Applying algebraic skills to linear equations
Simultaneous equations
I can…
…solve simultaneous equations graphically
…solve simultaneous equations algebraically by substitution
…solve simultaneous equations algebraically by elimination
…use context to create simultaneous equations
An introduction
Reminders
The equation
y = mx + c
can be represented by a straight line
which has a
gradient
of m and passes through the point
(0, c)
Systems of Equations
A System of Equations consists of two (or more) equations with at least two variables.
These are also referred to as
Simultaneous Equations
as their solution holds true
for both equations.
When the System consist of two equations, with two variables, there are three
methods of finding the solution: * by drawing graphs
* by substitution
* by elimination
…solve simultaneous equations graphically
If the lines representing the equations are drawn then the solution
is the coordinates of the point where the lines intersect (meet).
Drawing straight lines
Either: - set x = 0, find the y-coordinate from the formula, then set y = 0 and find x
- pick 2 values for x and find the corresponding values of y
- use the y-intercept, gradient and y = mx + c
Example
2– 2
4
6
8
Solve these equations simultaneously
2– 2
4
6
8
Line two - y = 7 – x
y = ½x + 1 & y = 7 – x
intercept = (0, 7)
Line one - y = ½x + 1
When x = 0, y =
When x = 2, y =
½×0+1 = 1
→ (0, 1)
½×2+1 = 2
→ (2, 2)
y
m=
-1
8
6
4
Draw the two lines with the given information
Lines intersect at (4, 3)
so solution is x =
y=
2
4
3
–2
–2
2
4
6
8 x
…solve simultaneous equations graphically
Simultaneous equations are often used to solve
problems and use letters other than x & y.
Solving simultaneous equations graphically will
often only give approximate solutions and relies on
accurate drawing of graphs.
In order to get precise solutions it is better to
use one of the other methods – substitution or
elimination.
(continued)
…solve simultaneous equations algebraically by substitution
At the point where the lines meet, the values of x and y are
the same
in both equations.
This allows the first equation to be substituted into the second.
Examples
1) y = x + 1
y = 4x – 5
x+1
= 4x – 5
Make x the subject
of the formula
1 = 3x – 5
6 = 3x
x = 2
x = 2, y = 2 + 1 = 3
the solution is
3 = 4×2–5
Replace the y in the second
equation with the first
(2, 3)
Substitute x into the
first equation to find y
Check by substituting into the
second equation, if it is true
the solution is correct
2) y = 4x + 1
2y – 5x + 4 = 0
2 (4x + 1) – 5x + 4 = 0
8x + 2 – 5x + 4 = 0
3x + 6 = 0
3x = –6
x = –2
x = -2,
y = 4 × (–2) + 1 = – 7
the solution is (–2, –7)
2 x (-7) - 5 x (-2) + 4 = 0
…solve simultaneous equations algebraically by substitution
Sometimes it is necessary to rearrange one of the equations first.
Example
3) y – 2x = 3
3y – 2x = 17
y =
3
Rearrange the first equation
2x + 3
(2x + 3) – 2x = 17
6x + 9 – 2x = 17
4x + 9 = 17
4x = 8
x = 2
x = 2, y – 2 × 2 = 3
y–4 = 3
y = 7
the solution is
(2, 7)
3 x 7 – 2 x 2 = 17
Replace the y in the second
equation with the first
Make x the subject
of the formula
Substitute x into the
first equation to find y
Check by substituting into the
second equation, if it is true
the solution is correct
…solve simultaneous equations algebraically by elimination
In this method the equations are added or subtracted so
that one of the variables will be eliminated.
Examples
Place the letters in
the same order
Add/subtract to
remove a letter
1)
2x + 3y = 35
7x – 3y = 1
2x + 3y + 7x + (-3y)
= 9x
35 + 1 = 36
2)
3x + 2y = 7
5x + 2y = 13
3x + 2y – 5x – 2y = –2x
7 - 13 = – 6
So -2x = -6
So 9x = 36
Solve for the
remaining letter
Substitute into
the first equation
to find y
x = 4
x = 3
2 x 4 + 3y = 35
3y = 27
y = 9
the solution is
3 × 3 + 2y = 7
2y = –2
y = –1
(4, 9)
the solution is
(3, –1)
…solve simultaneous equations algebraically by elimination
Sometimes it is necessary to multiply one or both of the
equations before adding or subtracting.
