© boardworks ltd 2005 1 of 43 a5 simultaneous equations ks4 mathematics

© Boardworks Ltd 2005 1 of 43

A5 Simultaneous equations

KS4 Mathematics


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AA5.1 Solving simultaneous equations graphically


Contents

A5.2 The elimination method

A5.3 The substitution method

A5.5 Problems leading to simultaneous equations

A5.4 Simultaneous linear and quadratic equations


Simultaneous equations

Equations in two unknowns have an infinite number of solution pairs. For example,

x + y = 3

is true when x = 1 and y = 2

x = 3 and y = 0

x = –2 and y = 5 and so on …

We can represent the set of solutions on a graph:

0

3

3 x

y

x + y = 3



Another equation in two unknowns will also have an infinite number of solution pairs. For example,

y – x = 1

is true when x = 1 and y = 2

x = 3 and y = 4

x = –2 and y = –1 and so on …

This set of solutions can also be represented in a graph:

0 x

y

3

3

y – x = 1



There is one pair of values that solves both these equations:

y – x = 1

We can find the pair of values by drawing the lines x + y = 3 and y – x = 1 on the same graph.

0 x

y

3

3

y – x = 1

x + y = 3

x + y = 3

The point where the two lines intersect gives us the solution to both equations.

This is the point (1, 2).

At this point x = 1 and y = 2.



y – x = 1

are called a pair of simultaneous equations.

x + y = 3

The values of x and y that solve both equations are x = 1 and y = 2, as we found by drawing graphs.

We can check this solution by substituting these values into the original equations.

1 + 2 = 3

2 – 1 = 1

Both the equations are satisfied and so the solution is correct.


Solving simultaneous equations graphically


Simultaneous equations with no solutions

Sometimes pairs of simultaneous equations produce graphs that are parallel.

Parallel lines never meet, and so there is no point of intersection.

When two simultaneous equations produce graphs which are parallel there are no solutions.

When two simultaneous equations produce graphs which are parallel there are no solutions.

How can we tell whether the graphs of two lines are parallel without drawing them?

Two lines are parallel if they have the same gradient.


Simultaneous equations with no solutions

We can find the gradient of the line given by a linear equation by rewriting it in the form y = mx + c.The value of the gradient is given by the value of m.

Show that the simultaneous equationsy – 2x = 3

2y = 4x + 1have no solutions.

Rearranging these equations in the form y = mx + c gives,y = 2x + 3y = 2x + ½

The gradient m is 2 for both equations and so there are no solutions.


Simultaneous equations with infinite solutions

Sometimes pairs of simultaneous equations are represented by the same graph. For example,

Notice that each term in the second equation is 3 times the value of the corresponding term in the first equation.

2x + y = 3

6x + 3y = 9

Both equations can be rearranged to give

y = –2x + 3

When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.


Special solutions


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Contents





A5.1 Solving simultaneous equations graphically


The elimination method

If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example, suppose

3x + y = 9

5x – y = 7

Adding these equations:

3x + y = 9

5x – y = 7+

8x = 16

The y terms have been eliminated.

divide both sides by 8: x = 2



Adding the two equations eliminated the y terms and gave us a single equation in x.

3x + y = 9

5x – y = 7

To find the value of y when x = 2 substitute this value into one of the equations.

Solving this equation gave us the solution x = 2.

Substituting x = 2 into the first equation gives us:

3 × 2 + y = 9

6 + y = 9

y = 3subtract 6 from both sides:



We can check whether x = 2 and y = 3 solves both:

3x + y = 9

5x – y = 7

by substituting them into the second equation.

5 × 2 – 3 = 7

10 – 3 = 7

This is true, so we have confirmed that

x = 2

y = 3

solves both equations.



