Free worked solutions · 2026 Edition

Chapter 7: Systems of equations worked solutions

Worked solutions to the practice questions in Chapter 7 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

Chapter 7 of 19
Volume 1
22 practice questions
4 Going further
Free, no sign-up

Start with Basics All 19 chapters

Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[4 marks]

Solve the system of equations

\begin{cases} x + 2y - z = -1 \\ y + 3z = 5 \\ 5z = 10 \end{cases}

by back-substitution.

[4]

Show solutionHide solution

from the third equation z = 2

substituting into the second equation: y + 6 = 5

y = -1

first equation: x - 2 - 2 = -1, so x = 3

Final answer

x=3,\ y=-1,\ z=2

Question 2

[5 marks]

Solve the system of equations

\begin{cases} x + 2y + z = 0 \\ 3x - y = 9 \\ 2x + y - z = -3 \end{cases}

showing your working.

[5]

Show solutionHide solution

adding the first and third equations to eliminate z

3x + 3y = -3, so x + y = -1

adding x + y = -1 and 3x - y = 9: 4x = 8, x = 2

y = -3

z = -x - 2y = -2 + 6 = 4

Note: accept any valid order of elimination.

Final answer

x=2,\ y=-3,\ z=4

Question 3

[6 marks]

Consider the system of equations

\begin{cases} 2x + 3y - z = -5 \\ 3x - 2y + 2z = 23 \\ 4x + y + 3z = 25 \end{cases}

(a) Explain why it is convenient to eliminate z first.

[1]

(b) Solve the system.

[5]

Show solutionHide solution

(a) the coefficient of z in the first equation is -1, so adding multiples of the first equation (E_2 + 2R_1, E_3 + 3R_1) removes z without fractions

(b) E_2 + 2R_1: 7x + 4y = 13 M1A1

E_3 + 3R_1: 10x + 10y = 10, so x + y = 1

7x + 4(1 - x) = 13 \Rightarrow x = 3, y = -2

z = 2x + 3y + 5 = 6 - 6 + 5 = 5

Note: accept any valid elimination order for (b).

Final answer

(a) z has coefficient -1 in E_1, so no fractions
(b) x=3,\ y=-2,\ z=5

Question 4

[3 marks]

Use technology to solve the system of equations

\begin{cases} 1.2x + 0.5y - 2.3z = 4.1 \\ -0.7x + 3.1y + 1.4z = 1.9 \\ 2.5x - 1.8y + 0.9z = 6.4 \end{cases}

giving each value correct to three significant figures.

[3]

Show solutionHide solution

x = 3.38

y = 1.26

z = 0.253

Note: exact values x = 3.37672\ldots, y = 1.26100\ldots, z = 0.253296\ldots

Final answer

x=3.38,\ y=1.26,\ z=0.253

Question 5

[7 marks]

Consider the following pairs of equations.

(a) Show that the following system has no solution.

[2]

\begin{cases} 5x - 3y = 8 \\ -10x + 6y = 9 \end{cases}

(b) Show that the following system has infinitely many solutions, and find its general solution using y = \lambda.

[3]

\begin{cases} 30x + 40y = 70 \\ 45x + 60y = 105 \end{cases}

(c) Solve the following system.

[2]

\begin{cases} 4x + 7y = 17 \\ 5x - 2y = -11 \end{cases}

Show solutionHide solution

(a) 2R_1 + E_2: (10x - 6y) + (-10x + 6y) = 16 + 9

0 = 25, which is false, so there is no solution

(b) 3R_1 - 2R_2: 0 = 210 - 210 = 0, so the equations are equivalent (both are 3x + 4y = 7) and there are infinitely many solutions M1R1

y = \lambda, x = \dfrac{7 - 4\lambda}{3}

(c) e.g. 2R_1 + 7R_2: 43x = -43, x = -1

y = 3

Final answer

(a) 0=25, no solution
(b) x=\frac{7-4\lambda}{3},\ y=\lambda
(c) x=-1,\ y=3

Question 6

[5 marks]

Consider the system of equations

\begin{cases} x + y - 2z = 3 \\ 2x - y + z = 1 \\ 4x + y - 3z = 4 \end{cases}

(a) Show that the system has no solution.

[4]

(b) The constant 4 on the right-hand side of the third equation is replaced by c. Write down the value of c for which the new system has infinitely many solutions.

[1]

Show solutionHide solution

(a) E_2 - 2R_1: -3y + 5z = -5 and E_3 - 4R_1: -3y + 5z = -8 M1A1

subtracting: 0 = -3

this is a contradiction, so the system has no solution (it is inconsistent)

(b) c = 7

Note: the left side of the third equation is 2R_1 + E_2, whose right side is 2(3) + 1 = 7.

