Skills Checklist | IB Math AAHL

Skill Checklist

Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.

5 Skills Available

Track your progress:

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Working on it

Confident

📖 = included in formula booklet • 🚫 = not in formula booklet

Track your progress:

Don't know

Working on it

Confident

📖 = included in formula booklet • 🚫 = not in formula booklet

Systems of equations

3 skills

Systems of equations with 2 unknowns

SL Core 2.1

We can solve a system of 2 equations and 2 unknowns with different methods.

{3y+2x−2=03y−3x+1=0

By substitution

Rearranging

3y+2x−2=0⇒y=−32x+32

Substituting this into 3y−3x+1=0:

(−2x+2)−3x+1=0

−5x=−3

So x=53, which implies y=−32⋅53+32=154. So the intersection is (53,154).

By elimination

We can eliminate y from the equations by subtracting the second from the first:

(3y+2x−2)−(2y−3x+1)2x−2+3x−15x=0=0=3

So x=53⇒y=154 and the intersection is again (53,154).

We can use either of these methods to systems of equations with 2 equations and 2 unknowns.

Solving systems of equations with 3 unknowns

AHL AA 1.16

Systems of 3 equations with 3 unknowns, for example

(1)(2)(3)⎩⎪⎨⎪⎧2x−3y+4z=85x+2y−z=63x+4y+2z=17

can be solved with a calculator, or by using substitution.

Worked solution

For the system above, equation (2) is a convenient place to start, because it contains just −z, so it is easy to rearrange for z:

z=5x+2y−6

Now substitute this expression for z into equations (1) and (3).

2x−3y+4(5x+2y−6)=8

22x+5y=32

Using equation (3):

3x+4y+2(5x+2y−6)=17

13x+8y=29

We know have a system of 2 equations with 2 unknowns. Eliminate y by taking

8(22x+5y)−5(13x+8y)=8⋅32−5⋅29

176x+40y−65x−40y=256−145

111x=111

x=1

Now substitute x=1 into 13x+8y=29:

13(1)+8y=29

y=2

Finally, substitute x=1 and y=2 into z=5x+2y−6:

z=5(1)+2(2)−6

z=3

So the solution is x=1,y=2,z=3.

Solution count for 3 by 3 systems of equations

AHL AA 1.16

A system of 3 equations with 3 unknowns can have

no solutions
a unique solution
infinitely many solutions

Example

Consider the system of equations

(1)(2)(3)⎩⎪⎨⎪⎧x+2y−z=1−x−y−z=−1x+ay+2z=2

Add equations (1) and (2):

y−2z=0

y=2z

Now add equations (2) and (3):

(a−1)y+z=1

Substitute y=2z:

(a−1)(2z)+z=1

(2a−1)z=1

If 2a−1=0, then we can solve for

z=2a−11

and then find y and z. This represents a unique solution.

But if 2a−1=0, ie a=21, then we have 0z=1, to which there are no solutions.

Partial fractions

2 skills

Linear partial fractions

AHL AA 1.11

When the denominator of a fraction can be factored, it is possible to split it into the sum of 2 pieces:

x2+2x−34x−8≡(x+3)(x−1)4x−8

(where ≡ means "equal for all x")

Remember that when we add two fractions, we cross multiply the denominators:

x+3A+x−1B≡(x+3)(x−1)A(x−1)+B(x+3)

Then we can write

4x−8≡A(x−1)+B(x+3)

The easiest way to solve this is to chose a value of x that eliminates A or B, eg x=1:

4⋅1−8=A⋅(1−1)+B(1+3)

−4=4B⇒B=−1

and therefore

4x−8=Ax−A−x−3⇒A=5

Non-linear partial fractions

AHL AA 1.11

Aside from expressions of the form (x−c)(x−d)ax+b, there are all sorts of types of partial fractions. For more complicated cases, exam questions will specify the form of the partial fractions decomposition, and you simply need to solve for the variables.

Solve for A and B in the equation x5−2x4+x−22x4+x≡x4+1A+x−2B

Cross multiplying:

x4+1A+x−2B=x5−2x4+x−2A(x−2)+B(x4+1)

A(x−2)+B(x4+1)=2x4+x

We notice immediately that B=2, so

Ax−2A+2x4+2=2x4+x⇒A=1

Skill Checklist

Track your progress across all skills in your objective. Mark your confidence level and identify areas to focus on.