Solve these equations: 3x + 7y = 223x + 4y = 10

Subtracting gives:3x + 7y = 223x + 4y = 10–

3y = 12

The x terms have been eliminated.

divide both sides by 3: y = 4

Substituting y = 4 into the first equation gives us,3x + 7 × 4 = 22

3x + 28 = 22

x = –2divide both sides by 3:

subtract 28 from both sides: 3x = –6



We can check whether x = –2 and y = 4 solves both,

3x + 7y = 22

3x + 4y = 10

by substituting them into the second equation.

3 × –2 + 7 × 4 = 22

–6 + 28 = 22

This is true and so,

x = –2

y = 4

solves both equations.


The elimination method 1



Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example,

4x – y = 29

3x + 2y = 19

We need to have the same number in front of either the x or the y before adding or subtracting the equations.

8x – 2y = 58

11x = 77

divide both sides by 11: x = 7

Call these equations 1 and 2 .

1

2

2 × 1 : 3

3x + 2y = 19+

3 + 2 :



To find the value of y when x = 7 substitute this value into one of the equations,

4x – y = 29 1

3x + 2y = 19 2

4 × 7 – y = 29

28 – y = 29

Substituting x = 7 into 1 gives,

subtract 28 from both sides: –y = 1

y = –1multiply both sides by –1:

Check by substituting x = 7 and y = –1 into 2 ,3 × 7 + 2 × –1 = 9

21 – 2 = 19



6x – 15y = 75Call these equations 1 and 2 .

Solve: 2x – 5y = 25

3x + 4y = 3

1

2

3 × 1– 6x + 8y = 62 × 2

– 23y =

3

4

3 – 4 ,y = –3divide both sides by –23:

Substitute y = –3 in 1 , 2x – 5 × –3 = 252x + 15 = 25

2x = 10subtract 15 from both sides:

x = 5divide both sides by 2:

69


The elimination method 2


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The substitution method

Two simultaneous equations can also be solved by substituting one equation into the other. For example,


y = 2x – 32x + 3y = 23

1

2

Substitute equation 1 into equation 2 .

y = 2x – 3

2x + 3(2x – 3) = 23

expand the brackets: 2x + 6x – 9 = 23

simplify: 8x – 9 = 23

add 9 to both sides: 8x = 32

x = 4divide both sides by 8:



To find the value of y when x = 4 substitute this value into one of the equations,

y = 2x – 3 1

2x + 3y = 23 2

y = 2 × 4 – 3

y = 5

Substituting x = 4 into 1 gives

Check by substituting x = 4 and y = 5 into 2 ,2 × 4 + 3 × 5 = 23

8 + 15 = 23

This is true and so the solutions are correct.



How could the following pair of simultaneous equations be solved using substitution?


3x – y = 9 8x + 5y = 1

1

2

One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation.

Rearrange equation 1 . 3x – y = 9

add y to both sides: 3x = 9 + y subtract 9 from both sides: 3x – 9 = y

y = 3x – 9



3x – y = 9 8x + 5y = 1

1

2

Now substitute y = 3x – 9 into equation 2 . 8x + 5(3x – 9) = 1

expand the brackets: 8x + 15x – 45 = 1

simplify: 23x – 45 = 1

add 45 to both sides: 23x = 46divide both sides by 23: x = 2

Substitute x = 2 into equation 1 to find the value of y. 3 × 2 – y = 9

6 – y = 9–y = 3y = –3



3x – y = 9

8x + 5y = 1

1

2

Check the solutions x = 2 and y = –3 by substituting them into

equation 2 .

8 × 2 + 5 × –3 = 1

16 – 15 = 1

This is true and so the solutions are correct.

Solve these equations using the elimination method to see if you get the same solutions for x and y.