Final answer

(a) 0=-3, no solution
(b) c=7

Question 7

[8 marks]

Consider the system of equations

\begin{cases} x + y + z = 3 \\ 2x + y - z = 0 \\ 5x + 3y - z = 3 \end{cases}

(a) Use elimination to show that the system has infinitely many solutions.

[3]

(b) Let z = \lambda. Find the general solution of the system.

[3]

(c) Verify that your general solution satisfies the third equation for every value of \lambda.

[2]

Show solutionHide solution

(a) E_2 - 2R_1: -y - 3z = -6 M1A1

E_3 - 5R_1: -2y - 6z = -12, which is twice the previous equation, so eliminating y gives 0 = 0; two independent equations in three unknowns, so infinitely many solutions

(b) y = 6 - 3\lambda (M1)A1

x = 3 - y - z = -3 + 2\lambda, so x = -3 + 2\lambda, y = 6 - 3\lambda, z = \lambda

Note: accept any equivalent parametrisation, e.g. x = \mu, y = \dfrac{3 - 3\mu}{2}, z = \dfrac{3 + \mu}{2}.

(c) 5(-3 + 2\lambda) + 3(6 - 3\lambda) - \lambda = -15 + 10\lambda + 18 - 9\lambda - \lambda

= 3 for every \lambda

A1AG

Final answer

(a) 0=0
(b) (-3+2\lambda,\ 6-3\lambda,\ \lambda)
(c) shown

Question 8

[6 marks]

Consider the system of equations, where k \in \mathbb{R},

\begin{cases} x + 4y - z = -1 \\ 2x + 3y + z = 4 \\ 3x + 7y = k \end{cases}

(a) Find the value of k for which the system has solutions.

[3]

(b) For this value of k, find the general solution of the system.

[3]

Show solutionHide solution

(a) E_2 - 2R_1: -5y + 3z = 6

E_3 - 3R_1: -5y + 3z = k + 3

the two equations agree only if k + 3 = 6 (otherwise 0 = k - 3 \ne 0), so k = 3

(b) let z = \lambda: y = \dfrac{3\lambda - 6}{5} M1A1

x = -1 - 4y + z = \dfrac{19 - 7\lambda}{5}

Note: accept any equivalent parametrisation, e.g. x = 1 - 7\mu, y = 3\mu, z = 2 + 5\mu.

Final answer

(a) k=3
(b) \left(\frac{19-7\lambda}{5},\ \frac{3\lambda-6}{5},\ \lambda\right)

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[6 marks]

At a market, apples cost a dollars per kg, bananas b dollars per kg and cherries c dollars per kg. Three customers buy the following amounts, in kg.

\begin{array}{lccc} \hline \text{Customer} & \text{Apples} & \text{Bananas} & \text{Cherries} \\ \hline \text{Mia} & \text{2} & \text{1} & \text{3} \\ \text{Noor} & \text{1} & \text{3} & \text{2} \\ \text{Omar} & \text{4} & \text{2} & \text{1} \\ \hline \end{array}

Mia pays \$23, Noor pays \$19 and Omar pays \$21.

(a) Write down a system of linear equations for a, b and c.

[2]

(b) Solve the system by elimination, and state the price per kg of each fruit.

[4]

Show solutionHide solution

(a)

\begin{aligned} 2a+b+3c&=23\\ a+3b+2c&=19\\ 4a+2b+c&=21 \end{aligned}

Two correct equations [1]; all three correct.

[1]

(b) Call the equations E_1, E_2, E_3. Remove a with E_2, whose a has coefficient 1:

[1]

E_1-2E_2: -5b-c=-15, so c=15-5b.

E_3-4E_2: -10b-7c=-55.

Substitute: -10b-7(15-5b)=-55, so 25b=50, b=2 and c=5.

[2]

Back-substitute in E_2: a=19-3(2)-2(5)=3.

[1]

Apples \$3 per kg, bananas \$2 per kg, cherries \$5 per kg.

[1]

Final answer

(a) 2a+b+3c=23, a+3b+2c=19, 4a+2b+c=21
(b) apples $3, bananas $2, cherries $5 per kg

Question 10

[6 marks]

At a school café, a coffee costs x dollars, a muffin costs y dollars and a sandwich costs z dollars. Three receipts are shown.

\begin{array}{lccc} \hline \text{Receipt} & \text{Coffees} & \text{Muffins} & \text{Sandwiches} \\ \hline \text{1} & \text{2} & \text{3} & \text{1} \\ \text{2} & \text{1} & \text{2} & \text{4} \\ \text{3} & \text{3} & \text{5} & \text{5} \\ \hline \end{array}

Receipts 1 and 2 have totals \$13.90 and \$19.00. Receipt 3 shows a total of \$33.40.

