Section 3: Equations and Inequalities

Introduction

This section will cover how to:

Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page.

Remember that you are NOT allowed to use calculators in this topic.

 

Simultaneous Equations

Basic Principles
Simultaneous equation questions have two equations, both of which have two unknown variables. The solution of a pair of simultaneous equation is the values of both variables which make both equations work at the same time.

For example, consider the equations: x + y = 8 and xy = 2.
x = 1, y = 7 is a solution of the first equation but not the second.
x = 4, y = 2 is a solution of the second equation but not the first.
x = 5, y = 3 is a solution of both equations at the same time, so it is the solution of these simultaneous equations.

The simultaneous equations in this section are linear because the variables are not raised to any higher powers e.g x².
In this section the examples will mostly use x and y but the principles are the same for any pair of letters.

Notes
  • There will normally be one solution (a pair of values) to a pair of simultaneous linear equations.
  • Simultaneous equations can be solved using algebraic methods or graphically.
  • You can always check your answer by substituting the answer into both your original equations.

 

Solving Simultaneous Linear Equations by Elimination

Elimination Method
This example will illustrate the elimination method.
The steps are shown below and the working on the right.
2x = 16 – 5y
7x – 3y = 15
1. Number the equations and make sure matching letters are above each other in the pair of equations (usually we get the letters on the left-hand side and the numbers on the right-hand side. (1)     2x + 5y = 16
(2)     7x – 3y = 15
2. Get the coefficients of one of the variables to be the same in both equations by multiplying one or both of the equations by numbers (it doesn't matter if the signs of the coefficients are different). Multiply equation (1) by 7 and equation (2) by 2:
(1)     14x + 35y = 112
(2)     14x   – 6y = 30
3. Eliminate the variable with the matching coefficients by adding or subtracting the two equations (add if the signs on the matching coefficients are different, or subtract if the signs are the same). Signs on the matching terms are the same,
so subtract equation (2) from equation (1):
                    41y = 82
(always be careful with negative signs)
4. Solve the resulting equation which now only has one variable to get the first part of the solution. Divide by 41 on both sides:
                        y = 2
5. Substitute the first half of the solution into one of the original equations and solve to get the other part of the solution. Substitute y = 2 into equation (1):
(1)     2x + 10 = 16
                  2x = 6
                    x = 3
6. Substitute both parts of the solution into the other original equation to check your answer. Substitute x=3, y = 2 into equation (2):
(2)     21 – 6 = 15
7. Write the both parts of the solution clearly. It works, so the solution is:
                    x = 3
                    y = 2
Further Examples
Make sure you understand these examples before moving on:
Example 1Example 2
10x = 3y + 4
5x – 6y + 1 = 0
3x + y = 1
5x – 4y = 47
(1)     10x – 3y = 4
(2)       5x – 6y = –1
(1)     3x   + y = 1
(2)     5x – 4y = 47
Multiply (2) by 2 so the x terms match:                
(1)   10x   – 3y = 4
(2)   10x – 12y = –2
Multiply (1) by 4 so the y terms match:
(1)   12x + 4y = 4
(2)     5x – 4y = 47
Do equation (1) – equation (2):
                    9y = 6
Signs different so do equation (1) + equation (2):
                17x = 51
Divide by 9 on both sides:
                    y  2 
 3 
Divide by 17 on both sides:
                     x = 3
Substitute y  2   into equation (1):
 3 
(1)     10x – 2 = 4
                10x = 6
                    x  3 
 5 
Substitute x = 3 into equation (1):

(1)     9 + y = 1
                y = –8

Check x 3   , y 2    in equation (2):
 5  3 
(2)     3 – 4 = –1
Check x=3, y = –8 in equation (2):

(2)     15 – 32 = 47

It works, so the solution is:
       x 3   , y 2 
 5  3 
It works, so the solution is:
                    x = 3
                    y = –8
Note that there are other ways to solve these problems, but they will all give the same answers if done correctly.
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Solve these simultaneous equations using the elimination method:

(a) 4x = 10 – y
      8x – 3y = 0

(1)     4x   + y = 10
(2)     8x – 3y = 0
Multiply (1) by 3:
(1)   12x + 3y = 30
(2)     8x – 3y = 0
Add equations (1) and (2):
        20x        = 30
                   x  3 
 2 
Substitute x  3   into equation (1):
 2 
(1)     6 + y = 10
                y = 4
Check x 3   , y = 4 in equation (2):
 2 
(2)     12 – 12 = 0
It works, so the solution is:
       x 3   , y = 4
 2 

