1. Simplify √5
Simplify 4 2−√ √5
Solve 1+ 𝑥 = 2𝑥 +1

2. 7 −√5
−12−2 5
x=0, x=4

3. Simultaneous equations
Chapter 4.1
Simultaneous equations
“It’s not that I’m so smart, it’s just that I stay with problems longer.”
- Albert Einstein

4. Simultaneous equations
Solution by substitution
Solve the simultaneous equations
𝑥+ 𝑦=4
𝑦=2𝑥+1
Chapter 4.1
𝑥+𝑦=4
Take the expressions for y from the second equation and substitute it into the first
⇒ 𝑥+ 2𝑥+1 =4
Substitution
Elimination
Linear
Quadratic
⇒ 3𝑥=3
⇒ 𝑥=1
Substituting 𝑥=1 into 𝑦=2𝑥+1 gives 𝑦=3 so the solution is
𝑥=1, 𝑦=3.

5. Simultaneous equations
Solution by elimination
Solve the simultaneous equations
3𝑥+ 5𝑦=12
2𝑥+ 3𝑦=7
Chapter 4.1
Multiplying the first equation by 2 and the second by 3 will give two equations each containing the term 6𝑥:
Substitution
Elimination
Linear
Quadratic
This gives two equations
containing 6𝑥
Multiplying 3𝑥+5𝑦=12 by 2
⇒6𝑥+10𝑦=24
Multiplying 2𝑥+3𝑦=7 by 3
⇒6𝑥+ 9𝑦=21
Subtracting
𝑦=3
Substituting this into either of the original equations gives 𝑥=−1.
The solution is
𝑥=−1, 𝑦=3

6. Simultaneous equations
Solution by elimination
Solve the simultaneous equations
9𝑥− 2𝑦=5
5𝑥+3𝑦=11
Chapter 4.1
Multiplying 9𝑥 −2𝑦=5 by 3
⇒27𝑥 −6𝑦=15
This gives two equations each containing 6𝑦
Multiplying 5𝑥+3𝑦=11 by 2
⇒10𝑥+ 6𝑦=22
Substitution
Elimination
Linear
Quadratic
Adding
37𝑥 =37
Hence
𝑥 =1
Substituting this into 5𝑥+3𝑦=11 gives 3𝑦=6 and hence 𝑦=2.
The solution is
𝑥=1, 𝑦=2

7. Simultaneous equations
Solve the simultaneous equations
2𝑥+3𝑦=17 and
5𝑥 −4𝑦=4
using the elimination method.
Chapter 4.1
Multiplying 2𝑥+3𝑦=17 by 5
⇒10𝑥+15𝑦=85
Substitution
Elimination
Linear
Quadratic
Multiplying 5𝑥 −4𝑦=4 by 2
⇒ 10𝑥 − 8𝑦=8
Subtracting
⇒ 𝑦=77
𝑦= = (= 3.35 to 2 d.p.)
Hence
𝑥= = (= 3.48 to 2 d.p.)
Substituting gives

8. Simultaneous equations
Solve the simultaneous equations
𝑦= 𝑥 2 + 𝑥
2𝑥+ 𝑦= 4
using the substitution method.
Chapter 4.1
From the linear equation, 𝑦=4 −2𝑥.
Substituting this into the quadratic equation:
Collect all terms on the rhs (this makes the quadratic positive and easier to factorise).
Substitution
Elimination
Linear
Quadratic
4 −2𝑥= 𝑥 2 + 𝑥
⇒0= 𝑥 2 +3𝑥 −4
⇒ 𝑥+4 𝑥 −1 =0
⇒ 𝑥=−4 or 𝑥=1
Substituting into 2𝑥+𝑦=4
𝑥=−4
⇒−8+𝑦=4
⇒ 𝑦=12
⇒ 𝑦=2
𝑥=1
⇒ 2+𝑦=4
The solution is
𝑥=−4, 𝑦=12
and
𝑥=1, 𝑦=2

9. Solve the following simultaneous equations

10. Select menu MODE A: Equation/Func
Step 1: Select simultaneous equation solver.
Simultaneous equations are where you have multiple equations with multiple variables, e.g. and .
“Solve the simultaneous equations:
Choose simultaneous equations, 2 “unknowns” (as we have and ).
Enter 2, -1, 4, 3, 2, 5, pressing = after each number. Use down arrows to scroll through solutions.

11. Chapter 4.2
Inequalities
“It’s not that I’m so smart, it’s just that I stay with problems longer.”
- Albert Einstein

12. Inequalities Solve 2𝑥 −15 ⩽ 5𝑥 −3.
Add 3 to, and subtract 2x from both sides
Chapter 4.2
⇒−15+3 ⩽ 5𝑥 −2𝑥
Linear
Quadratic
Tidy up
⇒ −12 ⩽ 3𝑥
Divide both sides by 3
⇒−4 ⩽ 𝑥
Note that this implies that x is the larger of the two terms.
Make x the subject
⇒ 𝑥 ⩾ −4

13. Inequalities
Solve the inequality 8 <3𝑥 −1 ⩽ 14 and illustrate the solution on a number line.
8 <3𝑥 −1 ⩽ 14
Chapter 4.2
Add 1 throughout:
⇒ 9 <3𝑥 ⩽ 15
Divide by 3:
⇒ 3 <𝑥 ⩽ 5
Linear
Quadratic
This is illustrated as
Notice how the different circles show that 5 is included in the solution but 3 is not.
It is useful to emphasise the line segment that you want as shown.