(a) Write down three equations for x, y and z.

[2]

(b) Show that these three equations have no solution.

[2]

The cashier says that the total on Receipt 3 should have been \$32.90. With this correction the system has infinitely many solutions, and a sandwich costs \$3.10.

(c) Find the price of a coffee and the price of a muffin.

[2]

Show solutionHide solution

(a) 2x+3y+z=13.9, x+2y+4z=19, 3x+5y+5z=33.4.

[2]

(b) Call the equations E_1, E_2, E_3. E_1+E_2 gives 3x+5y+5z=32.9, which has the same left side as E_3.

[1]

So E_3-E_1-E_2 reads 0=33.4-32.9=0.5: a contradiction, so no solution. (Technology: \mathrm{rref} of the augmented matrix ends with the row 0,\,0,\,0\mid 1, the same contradiction.)

[1]

(c) With z=3.1, E_1 and E_2 give 2x+3y=10.8 and x+2y=6.6.

[1]

Subtracting twice the second from the first: -y=10.8-13.2=-2.4, so y=2.4 and x=6.6-4.8=1.8: a coffee costs \$1.80 and a muffin \$2.40.

[1]

Final answer

(a) 2x+3y+z=13.9, x+2y+4z=19, 3x+5y+5z=33.4
(b) no solution: E_1+E_2 has the left side of E_3 but 32.90\ne33.40
(c) coffee $1.80, muffin $2.40

Question 11

[6 marks]

Consider the planes

\begin{gathered}\pi_1: x+y+z=2,\\ \pi_2: x-y+2z=1,\\ \pi_3: 2x+3z=5.\end{gathered}

(a) Show that no two of the planes are parallel.

[2]

(b) Show that the system formed by the three equations has no solution.

[2]

(c) Describe the arrangement of the three planes, and justify your description.

[2]

Show solutionHide solution

(a) The coefficient triples are (1,1,1), (1,-1,2) and (2,0,3).

[1]

No triple is a multiple of another: (1,1,1) and (1,-1,2) differ in the ratio of their first two entries, and (2,0,3) has y-coefficient 0 while the other two do not. So no two planes are parallel.

[1]

(b) Call the equations E_1, E_2, E_3. The left side of E_3 is the left side of E_1 plus the left side of E_2, so form E_3-E_1-E_2:

[1]

(2x+3z)-(x+y+z)-(x-y+2z)=0 and 5-2-1=2, so the line reads 0=2, a contradiction: no solution.

[1]

(c) A triangular prism.

[1]

No two planes are parallel, so the planes meet pairwise in three lines, but there is no point common to all three; the three lines are therefore parallel and the planes form the three sides of a prism.

[1]

Final answer

(a) shown
(b) 0=2
(c) triangular prism

Question 12

[4 marks]

Find the general solution of the system

x+2y-z=4, \qquad 2x+y+z=5.

Give your answer in the form (x,y,z)=\mathbf{a}+t\,\mathbf{d}, where t\in\mathbb{R}.

[4]

Show solutionHide solution

Two equations in three unknowns: expect one free parameter. Adding the equations eliminates z: 3x+3y=9, so x+y=3.

[1]

Let z=t. The first equation gives x+2y=4+t; with x+y=3 subtraction gives y=1+t.

[1]

Then x=3-y=2-t.

[1]

(x,y,z)=(2-t,\ 1+t,\ t)=(2,1,0)+t(-1,1,1), t\in\mathbb{R}.

[1]

Check in the second equation: 2(2-t)+(1+t)+t=5 ✓

Final answer

(2,1,0)+t(-1,1,1)

Question 13

[6 marks]

Consider the planes

\begin{gathered}\pi_1: 2x-y+z=1,\\ \pi_2: x+y-z=2,\\ \pi_3: 4x+y-z=5.\end{gathered}

(a) Use elimination to show that the three planes meet in a line.

[2]

(b) Find the general solution of the system.

[2]

(c) Name the arrangement of the three planes, and find the coordinates of the point of the common line at which z=3.

[2]

Show solutionHide solution

(a) Call the equations E_1, E_2, E_3. E_2 has x with coefficient 1, so remove x using E_2:

[1]

E_1-2E_2: -3y+3z=-3.