(b) 3x – 5y = 4
      3y – 4x = 2

(1)      3x – 5y = 4
(2)   – 4x + 3y = 2
Multiply (1) by 4 and (2) by 3:
(1)    12x – 20y = 16
(2) – 12x   + 9y = 6
Add equations (1) and (2):
                 –11y = 22
                       y = –2
Substitute y = –2 into equation (1):
(1)     3x + 10 = 4
                  3x = –6
                    x = –2
Check x = –2, y = –2 in equation (2):
(2)     8 + – 6 = 2
It works, so the solution is:
       x = –2, y = –2

 

Solving Simultaneous Linear Equations by Substitution

Substitution Method
This example will illustrate the substitution method.
The steps are shown below and the working on the right.
3x = 9 – 2y
2xy = 20
1. Number the equations (there is no need to rearrange them yet). (1)     3x = 9 – 2y
(2)     2xy = 20
2. Rearrange one of the equations so that one of the letters is the subject (on its own). Pick the one which is easiest to rearrange! Rearrange equation (2) to make y the subject:
(2)     2x – 20 = y
3. Substitute the newly rearranged equation into the other original equation. Substitute y = 2x – 20 into equation (1):
(1)     3x = 9 – 2( 2x – 20 )
4. Solve the resulting equation which now only has one variable to get the first part of the solution. Multiply out the brackets:
         3x = 9 – 4x + 40
Add 4x to both sides:
         7x = 49
Divide by 7 both sides:
           x = 7
5. Substitute the first half of the solution into the rearranged version of the equation from earlier, giving the second part of the solution. Substitute x = 7 into y = 2x – 20:
           y = 14 – 20
           y = –6
6. Substitute both parts of the solution into the original equation which you didn't rearrange, to check your answer. Substitute x = 7, y = –6 into equation (1):
(1)     21 = 9 – 12
7. Write the both parts of the solution clearly. It works, so the solution is:
           x = 7
           y = –6
Further Examples
Make sure you understand these examples before moving on:
Example 1Example 2
2x = 3y + 2
xy = 2
y = 4x – 5
y = 2x + 2
(1)     2x = 3y + 2
(2)     xy = 2
(1)     y = 4x – 5
(2)     y = 2x + 2
Make y the subject of (2):
(2)     x = 2 + y
y is already the subject of (1)
Substitute y = 2 + y into (1):
(1)     2( 2 + y ) = 3y + 2
Substitute y = 4x – 5 into (2):
(2)     4x – 5 = 2x + 2
              4 + 2y = 3y + 2
              4         =  y + 2
              2         =  y
          2x – 5 =         2
          2x       =         7
            x       =         3.5
Substitute y = 2 into x = 2 + y:
           x = 2 + 2
           x = 4
Substitute x = 3.5 into y = 4x – 5:
           y = 14 – 5
           y = 9
Substitute x = 4, y = 2 into equation (1):
(1)     8 = 6 + 2
Substitute x = 3.5, y = 9 into equation (2):
(2)     9 = 7 + 2
It works, so the solution is:
           x = 4
           y = 2
It works, so the solution is:
           x = 3.5
           y = 9
Note that both these problems could have been solved in other ways, including by elimination.
Choosing Elimination or Substitution
It doesn't matter which method you use for linear simultaneous equations, although one will usually be easier than the other for a particular question. If the corresponding terms are already lined up and maybe already matching, then elimination is usually easier. If there is already a letter on its own, maybe already the subject of one of the equations, then substitution should be easier.

In the A-Level Core 1 module you will need to do simultaneous equations where one is linear and the other is quadratic (it includes squared terms). In this case the only option is to use substitution, so you definitely need to know this skill for that situation.