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Simultaneous linear and quadratic equations

When one of the equations in a pair of simultaneous equations is quadratic, we often end up with two pairs of solutions. For example,

y = x + 3

y = x2 + 1 1

2

x2 + 1 = x + 3

We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax2 + bx + c = 0.

x2 – x – 2 = 0

factorize: (x + 1)(x – 2) = 0

x = –1 or x = 2

Substituting equation 1 into equation 2 ,



We can substitute these values of x into one of the equations

y = x + 3

y = x2 + 1 1

2

When x = –1 we have,

It is easiest to substitute into equation 2 because it is linear.

to find the corresponding values of y.

y = –1 + 3

y = 2

When x = 2 we have,

y = 2 + 3

y = 5

The solutions are x = –1, y = 2 and x = 2, y = 5.


Using graphs to solve equations

–1–2–3–4 0 1 2 3 4–2

2

4

6

8

10We can also show the solutions to

using a graph.

y = x2 + 1

y = x + 3

y = x2 + 1

y = x + 3

(–1,2)

(2, 5)

The points where the two graphs intersect give the solution to the pair of simultaneous equations.

It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not exact.


Linear and quadratic graphs



Look at this pair of simultaneous equations:

x2 + y2 = 13

y = x + 1 1

2

The graph of x2 + y2 = 13 is a circular graph with its centre at the origin and a radius of √13.

What shape is the graph given by x2 + y2 = 13?

We can solve this pair of simultaneous equations algebraically using substitution.

We can also plot the graphs of the equations and observe where they intersect.



x2 + y2 = 13

y = x + 1 1

2

x2 + (x + 1)2 = 13

factorize: (x + 3)(x – 2) = 0

x = –3 or x = 2

Substituting equation 1 into equation 2 ,

x2 + x2 + 2x + 1 = 13expand the bracket:

simplify: 2x2 + 2x + 1 = 13

2x2 + 2x – 12 = 0subtract 13 from both sides:

x2 + x – 6 = 0divide through by 2:



We can substitute these values of x into one of the equations

x2 + y2 = 13

y = x + 1 1

2

When x = –3 we have,

It is easiest to substitute into equation 1 because it is linear.

to find the corresponding values of y.

y = –3 + 1

y = –2

When x = 2 we have,

y = 2 + 1

y = 3

The solutions are x = –3, y = –2 and x = 2, y = 3.


Linear and circular graphs


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Solving problems

The sum of two numbers is 56 and the difference between the two numbers is 22. Find the two numbers.

a + b = 56

a – b = 22

Adding these equations gives:

Let’s call the unknown numbers a and b.

We can use the given information to write a pair of simultaneous equations in terms of a and b,

2a = 78

a = 39


Solving problems

39 + b = 56

So the two numbers are 39 and 17.

39 – 17 = 22

This is true and so our solution is correct.

Substituting a = 39 into the first equation gives,

b = 17subtract 39 from both sides:

We can check these solutions by substituting them into the second equation, a – b = 22:


Solving problems

The cost of theatre tickets for 4 adults and 3 children is £47.50. The cost for 2 adults and 6 children is £44.

How much does each adult and child ticket cost?

Let’s call the cost of an adult’s ticket a and the cost of a child’s ticket c. We can write,

4a + 3c = 47.50 1

2a + 6c = 44 2

Dividing equation 2 by 2 gives,a + 3c = 22 3

We can now subtract equation 3 from equation 1 to eliminate the terms containing c.


Solving problems

4a + 3c = 47.50 1

a + 3c = 22 3

The cost of an adult’s ticket is £8.50 and the cost of a child’s ticket is £4.50.

–

3a1 – 3 ,a = 8.50divide both sides by 3:

Substitute a = 8.50 in 3 :

8.50 + 3c = 22

3c = 13.50subtract 8.50 from both sides:

c = 4.50divide both sides by 3:

= 25.50


Solving problems

Remember, when using simultaneous equations to solve problems:

1) Decide what letters to use to represent each of the unknown values.

2) Use the information given in the problem to write down two equations in terms of the two unknown values.

3) Solve the simultaneous equations using the most appropriate method.

4) Check the values by substituting them back into the original problem.

© boardworks ltd 2005 1 of 43 a5 simultaneous equations ks4 mathematics

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