E_3-4E_2: -3y+3z=-3.

The same equation twice, so subtracting gives 0=0: the system is consistent with a free unknown, so there are infinitely many solutions and the planes meet in a line.

[1]

(b) Let z=t. Then -3y+3t=-3, so y=1+t.

[1]

From E_2: x=2-y+z=2-(1+t)+t=1. The general solution is (x,y,z)=(1,\ 1+t,\ t), t\in\mathbb{R}.

[1]

(c) The planes share a common line (no two are parallel, since the coefficient triples are pairwise different).

[1]

On the line (1,\ 1+t,\ t) the condition z=3 gives t=3, so the point is (1,\,4,\,3).

[1]

Final answer

(a) shown
(b) (1,\ 1+t,\ t)
(c) common line (sheaf); (1,4,3)

Question 14

[6 marks]

(a) Show that the solutions of the system

x+2y-3z=4, \qquad 2x+4y-6z=8

form a plane, and give the general solution using two parameters.

[4]

(b) The second equation is changed to 2x+4y-6z=9. Explain why the system now has no solution.

[2]

Show solutionHide solution

(a) E_2 is 2\times E_1, so E_2-2E_1 reads 0=0 and the system contains only one independent equation.

[2]

Two unknowns are free. Let y=s and z=t; then x=4-2s+3t.

[1]

(x,y,z)=(4-2s+3t,\ s,\ t), s,t\in\mathbb{R}: every point of the plane x+2y-3z=4.

[1]

(b) Now E_2-2E_1 reads 0=9-8=1.

[1]

The two planes are parallel (their coefficient triples are proportional) but distinct, so they have no point in common.

[1]

Final answer

(a) (4-2s+3t,\ s,\ t), a plane
(b) 0=1: parallel distinct planes

Question 15

[7 marks]

Consider the system, where a,b\in\mathbb{R},

\begin{aligned} x+y&=4\\ x+2y+z&=9\\ 2x+3y+az&=b \end{aligned}

(a) Find the value of a and the value of b for which the system has infinitely many solutions.

[4]

(b) For these values, find the general solution of the system in terms of a parameter.

[3]

Show solutionHide solution

(a) Call the equations E_1, E_2, E_3. Remove x from E_2 and E_3:

[1]

E_2-E_1: y+z=5.

E_3-2E_1: y+az=b-8.

Subtract: (a-1)z=b-13.

[1]

Infinitely many solutions need this to be 0=0: a-1=0 and b-13=0.

[1]

So a=1 and b=13.

[1]

(b) Let z=t. Then y=5-t.

[1]

From E_1: x=4-y=4-(5-t)=t-1.

[1]

(x,y,z)=(t-1,\ 5-t,\ t), that is (-1,5,0)+t(1,-1,1), t\in\mathbb{R}.

[1]

Final answer

(a) a=1,\ b=13
(b) (t-1,\ 5-t,\ t)

Question 16

[8 marks]

The solutions of the system

x+2y=-3, \qquad 3x-2z=3

are the points of the line L with equation (x,y,z)=(1,-2,0)+t(2,-1,3).

(a) Show that for every value of t the point (1+2t,\ -2-t,\ 3t) satisfies both equations.

[3]

A third plane has equation x-y+kz=c, where k,c\in\mathbb{R}.

(b) By substituting the coordinates of a general point of L into the equation of the third plane, show that the three planes meet at a single point unless k=-1.

[2]

(c) Find the value of c for which, when k=-1, the line L lies in the third plane. State what happens when k=-1 and c takes any other value.

[2]

(d) Find the point where the three planes meet when k=1 and c=9.

[1]

Show solutionHide solution

(a) First equation: (1+2t)+2(-2-t)=1+2t-4-2t=-3.

[2]

Second equation: 3(1+2t)-2(3t)=3+6t-6t=3. The parameter cancels in both, so every point of L satisfies both equations.

[1]

(b) Substitute into x-y+kz=c: (1+2t)-(-2-t)+k(3t)=c, that is 3+3(1+k)t=c.

[1]

If k\ne-1 the coefficient of t is non-zero, so there is exactly one value t=\dfrac{c-3}{3(1+k)}, hence exactly one common point. A unique solution fails only when k=-1.

[1]

(c) When k=-1 the equation becomes 3=c for all t: if c=3 it is true for every t, so L lies in the third plane (infinitely many solutions).

[1]

If c\ne3 it is false for every t: no solution, because the line is parallel to the third plane and does not lie in it.