Fraction or Decimal Answers
It doesn't matter whether you give your answers as fractions or decimals, but if you give your answers as fractions you should cancel them down and use top-heavy fractions rather than mixed numbers. Your answers should always be exact so avoid rounding decimals or using recurring decimals – for this reason it is best to mainly use fractions.
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Solve these simultaneous equations using the substitution method:

(a) 4x = 10 – y
      8x – 3y = 0

(1)     4x = 10 – y
(2)     8x – 3y = 0
Make y the subject in (1):
(1)     y = 10 – 4x
Substitute y = 10 – 4x into (2):
(2)     8x – 3( 10 – 4x ) = 0
Solve for x:
  8x – 30 + 12x = 0
          20x – 30 = 0
                  20x = 30
                      x  3 
 2 
Substitute x 3    in y = 10 – 4x:
 2 
                      y = 10 – 6
                      y = 4
Check x 3   , y = 4 in equation (2):
 2 
(2)     12 – 12 = 0
It works, so the solution is:
       x 3   , y = 4
 2 

(b) x = 5 + y
      y = 6 – x

(1)     x = 5 + y
(2)     y = 6 – x
Substitute y = 6 – x into (1):
(1)     x = 5 + ( 6 – x )
         x = 11 – x
       2x = 11
         x = 5.5
Substitute x = 5.5 in y = 6 – x:
         y = 6 – 5.5
         y = 0.5
Check x = 5.5, y = 0.5 in equation (1):
(2)     5.5 = 5 + 0.5
It works, so the solution is:
           x = 5.5
           y = 0.5

 

Inequalities

Inequality Symbols
<less thanfurther left on the number line
>greater thanfurther right on the number line
less than or equal tofurther left or in the same position on the number line
greater than or equal tofurther right or in the same position on the number line
not equal toin a different position on the number line
The following statements are all true – check through them to make sure you agree with them:
3 < 5 –6 < –2 –4 ≤ 3 6 ≤ 6
4 ≠ –4 4 ≥ –4 3.4 > 2.7 0 ≥ –5
It's worth noting that when we write the ≤ and ≥ symbols by hand, we usually make the additional line parallel with the lower half of the < or > symbol. In the exercises below you will see examples of them written this way.
Representing Single Inequalities on a Number Line
The single inequality x > 5 means that x can take any value from 5 upwards, but not 5 itself. This includes all the decimals as well as whole numbers. We represent this on a number line as follows:

If we wanted to also include the number 5 we would use the inequality x ≥ 5, and would represent it as follows:

The filled circle above the 5 indicates that 5 is included in the inequality. Similarly, we can represent x ≤ 0 as follows:

The number line below represents the inequality x < 2:

Representing Double Inequalities on a Number Line
Sometimes we want to represent two inequalities at the same time. For example, if we wanted to represent all the values between –3 and +3 including the end values, we would use the inequality –3 ≤ x ≤ 3. This is a shortform for the following statement composed of two inequalities: " x ≥ –3 and x ≤ 3 " and would be represented as follows:

Imagine instead we wanted to represent all the values which are less than 0 or greater than 2. It's not possible in this situation to use just one statement, so we need to write both inequalities: " x < 0 or x > 2 ". We can still represent this on a number line as seen here:

It is not necessary for both ends to be both filled or both not filled. The number line below represents the double inequality 1 ≤ x < 3, where x can take any value between 1 and 3, including 1 but not including 3:

As before, 1 ≤ x < 3 is a shortform for the following statement composed of two inequalities: " x ≥ 1 and x < 3 "

Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Represent these inequalities on a number line:

(a) x < 4

(b) 2 ≤ x < 7

State the inequalities represented by these number lines:

(a)

x ≥ 7

(b)

x < 2 or x > 6

 

Solving Algebraic Inequalities

Basic Principles
Solving inequalities is very similar to solving simple equations with two important differences:
  • You should avoid multiplying or dividing by negative numbers, or if if you do multiply or divide by a negative number you should then reverse the direction of the inequality.
  • As with equations we do operations to both sides of the inequality, but if we have a double inequality with three parts then we need to do the same operation to all three parts.
The main aim is still to get the unknown letter (e.g. x) on its own, leaving a simple inequality which can be represented easily on a number line.
Examples
Look through these and make sure you understand them before moving on:
e.g.
2x + 3 ≥ 11
2x       ≥ 8
  x       ≥ 4
e.g.
3x + 1 < x + 7
2x + 1 <       7
2x       <       6
  x       <       3
e.g.
5 < 2x – 1 < 13
6 <    2x    < 14
3 <     x     < 7
                         
Combining Inequalities
Sometimes we are asked to find all the values which satisfy both of two inequalities. In this case it is helpful to represent both inequalities above each other on a number line and then look for those parts of the number line on which both inequalities are true. It is worth noting that sometimes there are no values which satisfy both inequalities. For example, there are no values which satisfy both " x ≤ 2 " and " x ≥ 5 ".
Examples
e.g. Find all the values which satisfy both
" x < 5 " and " x > 2 "
Start by representing both above the same number line:
Looking at these, we can see that both inequalities are true at the following points on the number line:
The numbers 2 and 5 are not included as they do not satisfy both inequalities.

