Download Simultaneous linear equations Elimination method

List six positive integer solutions
for each of these equations and
comment on your results. Two
have been done for you.
x – y = 2
x + 2y = 14
x = 2, y = 0
x = 0, y = 7
x = 3, y = 1
x = 2, y = 6
x = 4, y = 2
x = 4, y = 5
x = 5, y = 3
x = 6, y = 4
x = 6, y = 4
x = 8, y = 3
x = 7, y = 5
x = 10, y = 2
© teachitmaths.co.uk 2013
We can see this clearly
on a graph:
If we considered all
solutions, not just
positive integers, there
would be an infinite
number of answers.
However, there is one
solution which is
correct for both of these
equations
simultaneously.
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x – y = 2
x + 2y = 14
x = 2, y = 0
x = 0, y = 7
x = 3, y = 1
x = 2, y = 6
x = 4, y = 2
x = 4, y = 5
x = 5, y = 3
x = 6, y = 4
x = 6, y = 4
x = 8, y = 3
x = 7, y = 5
x = 10, y = 2
© teachitmaths.co.uk 2013
Elimination method
Simultaneous linear equations
Comparing the coefficients between each equation tells us
how difficult the simultaneous equations will be to solve.
Solving two simultaneous equations involves finding the
unique solution which satisfies both of them.
6x + y = 20
3x + y = 11
There are two algebraic methods you could use:
Elimination method
Substitution method
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go
3
Elimination method – beginner
3x
2
= 9
x = 3
3 × 3 + y = 11
y = 2
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1. Label the equations.
2. Compare the coefficients – both y
terms have a coefficient of 1.
3. Compare the signs – both y terms
are positive; subtract 2 from 1 to
eliminate y.
2
4. Solve for x.
Why should we
subtract, not add?
5. With an original equation,
substitute x and solve for y.
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Beginner
Matching coefficients
go
3p + 4q = 24
p + 5q = 19
Intermediate
One coefficient is a multiple of
the other
go
3x + 8y = 32
2x + 7y = 23
Expert
Coefficients are not multiples
go
go
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6x + y = 20
3x + y = 11
2
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Elimination method – beginner
6x + y = 20
3x + y = 11
3x
2
You can check your solution by
substituting the values into both
solutions:
6x + y = 20
6 × 3 + 2 = 20
= 9
x = 3
3 × 3 + y = 11
y = 2
© teachitmaths.co.uk 2013
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3x + y = 11
3 × 3 + 2 = 11
2
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2
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Elimination method – beginner
3p + 2q = 19
3p + 6q = 27
1
2
4q = 8
q = 2
3p + 2 × 2 = 19
3p = 15
p = 5
2
1. Label the equations.
2. Compare the coefficients – both p
terms have a coefficient of 3.
3. Compare the signs – both p terms
are positive; subtract 1 from 2 to
eliminate p.
Why is this easier
than subtracting
4. Solve for q.
2 from 1 ?
5. With an original equation,
substitute q and solve for p.
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Elimination method – beginner
Solve these pairs of simultaneous equations:
x + 2y = 30
x – 2y = 14
4x + y = 23
3x + y = 18
x = 22, y = 4
x = 5, y = 3
Elimination method – beginner
2x + y = 21
x – y = 6
3x
1
5x + 4y = 23
12x – 4y = 28
1
17x
2
3p + 4q = 24
3p + 15q = 57
1
2
2
1. Label the equations.
2. Compare the coefficients – no
coefficients match.
3. Multiply all terms in 2 by 3, to
match the p coefficients.
Why is it easier to
eliminate p instead of q?
4. Compare the signs – signs match;
subtract.
2
5. Solve for q.
6. Substitute q and solve for p.
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Elimination method – intermediate
Solve these pairs of simultaneous equations:
2. Compare the coefficients – no
coefficients match.
3. Multiply all terms in 2 by 4, to
match the y coefficients.
4. Compare the signs – signs are
different; add.
= 51
x = 3
3 × 3 – y = 7
y = 2
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1. Label the equations.
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p + 5 × 3 = 19
p = 4
5x + 4y = 23
3x – y = 7
Why should we
add?
4. Solve for x.
5. Substitute x into an original
equation and solve for y.