[1]

(d) With k=1, c=9: 3+6t=9, so t=1, and the point is (3,\,-3,\,3).

[1]

Final answer

(a) shown
(b) 3+3(1+k)t=c
(c) c=3: L lies in the plane; c\ne3: no solution
(d) (3,-3,3)

Question 17

[6 marks]

Consider the system, where k\in\mathbb{R},

\begin{aligned} x+2y-z&=0\\ 2x-y+kz&=0\\ x+y+z&=0 \end{aligned}

(a) Show that (x,y,z)=(0,0,0) is a solution for every value of k.

[1]

(b) Find the value of k for which the system has other solutions, and find all solutions for this value of k.

[5]

Show solutionHide solution

(a) Substituting x=y=z=0 makes the left-hand side of every equation 0, which equals the right-hand side.

[1]

(b) Call the equations E_1, E_2, E_3. Remove x:

[1]

E_2-2E_1: -5y+(k+2)z=0.

E_3-E_1: -y+2z=0.

Remove y: (E_2-2E_1)-5(E_3-E_1) gives (k+2-10)z=0, that is (k-8)z=0.

[1]

The system is consistent for every k (the constants are all 0), so the only way to lose uniqueness is k-8=0: other solutions exist exactly when k=8.

[1]

Then let z=t. -y+2t=0 gives y=2t, and E_1 gives x=-2y+z=-4t+t=-3t.

[1]

(x,y,z)=t(-3,2,1), t\in\mathbb{R}.

[1]

Final answer

(a) shown
(b) k=8; t(-3,2,1)

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 18

[5 marks]

Consider the system, where k,m\in\mathbb{R},

\begin{aligned} 2x-y+z&=4\\ x+y-z&=-1\\ 4x+y+kz&=m \end{aligned}

Find the value of k and the value of m for which the system has infinitely many solutions.

[5]

Show solutionHide solution

Add the first two equations: 3x=3, so x=1.

[2]

Then the second equation gives y-z=-2, so y=z-2.

[1]

Substitute into the third equation: 4+(z-2)+kz=m, that is (k+1)z=m-2.

[1]

For infinitely many solutions this equation must be 0=0, so k=-1 and m=2.

[1]

(Then z is free and the solutions are (1,\ z-2,\ z).)

Final answer

k=-1,\ m=2

Question 19

[10 marks]

Consider the system of equations, where k \in \mathbb{R},

\begin{cases} x + 3y - 2z = 2 \\ 2x + 7y - z = 1 \\ x + 4y + k^2 z = k \end{cases}

(a) Show that the system reduces to y + 3z = -3 and (k^2 - 1)z = k + 1.

[3]

(b) Write down the values of k for which the system does not have a unique solution.

[2]

(c) For each value of k in (b), determine whether the system has no solution or infinitely many solutions.

[2]

(d) In the case with infinitely many solutions, find the general solution.

[3]

Show solutionHide solution

(a) E_2 - 2R_1: y + 3z = -3

E_3 - E_1: y + (k^2 + 2)z = k - 2

subtracting: (k^2 + 2 - 3)z = k - 2 + 3, i.e. (k^2 - 1)z = k + 1

M1AG

(b) no unique solution when k^2 - 1 = 0 (M1)

k = 1 or k = -1

(c) k = 1: the equation is 0 = 2, so no solution

k = -1: the equation is 0 = 0, so infinitely many solutions

(d) z = \lambda, y = -3 - 3\lambda M1A1

x = 2 - 3y + 2z = 11 + 11\lambda

Note: accept any equivalent parametrisation.

Final answer

(a) shown
(b) k=\pm1
(c) k=1: none (0=2); k=-1: infinitely many
(d) (11+11\lambda,\ -3-3\lambda,\ \lambda)

Question 20

[8 marks]

A juice company mixes three fruit concentrates A, B and C to make 100 kg of drink base. Concentrate A is 10\% sugar, B is 20\% sugar and C is 30\% sugar by mass, and the drink base must be 19\% sugar. The mixture uses x kg of A, y kg of B and z kg of C.

(a) Write down two equations for x, y and z.

[2]

(b) Use your GDC to find x and y in terms of z.

[2]

(c) Find the range of possible values of z.

[2]

Concentrate A costs \$1.20 per kg, B costs \$1.80 per kg and C costs \$2.50 per kg.

(d) Find the least possible cost of 100 kg of the drink base, and the amounts of A, B and C that give it.

[2]

Show solutionHide solution

(a) Total mass: x+y+z=100.

[1]

Sugar: 0.1x+0.2y+0.3z=0.19\times100=19.