The correct answer is " 2 < x < 5 ".

e.g. Find all the values which satisfy both
" x < 4 or x > 6 " and " x ≥ 2 "
Start by representing both above the same number line:
Looking at these, we can see that both inequalities are true at the following points on the number line:
Number 2 is included this time as it satisfies both inequalities. Numbers 4 and 6 do not satisfy both.

The correct answer is " 2 ≤ x < 4 or x ≥ 6 ".

Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Solve these inequalities and represent the solution on a number line:

(a) 4( x – 3 ) ≤ 2( x + 2 )

4( x – 3 ) ≤ 2( x + 2 )
   4x – 12 ≤ 2x + 4
           2x ≤ 16
             x ≤ 8

(b) 2 < 3x – 7 ≤ 8

9 < 3x ≤ 15
3 <  x  ≤ 5
State the values for which both the inequalities in the question are true:

(a) x < 7
     x > 3

Individually:

Combined:

The correct answer is " 3 < x < 7 ".

(b) x ≤ 5
     x < 3 or x > 6

Individually:

Combined:

The correct answer is " x < 3 ".

 


Exercise

Work out the answers to the questions below and fill in the boxes. Click on the
button to find out whether you have answered correctly. If you have then the answer will be ticked
and you should move on to the next question. If a cross
appears then your answer is wrong. Click on
to clear the incorrect answer and have another go, or you can click on
to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on
to see the solution.

 

Question 1

Solve the simultaneous equations below by elimination, simplifying your answers where necessary.

(a)

3x + 5y = 1
3x – 2y = 8
x =
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y =
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Subtract the second equation from the first as the x terms already match.
(1) 3x + 5y = 1
(2) 3x – 2y = 8
Do (1) – (2):
7y = –7
  y = –1
Substitute y = –1 into (1):
3x –5 = 1
     3x = 6
       x = 2
The solution is x = 2, y = –1

(b)

5y = 11 – 2x
3x – 10y + 36 = 0
x =
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y =
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Rearrange to get the terms of the same type above each other, then get the y terms matching.
(1) 2x   + 5y = 11
(2) 3x – 10y = –36
Multiply (1) by 2:
(1) 4x + 10y = 22
(2) 3x – 10y = –36
Do (1) + (2):
7x = –14
  x = –2
Substitute x = –2 into (1):
–4 + 5y = 11
        5y = 15
          y = 3
The solution is x = –2, y = 3

(c)

5x + 4y = 25
6y + 4x = 27
x =
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y =
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Player 8 or above. Click here.
Rearrange to get the terms of the same type above each other, then get the x or y terms matching by multiplying each equation by a different number.
(1) 5x + 4y = 25
(2) 4x + 6y = 27
Multiply (1) by 4 and multiply (2) by 5:
(1) 20x + 16y = 100
(2) 20x + 30y = 135
Do (2) – (1):
14y = 35
y 35  =  5 
 14  2 
Substitute y 5   into (1):
 2 
5x + 25 = 11
        5x = 15
          x = 3
The solution is x = 3, y 5 
 2 

Question 2

Solve the simultaneous equations below by substitution, simplifying your answers where necessary.