3p + 4q = 24
p + 5q = 19
k = 3, s = 2
Elimination method – intermediate
2. Compare the coefficients – both y
terms have a coefficient of 1.
Elimination method – intermediate
x = 3, y = 5
9
1
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8k – s = 22
k – s = 1
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1. Label the equations.
3. Compare the signs – each y sign
is different; add 1 and 2 to
eliminate y.
2 × 9 + y = 21
y = 3
2x + y = 11
3x + y = 14
© teachitmaths.co.uk 2013
2
= 27
x = 9
11q = 33
q = 3
Click to show answers
1
5. Solve for x.
2
6. Substitute x and solve for y.
3x + 7y = 47
2x – y = 3
4x + 9y = 13
3x + y = 4
x = 4, y = 5
x = 1, y = 1
3x + y = 39
5x + 7y = 113
4s + 3k = 48
2s + 8k = 26
x = 10, y = 9
k = 1, s = 9
Click to show answers
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Elimination method – expert
3x + 8y = 32
2x + 7y = 23
1. Label the equations.
1
2. Compare the coefficients – no
coefficients match.
2
6x + 16y = 64
6x + 21y = 69
3. Multiply 1 by 2 and 2 by 3, to
match the x coefficients.
1
2
We could have chosen to make the
y coefficients match instead – how
would you do this?
5y = 5
y = 1
2x + 7 = 23
x = 8 Elimination method – expert
11x + 9y = 42
2x + 6y = 12
1
22x + 18y = 84
6x + 18y = 36
1
16x
4. Compare the signs – signs match;
subtract.
2
5. Solve for y.
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3. Multiply 1 by 2 and 2 by 3, to
match the y coefficients.
2
Why did we multiply by
2 and 3 instead of 9
and 6?
= 48
x = 3
2 × 3 + 6y = 12
y = 1 6. Substitute y and solve for x.
1. Label the equations.
2. Compare the coefficients – no
coefficients match.
4. Compare the signs – signs match;
subtract.
2
6. Substitute x and solve for y.
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Elimination method – expert
5. Solve for x.
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Substitution method
If one variable can be expressed easily in terms of the
other, you may prefer to use the substitution method.
Solve these pairs of simultaneous equations:
9x + 3y = 30
6x – 2y = 16
7x + 2y = 23
5x + 3y = 29
x = 3, y = 1
x = 1, y = 8
4x + 9y = 35
3x + 7y = 27
8s + 4k = ‐16
3s + 6k = 3
x = 2, y = 3
k = 2, s = ‐3
6x + y = 19
4x + 3y = 15
y = 19 – 6x
4x + 3y = 35
7x + 5y = 59
y = ⅓(35 – 4x)
suitable for the substitution method
will be tricky with the substitution
method
Click to show answers
© teachitmaths.co.uk 2013
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© teachitmaths.co.uk 2013
Substitution method
3x + 2y = 18
2x – y = 5
y = 2x – 5
3x + 2(2x – 5) = 18
3x + 4x – 10 = 18
x = 4
y = 2 × 4 – 5
y = 3 © teachitmaths.co.uk 2013
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Substitution method
x + 3y = 19
3x – 2y = 2
1. Label the equations.
2
2. Rewrite 2
to give y in terms of x.
2
1
2
3. Substitute y into
1
4. Solve for x.
Remember to use
brackets and watch
out for signs!
.
5. Substitute x into an original
equation and solve for y.
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x = 19 – 3y
3(19 – 3y) – 2y = 2
57 – 9y – 2y = 2
y = 5
x = 19 – 3 × 5
x = 4 17
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1
1. Label the equations.
2
2. Rewrite 1
to give x in terms of y.
1
2
3. Substitute x into
2
.
4. Solve for y.
1
5. Substitute y into an original
equation and solve for x.
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Substitution method
Solve these pairs of simultaneous equations using the
substitution method:
3x + 5y = 19
2x – y = 4
4x + 9y = 13
3x + y = 4
x = 3, y = 2
x = 1, y = 1
3x + y = 13
5x + 7y = 27
s + 4k = 6
5s + 3k = 47
x = 4, y = 1
k = ‐1, s = 10
Click to show answers
© teachitmaths.co.uk 2013
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