[1]

(b) Call the equations E_1 and E_2. E_2-0.1E_1: 0.1y+0.2z=9, so y+2z=90 and y=90-2z.

[1]

x=100-y-z=10+z. (Technology: \mathrm{rref} of \left[\begin{smallmatrix}1&1&1&100\\0.1&0.2&0.3&19\end{smallmatrix}\right] gives the same two equations.)

[1]

(c) All masses must be non-negative: z\ge0, x=10+z\ge0 and y=90-2z\ge0.

[1]

So 0\le z\le45.

[1]

(d) Cost =1.2(10+z)+1.8(90-2z)+2.5z=174+0.1z.

[1]

This increases with z, so the least cost is at z=0: \$174.00, using 10 kg of A, 90 kg of B and none of C.

[1]

Final answer

(a) x+y+z=100, 0.1x+0.2y+0.3z=19
(b) x=10+z, y=90-2z
(c) 0\le z\le45
(d) $174.00: 10 kg of A, 90 kg of B, no C

Question 21

[7 marks]

Consider the system, where a,b,c\in\mathbb{R},

\begin{aligned} x+2y+z&=3\\ 2x-y+az&=b\\ 3x+y+2z&=c \end{aligned}

It is given that (x,y,z)=(2,1,-1) is a solution of the system.

(a) Find the value of c.

[1]

(b) Find b in terms of a.

[1]

It is also given that the system does not have a unique solution.

(c) Find the value of a and the value of b, and explain why the system cannot have no solution.

[3]

(d) Find the general solution of the system.

[2]

Show solutionHide solution

(a) c=3(2)+1+2(-1)=5.

[1]

(b) b=2(2)-1+a(-1)=3-a.

[1]

(c) Call the equations E_1, E_2, E_3. Remove x:

[1]

E_2-2E_1: -5y+(a-2)z=-3-a.

E_3-3E_1: -5y-z=-4.

Subtract: (a-1)z=1-a. No unique solution means a-1=0, so a=1 and b=3-1=2.

[1]

The system has a solution (2,1,-1), so it is consistent: it cannot have no solution. (Indeed the right-hand entry 1-a is 0 when a=1.)

[1]

(d) With a=1: -5y-z=-4. Let y=s, so z=4-5s.

[1]

Then x=3-2y-z=3-2s-4+5s=-1+3s. The general solution is (x,y,z)=(-1+3s,\ s,\ 4-5s), that is (-1,0,4)+s(3,1,-5) (the point (2,1,-1) is s=1).

[1]

Final answer

(a) c=5
(b) b=3-a
(c) a=1,\ b=2; a known solution means the system is consistent
(d) (-1+3s,\ s,\ 4-5s)

Question 22

[14 marks]

Consider the system, where a,b\in\mathbb{R},

\begin{aligned} x-y+2z&=5\\ 2x+y-z&=1\\ x+2y+az&=b \end{aligned}

(a) Show that eliminating x and then y leads to the equation (a+3)z=b+4.

[4]

(b) Determine a and b so that the system has infinitely many solutions. For these values, find the general solution.

[3]

(c) Given that a=-3 and b\ne-4, describe the arrangement of the three planes represented by the system.

[2]

Now let b=2 and a\ne-3.

(d) Find x, y and z in terms of a.

[3]

(e) Hence determine a given that x=y.

[2]

Show solutionHide solution

(a) Call the equations E_1, E_2, E_3. Remove x:

[1]

E_2-2E_1: 3y-5z=-9.

E_3-E_1: 3y+(a-2)z=b-5.

[2]

Remove y: subtracting gives (a-2+5)z=b-5+9, that is (a+3)z=b+4.

[1]

(b) Infinitely many solutions need this equation to be 0=0: a=-3 and b=-4.

[1]

Let z=3s. Then 3y-15s=-9, so y=5s-3.

[1]

From E_1: x=5+y-2z=5+5s-3-6s=2-s. The general solution is (x,y,z)=(2-s,\ 5s-3,\ 3s), s\in\mathbb{R}.

[1]

(c) With a=-3 the last equation is 0=b+4\ne0: no solution.

[1]

No two of the planes are parallel (the coefficient triples (1,-1,2), (2,1,-1) and (1,2,-3) are pairwise different directions), so the planes form a triangular prism.

[1]

(d) The last equation gives (a+3)z=6, so z=\dfrac{6}{a+3}.

[1]

From 3y-5z=-9: y=\dfrac{5z-9}{3}=\dfrac{10}{a+3}-3=\dfrac{1-3a}{a+3}.