(a)

2x + 3y = 11
y = 3x – 11
x =
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y =
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Player 8 or above. Click here.
y is already the subject of the second equation, so substitute that into the first equation.
(1) 2x + 3y = 11
(2) y = 3x – 11
y is already the subject of equation (2), so
substitute y = 3x – 11 into (1)
(1) 2x + 3( 3x – 11) = 11
2x + 9x – 33 = 11
              11x = 44
                  x = 4
Substitute x = 4 into y = 3x – 11
                  y = 12 – 11
                  y = 1
The solution is x = 4, y = 1

(b)

5yx = 20
2x + 4y = 2
x =
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Player 8 or above. Click here.
y =
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Player 8 or above. Click here.
Rearrange the first equation so that x is the subject and substitute it into the second equation.
(1) 5yx = 20
(2) 2x + 4y = 2
Make x the subject of equation (1):
(1) x = 5y – 20
Substitute x = 5y – 20 into (2):
(2) 2( 5y – 20 ) + 4y = 2
10y – 40 + 4y = 2
                14y = 42
                    y = 3
Substitute y = 3 into x = 5y – 20
                    x = 15 – 20
                    x = –5
The solution is x = –5, y = 3

(c)

y = 7x – 4
y = 2x – 1
x =
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y =
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Player 8 or above. Click here.
You can substitute this one either way around!
(1) y = 7x – 4
(2) y = 2x – 1
Substitute y = 7x – 4 into (2):
(2) 7x – 4 = 2x – 1
 5x = 3
   x 3 
 5 
Substitute x 3   into y = 7x – 4:
 5 
y 21  – 4:
 5 
y 1 
 5 
The solution is x 3   , y 1 
 5  5 

In the question below you will need to represent inequalities by clicking in the space above a number line
If it is only a single inequality then the first time you click you will set the "end value" of the inequality, and subsequent clicks in the same place will change the type and direction of the inequality. Try it here:
0123456789
You can start again at any time by clicking in a new position on the number line.
If the inequality has two parts you will need to set the left-hand "end value" and direction as described above, then click on the position of the right-hand "end value". Subsequent clicks on the right-hand end will change whether the point is included or not. Try it here:
0123456789
Once both ends have been set you can start again by clicking in a new position to set the left-hand end again.

Question 3

Represent the following inequalities on number lines.

(a)

x < 2
0123456789
You need to mark the points to the left of 2, but not including 2. The answer is shown on the left.

(b)

x ≥ 6
0123456789
You need to mark the points to the right of 6, including 6. The answer is shown on the left.

(c)

4 ≤ x < 7
0123456789
You need to mark the points between 4 and 7, including 4 but not including 7. The answer is shown on the left.

(d)

x ≤ 5 or x ≥ 8
0123456789
You need to mark the points outside 5 and 8, including both end values. The answer is shown on the left.

In the remaining questions you will need to complete answers including inequality symbols
In each case there will be a "less than" symbol in place already. If this is not the correct symbol you can click it to change it to a different symbol. Try clicking the inequality symbols in these examples to see how this works:
3              47

Question 4

State the inequalities represented on each of the following number lines.

(a)

The values indicated are greater than 8. The answer is x > 8.

(b)

The values indicated are less than or equal to 3. The answer is x ≤ 3.

(c)

The values indicated are greater than or equal to 6 and less than or equal to 8.. The answer is shown on the left.

(d)

 or 
The values indicated are less than or equal to 1 or greater than 2. The answer is x ≤ 1 or x > 2.

Question 5

Solve the following inequalities.

(a)

2x + 1 ≥ 13
Subtract one from each side and then divide by 2. 2x + 1 ≥ 13
      2x ≥ 12
        x ≥ 6

(b)

3( x + 1 ) < 2( 2x – 1)      
Multiply out the brackets and then solve as normal. 3( x + 1 ) < 2( 2x – 1)
     3x + 3 < 4x – 2
              5 < x     (so x > 5)

(c)

3 ≤ 5x – 7 < 28
Add 7 to all three parts and then divide by 5.   3 ≤ 5x – 7 < 28
10 ≤    5x    < 35
  2 ≤     x     < 7

Question 6

State the values for which both the inequalities given in each question are true.
You may find it helpful to represent both inequalities on a number line first.

(a)

x ≥ 4
x > 7
Here are the two inequalities:
Combined they give:

The answer is x > 7

(b)

3 < x < 8
x ≤ 6
Here are the two inequalities:
Combined they give:

The answer is 3 < x ≤ 6

(c)

12 < 4( x + 3 ) < 32         
12 ≤ 3x
Solve the two inequalities first. The two inequalities solve to:
0 < x < 5    and    x ≥ 4
Here they are on the number line:

Combined they give:

The answer is 4 ≤ x < 5


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