5 Skills Available

Track your progress:

Don't know

Working on it

Confident

📖 = included in formula booklet • 🚫 = not in formula booklet

Track your progress:

Don't know

Working on it

Confident

📖 = included in formula booklet • 🚫 = not in formula booklet

Systems of equations

3 skills

Systems of equations with 2 unknowns

SL Core 2.1

We can solve a system of 2 equations and 2 unknowns with different methods.

{3y+2x−2=03y−3x+1=0

By substitution

Rearranging

3y+2x−2=0⇒y=−32x+32

Substituting this into 3y−3x+1=0:

(−2x+2)−3x+1=0

−5x=−3

So x=53, which implies y=−32⋅53+32=154. So the intersection is (53,154).

By elimination

We can eliminate y from the equations by subtracting the second from the first:

(3y+2x−2)−(2y−3x+1)2x−2+3x−15x=0=0=3

So x=53⇒y=154 and the intersection is again (53,154).

We can use either of these methods to systems of equations with 2 equations and 2 unknowns.

Solving systems of equations with 3 unknowns

AHL AA 1.16

Systems of 3 equations with 3 unknowns, for example

(1)(2)(3)⎩⎪⎨⎪⎧2x−3y+4z=85x+2y−z=63x+4y+2z=17

can be solved with a calculator, or by using substitution.

Worked solution

For the system above, equation (2) is a convenient place to start, because it contains just −z, so it is easy to rearrange for z:

z=5x+2y−6

Now substitute this expression for z into equations (1) and (3).

2x−3y+4(5x+2y−6)=8

22x+5y=32

Using equation (3):

3x+4y+2(5x+2y−6)=17

13x+8y=29

We know have a system of 2 equations with 2 unknowns. Eliminate y by taking

8(22x+5y)−5(13x+8y)=8⋅32−5⋅29

176x+40y−65x−40y=256−145

111x=111

x=1

Now substitute x=1 into 13x+8y=29:

13(1)+8y=29

y=2

Finally, substitute x=1 and y=2 into z=5x+2y−6:

z=5(1)+2(2)−6

z=3

So the solution is x=1,y=2,z=3.

Solution count for 3 by 3 systems of equations

AHL AA 1.16

A system of 3 equations with 3 unknowns can have

no solutions
a unique solution
infinitely many solutions

Example

Consider the system of equations

(1)(2)(3)⎩⎪⎨⎪⎧x+2y−z=1−x−y−z=−1x+ay+2z=2

Add equations (1) and (2):

y−2z=0

y=2z

Now add equations (2) and (3):

(a−1)y+z=1

Substitute y=2z:

(a−1)(2z)+z=1

(2a−1)z=1

If 2a−1=0, then we can solve for

z=2a−11

and then find y and z. This represents a unique solution.

But if 2a−1=0, ie a=21, then we have 0z=1, to which there are no solutions.

Partial fractions

2 skills

Linear partial fractions

AHL AA 1.11

When the denominator of a fraction can be factored, it is possible to split it into the sum of 2 pieces:

x2+2x−34x−8≡(x+3)(x−1)4x−8

(where ≡ means "equal for all x")

Remember that when we add two fractions, we cross multiply the denominators:

x+3A+x−1B≡(x+3)(x−1)A(x−1)+B(x+3)

Then we can write

4x−8≡A(x−1)+B(x+3)

The easiest way to solve this is to chose a value of x that eliminates A or B, eg x=1:

4⋅1−8=A⋅(1−1)+B(1+3)

−4=4B⇒B=−1

and therefore

4x−8=Ax−A−x−3⇒A=5

Non-linear partial fractions

AHL AA 1.11

Solve for A and B in the equation x5−2x4+x−22x4+x≡x4+1A+x−2B

Cross multiplying:

x4+1A+x−2B=x5−2x4+x−2A(x−2)+B(x4+1)

A(x−2)+B(x4+1)=2x4+x

We notice immediately that B=2, so

Ax−2A+2x4+2=2x4+x⇒A=1