[1]

From E_1: x=5+y-2z=5+\dfrac{10}{a+3}-3-\dfrac{12}{a+3}=2-\dfrac{2}{a+3}=\dfrac{2a+4}{a+3}.

[1]

(e) x=y: 2a+4=1-3a.

[1]

5a=-3, so a=-\tfrac35.

[1]

Final answer

(a) shown
(b) a=-3,\ b=-4; (2-s,\ 5s-3,\ 3s)
(c) triangular prism
(d) x=\frac{2a+4}{a+3},\ y=\frac{1-3a}{a+3},\ z=\frac{6}{a+3}
(e) a=-\frac35

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 The circle through three points

G1 Worked example

[10 marks]

The circle x^2+y^2+Dx+Ey+F=0 passes through the points P(0,1), Q(2,5) and R(4,k), where k\in\mathbb{R}. A circle with this equation has centre \left(-\dfrac D2,-\dfrac E2\right) and radius \sqrt{\dfrac{D^2+E^2}{4}-F}.

(a) Write down three linear equations in D, E and F.

[2]

(b) Show that (k-9)E=41-k^2.

[3]

(c) For k=1, find the centre and the radius of the circle.

[3]

(d) Find the value of k for which no such circle exists, and give a geometric reason.

[2]

Show solutionHide solution

(a) P: E+F=-1; ; Q: 2D+5E+F=-29

R: 4D+kE+F=-16-k^2

Note: once a point is substituted, x^2+y^2 is a number, so each equation is linear in D, E, F.

(b) Second minus first: 2D+4E=-28, so D=-14-2E

Third minus first: 4D+(k-1)E=-15-k^2

Substitute: -56-8E+(k-1)E=-15-k^2, so (k-9)E=41-k^2

A1AG

(c) k=1: -8E=40, so E=-5, D=-4, F=4

centre \left(2,\dfrac52\right)

radius \sqrt{\dfrac{16+25}{4}-4}=\sqrt{\dfrac{25}{4}}=\dfrac52

Check: the distance from \left(2,\frac52\right) to P(0,1) is \sqrt{4+\frac94}=\frac52.

(d) k=9: the equation in (b) becomes 0=-40, so the system has no solution

R(4,9) lies on the line PQ, y=2x+1; no circle passes through three points on one line

Final answer

(c) centre \left(2,\frac52\right), radius \frac52
(d) k=9

G1 Investigation

[13 marks]

In this question, a circle has an equation of the form x^2+y^2+Dx+Ey+F=0. The points A(a,a^2), B(b,b^2) and C(c,c^2) lie on the curve y=x^2, where a, b and c are distinct.

(a) Explain why A, B and C do not lie on one straight line.

[2]

(b) Find the equation of the circle through the three points of y=x^2 with x=-1, x=1 and x=2. Find the coordinates of the fourth point where this circle meets the curve.

[4]

(c) Verify that the circle through the points of y=x^2 with x=0, x=1 and x=3 is x^2+y^2+12x-14y=0, and find the fourth point where it meets the curve.

[2]

(d) Let d be the x-coordinate of the fourth point. Conjecture a relation between a, b, c and d.

[1]

(e) Prove your conjecture.

[3]

(f) Write down the fourth point where the circle through (-3,9), (1,1) and (4,16) meets y=x^2.

[1]

Show solutionHide solution

(a) A vertical line meets y=x^2 once. A line y=mx+p meets it where x^2-mx-p=0

This quadratic has at most two roots, so no line contains three points of the curve

(b) Points (-1,1), (1,1), (2,4): -D+E+F=-2, D+E+F=-2, 2D+4E+F=-20

D=0, E=-6, F=4: the circle is x^2+y^2-6y+4=0

With y=x^2: x^4-5x^2+4=0, so (x^2-1)(x^2-4)=0

x=\pm1,\pm2: the fourth point is (-2,4)

(c) (0,0): 0=0; (1,1): 1+1+12-14=0; (3,9): 9+81+36-126=0

With y=x^2: x^4-13x^2+12x=x(x-1)(x-3)(x+4)=0, so the fourth point is (-4,16)

(d) -1+1+2-2=0 and 0+1+3-4=0: conjecture a+b+c+d=0

(e) Substituting y=x^2 gives x^4+(1+E)x^2+Dx+F=0, whose roots are the x-coordinates of the common points

a, b, c are roots, so the quartic is (x-a)(x-b)(x-c)(x-d) for some d

The coefficient of x^3 is -(a+b+c+d) on the right and 0 on the left, so a+b+c+d=0

Note: if d equals one of a, b, c, the circle touches the curve at that point.

