Download A5 Simultaneous equations

KS4 Mathematics
A5 Simultaneous
equations
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Contents
A5 Simultaneous equations
A A5.1 Solving simultaneous equations graphically
A A5.2 The elimination method
A A5.3 The substitution method
A A5.4 Simultaneous linear and quadratic equations
A A5.5 Problems leading to simultaneous equations
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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 …
y
We can represent the set
of solutions on a graph:
3
0
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x+y=3
3
x
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Simultaneous equations
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 …
y
This set of solutions can
also be represented in a
graph:
3
y–x=1
0
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3
x
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Simultaneous equations
There is one pair of values that solves both these equations:
x+y=3
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.
y y–x=1
The point where the two lines
intersect gives us the solution to
both equations.
3
This is the point (1, 2).
0
3
x+y=3
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x
At this point x = 1 and y = 2.
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Simultaneous equations
x+y=3
y–x=1
are called a pair of simultaneous equations.
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.
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Solving simultaneous equations graphically
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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.
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.
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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 equations
y – 2x = 3
2y = 4x + 1
have no solutions.
Rearranging these equations in the form y = mx + c gives,
y = 2x + 3
y = 2x + ½
The gradient m is 2 for both equations and so there are no
solutions.
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Simultaneous equations with infinite solutions
Sometimes pairs of simultaneous equations are represented
by the same graph. For example,
2x + y = 3
6x + 3y = 9
Notice that each term in the second equation is 3 times the
value of the corresponding term in the first equation.
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.
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Special solutions
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Contents
A5 Simultaneous equations
A A5.1 Solving simultaneous equations graphically
A A5.2 The elimination method
A A5.3 The substitution method
A A5.4 Simultaneous linear and quadratic equations
A A5.5 Problems leading to simultaneous equations
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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
divide both sides by 8:
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The y terms
have been
eliminated.
= 16
x=2
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The elimination method
Adding the two equations eliminated the y terms and gave us
a single equation in x.
Solving this equation gave us the solution x = 2.
To find the value of y when x = 2 substitute this value into one
of the equations.
3x + y = 9
5x – y = 7
Substituting x = 2 into the first equation gives us:
3×2+y=9
6+y=9
subtract 6 from both sides:
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y=3
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The elimination method
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.
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The elimination method
Solve these equations: 3x + 7y = 22
3x + 4y = 10
Subtracting gives:
3x + 7y = 22
– 3x + 4y = 10
3y = 12
y=4
divide both sides by 3:
The x terms
have been
eliminated.
Substituting y = 4 into the first equation gives us,
3x + 7 × 4 = 22
3x + 28 = 22
3x = –6
subtract 28 from both sides:
x = –2
divide both sides by 3:
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The elimination method
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.
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The elimination method 1
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The elimination method
Sometimes we need to multiply one or both of the equations
before we can eliminate one of the variables. For example,
4x – y = 29
1
3x + 2y = 19
2
We need to have the same number in front of either the x or
the y before adding or subtracting the equations.
Call these equations 1 and 2 .
2× 1 :
3 + 2 :
divide both sides by 11:
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8x – 2y = 58
+ 3x + 2y = 19
11x
3
= 77
x=7
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The elimination method
To find the value of y when x = 7 substitute this value into one
of the equations,
1
4x – y = 29
3x + 2y = 19
2
Substituting x = 7 into 1 gives,
4 × 7 – y = 29
28 – y = 29
subtract 28 from both sides:
multiply both sides by –1:
–y = 1
y = –1
Check by substituting x = 7 and y = –1 into 2 ,
3 × 7 + 2 × –1 = 9
21 – 2 = 19
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The elimination method
Solve:
2x – 5y = 25
1
3x + 4y = 3
2
Call these equations 1 and 2 .
3× 1
2× 2
3 – 4 ,
6x – 15y = 75
– 6x + 8y = 6
divide both sides by –23:
3
4
– 23y = 69
y = –3
Substitute y = –3 in 1 , 2x – 5 × –3 = 25
2x + 15 = 25
2x = 10
subtract 15 from both sides:
x=5
divide both sides by 2:
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The elimination method 2
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Contents
A5 Simultaneous equations
A A5.1 Solving simultaneous equations graphically
A A5.2 The elimination method
A A5.3 The substitution method
A A5.4 Simultaneous linear and quadratic equations
A A5.5 Problems leading to simultaneous equations
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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 – 3
2x + 3y = 23
1
2
Call these equations 1 and 2 .
Substitute equation 1 into equation 2 .