14. Inequalities Solve 𝑖 𝑥 2 −4𝑥+3 >0 (𝑖𝑖) 𝑥 2 −4𝑥+3 ⩽ 0.
𝑥 2 −4𝑥+3 =(𝑥 −1)(𝑥 −3)
So the graph of 𝑦= 𝑥 2 −4𝑥+3 is a -shaped quadratic (shown by the positive coefficient of 𝑥 2 ) which crosses the 𝑥 axis at 𝑥=1 and 𝑥=3.
Chapter 4.2
Linear
Quadratic
𝑖 𝑥 2 −4𝑥+3 >0
Values of 𝑥 correspond to
where the curve is above
the axis (i.e. 𝑦 >0).

15. Inequalities 𝑖𝑖 𝑥 2 −4𝑥+3 ⩽ 0 Chapter 4.2
𝑖𝑖 𝑥 2 −4𝑥+3 ⩽ 0
Chapter 4.2
Values of 𝑥 correspond to where the curve is on or below the axis (i.e. 𝑦 ≤0).
Linear
Quadratic
There are various ways of expressing these results:
𝑖 𝑥 <1 or 𝑥 >3
{𝑥 : 𝑥 <1} ∪{𝑥 : 𝑥 >3}
𝑖𝑖 1 ⩽ 𝑥 ⩽ 3
{𝑥 : 𝑥 ⩽ 3} ∩ {𝑥 : 𝑥 ⩾ 1}

16. Inequalities Solve (i) 𝑥 2 −4𝑥 −5 <0 (ii) 𝑥 2 −4𝑥 −5 ⩾ 0
Start by factorising the quadratic expression and finding the values of x which make the expression zero.
𝑥 2 −4𝑥 −5=(𝑥+1)(𝑥 −5)
Chapter 4.2
𝑥+1 𝑥 −5 =0 ⇒ 𝑥=−1 or 𝑥=5
Now think about the sign of each factor at these values and in the three intervals into which they divide the number line:
Linear
Quadratic

17. Inequalities
Chapter 4.2
Linear
Quadratic
(i) From the table the solution to 𝑥+1 𝑥 −5 =0 is −1 <𝑥 <5
which can also be written as {𝑥 : 𝑥 >−1} ∩{𝑥 : 𝑥 <5}.
(ii) The solution to (𝑥 + 1) (𝑥 − 5) ⩾ 0 is 𝑥 ⩽ −1 or 𝑥 ⩾ 5
which can also be written as {𝑥 : 𝑥 ⩽ −1} ∪ {𝑥 : 𝑥 ⩾ −5}.

18. Inequalities
Show the inequalities (𝑖) 𝑦 ≤ 𝑥 − (𝑖𝑖) 𝑦 > 𝑥 graphically.
Chapter 4.2
Linear
Quadratic

19. Inequalities
Chapter 4.2
Linear
Quadratic

20. Inequalities (ii) Determine the range of values of which satisfy the
 (i) Illustrate the inequalities 𝑦≥2 𝑥 2 +3𝑥−2 and 𝑦<2𝑥+4 on the same graph.
(ii) Determine the range of values of which satisfy the
inequality 2 𝑥 2 +3𝑥−2<2𝑥+4
Chapter 4.2
First work out where the curve and the line cut the axes.
For the curve 𝑦=2 𝑥 2 +3𝑥−2, when 𝑥=0, 𝑦=−2
Linear
Quadratic
and when 𝑦=0, 2 𝑥 2 +3𝑥−2=0
2𝑥−1 𝑥+2 =0
So when 𝑦=0 then 𝑥= 1 2 or −2
For the line 𝑦=2𝑥+4, when 𝑥=0, then 𝑦=4 and when 𝑦=0 then 𝑥=−2
The region that satisfies both inequalities is shaded and labelled R.

21. Inequalities
Chapter 4.2
Linear
Quadratic

22. Inequalities
(ii) The range of values of 𝑥 which satisfy the inequality
2 𝑥 2 +3𝑥−2<2𝑥+4 are the values of 𝑥 for which the
curve is below the line.
You need to solve 2 𝑥 2 +3𝑥−2=2𝑥+4 to find where the curve and the line intersect.
Chapter 4.2
2 𝑥 2 +3𝑥−2=2𝑥+4
Linear
Quadratic
⇒2 𝑥 2 +𝑥−6=0
⇒ 2𝑥−3 𝑥+2 =0
⇒𝑥= 3 2 or 𝑥=−2
The curve is below the line when −2<𝑥 < 3 2

23. Solve the following inequalities

24. Select menu MODE B: Inequality
This mode allows you to solve quadratic inequalities, e.g.:
Step 1: Select the ‘degree’ of your polynomial.
A polynomial is an expression of the form where the are constants, for example, or .
The degree of a polynomial is the highest power, so the degree of a cubic, e.g. , is 3. So for quadratic inequalities, use 2.
Step 2: Select vs vs vs
Step 3: Enter your coefficients.
A coefficient is the number/constant in front of each term. Press = after entering each number.
Your calculator will say “All real numbers” if any number is possible for , or “No real solutions” if there are no solutions for .