 0. With y = t, (1) gives x + z = 3 - 2t; together with 5x + 7z = 5 this gives z = 5t - 5 and x = 8 - 7t.
- No solution. Equation (2) minus twice (1) leaves 0 = 7 - 10 = -3, which is false. The first two planes are parallel.
- Infinitely many, a whole plane: (3 + s - 2t,\ s,\ t) for all real s and t. The second equation is twice the first and the third is -3 times it, so all three are the same plane x = 3 + y - 2z.
- k = 1, and then there is no solution. Three times the first equation minus the second gives 2x + 4y + z = 0, so with k = 1 the third equation asks the same left side to equal 5. For any other k, removing x and y leaves (k - 1)z equal to a number, so there is exactly one solution.
- 18 chickens, 8 goats and 4 cows. With c + g + w = 30, 2c + 4g + 4w = 84 and g = 2w: the legs equation minus four times the first gives -2c = -36, so c = 18, g + w = 12, 3w = 12, w = 4, g = 8.
- 40^\circ, 55^\circ and 85^\circ. With a + b + c = 180, c = a + b - 10 and b = a + 15: the first two give 2c = 170, so c = 85 and a + b = 95; then 2a + 15 = 95, so a = 40 and b = 55.
- 5000 dollars at 2\%, 3000 at 3\% and 2000 at 5\%. With a + b + c = 10\,000 and a = b + c, a = 5000 and b + c = 5000. The interest gives 0.02(5000) + 0.03b + 0.05c = 290, so 0.03b + 0.05c = 190; with b = 5000 - c, 150 + 0.02c = 190, so c = 2000 and b = 3000.
- x^2 + y^2 + 2x - 6y - 15 = 0, center (-1, 3), radius 5. The points give 2D + 7E + F = -53, 3D + F = -9 and -6D + 3E + F = -45, so D = 2, E = -6, F = -15. Completing the square, (x + 1)^2 + (y - 3)^2 = 25.
- Both new equations are built from (1) and (2) only, so equation (3) is never used, and the information in it is lost. Worse, any combination of (1) and (2) with no z is a multiple of any other such combination (the multipliers have to stand in the same ratio to cancel z), so her second equation is a multiple of her first. One equation in x and y has infinitely many solutions.
- The equation missing a letter is already an equation in the other two letters. Removing the same letter from the remaining pair gives a second equation in those same two letters, and the two together form a two-letter system. Removing a different letter would leave the two results with different letters, and more work to bring them together.
- Write one equation as L(x, y, z) = d, where L(x, y, z) = ax + by + cz. Because L only multiplies each coordinate by a number and adds, it can be applied to P and Q separately: L((1 - s)P + sQ) = (1 - s)L(P) + sL(Q). Both L(P) and L(Q) equal d, so this is (1 - s)d + sd = d. The same holds for every equation. Different values of s give different triples, since P and Q differ, so two solutions bring infinitely many with them: the whole line through P and Q.
- Each equation is a plane. Two planes are either parallel (no common point), the same plane (infinitely many), or meet in a line (infinitely many). In algebra: removing one letter leaves at most one equation in two letters, and one such equation always has either no solution or infinitely many.