(f) d=-(-3+1+4)=-2: the point (-2,4)

Final answer

(b) x^2+y^2-6y+4=0; (-2,4)
(c) (-4,16)
(d) a+b+c+d=0
(f) (-2,4)

G2 Magic squares as a linear system

G2 Worked example

[9 marks]

In a 3\times3 magic square, the three rows, the three columns and the two diagonals all have the same sum S. The entries are labelled

\begin{array}{ccc} a & b & c\\ d & e & f\\ g & h & k\end{array}.

(a) By adding the middle row, the middle column and the two diagonals, show that e=\dfrac S3.

[3]

(b) Deduce that a+k=c+g=b+h=d+f=\dfrac{2S}{3}.

[1]

(c) A magic square has a=10, b=3 and k=4. Find S, and complete the square.

[4]

(d) Explain why no magic square with integer entries has S=20.

[1]

Show solutionHide solution

(a) (d+e+f)+(b+e+h)+(a+e+k)+(c+e+g)=4S

The left side is the sum of all nine entries plus 3e, and the nine entries are the three rows, 3S

3S+3e=4S, so e=\dfrac S3

A1AG

(b) Each of these pairs lies on a line through the centre: a+e+k=S gives a+k=S-\dfrac S3=\dfrac{2S}{3}, and the same for the other three lines

(c) a+k=14=\dfrac{2S}{3}

S=21, e=7

c=21-10-3=8, g=14-8=6, h=14-3=11

d=21-10-6=5, f=14-5=9; the square is \begin{array}{ccc}10&3&8\\5&7&9\\6&11&4\end{array}

Check: bottom row 6+11+4=21, third column 8+9+4=21.

(d) The centre would be \dfrac{20}{3}, which is not an integer

Final answer

G2 Investigation

[11 marks]

In a 3\times3 magic square the three rows, the three columns and the two diagonals all have the same sum S. You may use the results that the centre is \dfrac S3 and that two entries opposite each other through the centre add to \dfrac{2S}{3}.

(a) A magic square has centre m, top left entry m+p and top right entry m+q. Express all nine entries in terms of m, p and q.

[3]

(b) Show that, for every choice of the numbers m, p and q, the square in (a) is a magic square.

[2]

(c) Now the entries are 1,2,\dots,9, each used once. Show that m=5, and that |p|, |q|, |p+q| and |p-q| are 1, 2, 3 and 4 in some order.

[3]

(d) Hence find the number of magic squares that use 1,2,\dots,9 once each, and write one of them down.

[3]

Show solutionHide solution

(a) S=3m, and opposite entries add to 2m: bottom right m-p, bottom left m-q

top middle 3m-(m+p)-(m+q)=m-p-q, so bottom middle m+p+q

middle left 3m-(m+p)-(m-q)=m-p+q, so middle right m+p-q

\begin{array}{ccc} m+p & m-p-q & m+q\\ m-p+q & m & m+p-q\\ m-q & m+p+q & m-p\end{array}

(b) Rows: 3m, 3m, 3m; columns: (m+p)+(m-p+q)+(m-q)=3m, 3m, (m+q)+(m+p-q)+(m-p)=3m

Diagonals: (m+p)+m+(m-p)=3m and (m+q)+m+(m-q)=3m. All eight sums are 3m

(c) 1+2+\dots+9=45=3S, so S=15 and m=5

The other eight entries come in pairs 5\pm|p|, 5\pm|q|, 5\pm|p+q|, 5\pm|p-q|

They must be the pairs 5\pm1, 5\pm2, 5\pm3, 5\pm4, so |p|,|q|,|p+q|,|p-q| are 1,2,3,4 in some order

A1AG

(d) With x=|p|, y=|q|: \{|p+q|,|p-q|\}=\{x+y,|x-y|\}, and the largest of the four numbers is x+y=4

x\ne y (else |x-y|=0), so \{x,y\}=\{1,3\}, with |x-y|=2

(p,q) is (\pm1,\pm3) or (\pm3,\pm1): 8 squares. For example p=3, q=1: \begin{array}{ccc}8&1&6\\3&5&7\\4&9&2\end{array}

Note: the eight squares are the rotations and reflections of one square.

Final answer

(c) m=5
(d) 8 squares, e.g. rows 8,1,6 / 3,5,7 / 4,9,2

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Linear system	`2nd x⁻¹ (MATRIX), enter the augmented matrix, then MATH, rref(`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 7 teaches systems of equations from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book