2x + 3(2x – 3) = 23
2x + 6x – 9 = 23
expand the brackets:
8x – 9 = 23
simplify:
8x = 32
add 9 to both sides:
x=4
divide both sides by 8:
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The substitution method
To find the value of y when x = 4 substitute this value into one
of the equations,
1
y = 2x – 3
2x + 3y = 23
2
Substituting x = 4 into 1 gives
y=2×4–3
y=5
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.
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The substitution method
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.
Call these equations 1 and 2 .
Rearrange equation 1 .
add y to both sides:
subtract 9 from both sides:
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3x – y = 9
3x = 9 + y
3x – 9 = y
y = 3x – 9
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The substitution method
3x – y = 9
8x + 5y = 1
1
2
Now substitute y = 3x – 9 into equation 2 .
8x + 5(3x – 9) = 1
8x + 15x – 45 = 1
expand the brackets:
23x – 45 = 1
simplify:
23x = 46
add 45 to both sides:
x=2
divide both sides by 23:
Substitute x = 2 into equation 1 to find the value of y.
3×2–y=9
6–y=9
–y = 3
y = –3
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The substitution method
3x – y = 9
1
8x + 5y = 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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Contents
A5 Simultaneous equations
A A5.1 Solving simultaneous equations graphically
A A5.2 The elimination method
A A5.3 The substitution method
A A5.4 Simultaneous linear and quadratic equations
A A5.5 Problems leading to simultaneous equations
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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,
1
y = x2 + 1
2
y=x+3
Substituting equation 1 into equation 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
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or
x=2
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Simultaneous linear and quadratic equations
We can substitute these values of x into one of the equations
y = x2 + 1
y=x+3
1
2
to find the corresponding values of y.
It is easiest to substitute into equation 2 because it is linear.
When x = –1 we have,
When x = 2 we have,
y = –1 + 3
y=2
y=2+3
y=5
The solutions are x = –1, y = 2 and x = 2, y = 5.
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Using graphs to solve equations
We can also show the
solutions to
y = x2 + 1
y=x+3
using a graph.
10
y = x2 + 1
y=x+3
8
6
(2, 5)
4
(–1,2)
2
The points where the two
graphs intersect give the
solution to the pair of
–4 –3 –2 –1 0
simultaneous equations.
–2
1
2
3
4
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.
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Linear and quadratic graphs
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Simultaneous linear and quadratic equations
Look at this pair of simultaneous equations:
y=x+1
1
x2 + y2 = 13
2
What shape is the graph given by x2 + y2 = 13?
The graph of x2 + y2 = 13 is a circular graph with its centre at
the origin and a radius of √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.
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Simultaneous linear and quadratic equations
y=x+1
x2 + y2 = 13
1
2
Substituting equation 1 into equation 2 ,
x2 + (x + 1)2 = 13
expand the bracket:
simplify:
subtract 13 from both sides:
divide through by 2:
factorize:
x2 + x2 + 2x + 1 = 13
2x2 + 2x + 1 = 13
2x2 + 2x – 12 = 0
x2 + x – 6 = 0
(x + 3)(x – 2) = 0
x = –3
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or
x=2
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Simultaneous linear and quadratic equations
We can substitute these values of x into one of the equations
y=x+1
x2 + y2 = 13
1
2
to find the corresponding values of y.
It is easiest to substitute into equation 1 because it is linear.
When x = –3 we have,
When x = 2 we have,
y = –3 + 1
y = –2
y=2+1
y=3
The solutions are x = –3, y = –2 and x = 2, y = 3.
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Linear and circular graphs
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Contents
A5 Simultaneous equations
A A5.1 Solving simultaneous equations graphically
A A5.2 The elimination method
A A5.3 The substitution method
A A5.4 Simultaneous linear and quadratic equations
A A5.5 Problems leading to simultaneous equations
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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.
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,
a + b = 56
a – b = 22
Adding these equations gives:
2a = 78
a = 39
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Solving problems
Substituting a = 39 into the first equation gives,
39 + b = 56
subtract 39 from both sides:
b = 17
So the two numbers are 39 and 17.
We can check these solutions by substituting them into the
second equation, a – b = 22:
39 – 17 = 22
This is true and so our solution is correct.
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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.
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Solving problems
4a + 3c = 47.50
–
1 – 3 ,
a + 3c = 22
3a
divide both sides by 3:
1
3
= 25.50
a = 8.50
Substitute a = 8.50 in 3 :
8.50 + 3c = 22
subtract 8.50 from both sides:
divide both sides by 3:
3c = 13.50
c = 4.50
The cost of an adult’s ticket is £8.50 and the cost of a child’s
ticket is £4.50.
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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.
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