Examples
Place the letters in
the same order
Multiply as required
Add/subtract to
remove a letter
3) 3x + 4y = 26
6x – y = 7
4) 6x + 2y = 38
2x – 3y = 20
4(6x – y) = 4 x 7
3(2x – 3y) = 3 x 20
24x – 4y = 28
6x – 9y = 60
3x + 4y + 24x + (-4y)
= 27x
26 + 28 = 54
6x + 2y - 6x – (-9y)
= 2y + 9y = 11y
38 – 60 = -22
So 27x = 54
So 11y = -22
Solve for the
remaining letter
x= 2
y = -2
Substitute into
the first equation
to find y
3 x 2 + 4y = 26
6x + 2 × (–2) = 38
4y = 20
y = 5
6x = 42
x = 7
the solution is (7, –2)
the solution is (2, 5)
…solve simultaneous equations algebraically by elimination
Example
Place the letters in
the same order
Multiply as required
5) 3x + 2y = 7
4x + 3y = 9
3(3x + 2y) = 3 x 7
9x + 6y = 21
Add/subtract to
remove a letter
9x + 6y - 8x – 6y = x
21 – 18 = 3
So x = 3
Substitute into
the first equation
to find y
3 × 3 + 2y = 7
2y = –2
y = –1
the solution is (3, –1)
2(4x + 3y) = 2 x 9
8x + 6y = 18
…use context to create simultaneous equations
Examples
1) A jug and two glasses hold 1·6 litres altogether.
Two jugs and three glasses hold 2·9 litres altogether.
How much does each hold?
Choose appropriate
letters
j + 2g = 1·6
2j + 3g = 2·9
Multiply as required
2(j + 2g) = 2 x 1.6
2j + 4g = 3·2
Add/subtract to
remove a letter
Substitute into
the first equation
to find y
2j + 4g – 2j – 3g = g
3.2 – 2.9 = 0·3
So g = 0.3
j + 2 × 0·3 = 1·6
j + 0.6 = 1.6
j=1
So a Jug holds 1 litre and
a glass holds 0·3 litres
Some problems can be solved
using simultaneous equations
by turning the problem into a
set of equations.
If answering a question set in
a particular context you must
write your final answer in
context.
Applying algebraic skills to linear equations
Simultaneous equations
I can…
…solve simultaneous equations graphically
…solve simultaneous equations algebraically by substitution
…solve simultaneous equations algebraically by elimination
…use context to create simultaneous equations
An introduction
Reminders
The equation
can be represented by a straight line
which has a
of m and passes through the point
Systems of Equations
A System of Equations consists of two (or more) equations with at least two variables.
These are also referred to as
as their solution holds true
for both equations.
When the System consist of two equations, with two variables, there are three
methods of finding the solution: * by drawing graphs
* by substitution
* by elimination
…solve simultaneous equations graphically
If the lines representing the equations are drawn then the solution
is the coordinates of the point where the lines intersect (meet).
Drawing straight lines
Either: - set x = 0, find the y-coordinate from the formula, then set y = 0 and find x
- pick 2 values for x and find the corresponding values of y
- use the y-intercept, gradient and y = mx + c
Example
2– 2
4
6
8
Solve these equations simultaneously
2– 2
4
6
8
Line two - y = 7 – x
y = ½x + 1 & y = 7 – x
intercept =
Line one - y = ½x + 1
When x = 0, y =
→ (0, 1)
y
m=
8
When x = 2, y =
→ (2, 2)
6
4
Draw the two lines with the given information
Lines intersect at
so solution is x =
y=
2
–2
–2
2
4
6
8 x
…solve simultaneous equations graphically
Simultaneous equations are often used to solve
problems and use letters other than x & y.
Solving simultaneous equations graphically will
often only give approximate solutions and relies on
accurate drawing of graphs.
In order to get precise solutions it is better to
use one of the other methods – substitution or
elimination.
(continued)
…solve simultaneous equations algebraically by substitution
At the point where the lines meet, the values of x and y are
in both equations.
This allows the first equation to be substituted into the second.
Examples
1) y = x + 1
y = 4x – 5
= 4x – 5
Replace the y in the second
equation with the first
2) y = 4x + 1
2y – 5x + 4 = 0
2
– 5x + 4 = 0
Make x the subject
of the formula
x = 2, y =
the solution is
Substitute x into the
first equation to find y
Check by substituting into the
second equation, if it is true
the solution is correct
x = -2,
y =
the solution is
…solve simultaneous equations algebraically by substitution
Sometimes it is necessary to rearrange one of the equations first.
Example
3) y – 2x = 3
3y – 2x = 17
Rearrange the first equation
y =
3
– 2x = 17
Replace the y in the second
equation with the first
Make x the subject
of the formula
x = 2,
Substitute x into the
first equation to find y
the solution is
Check by substituting into the
second equation, if it is true
the solution is correct
…solve simultaneous equations algebraically by elimination
In this method the equations are added or subtracted so
that one of the variables will be eliminated.
Examples
Place the letters in
the same order
1)
2x + 3y = 35
7x – 3y = 1
2)
3x + 2y = 7
5x + 2y = 13
Add/subtract to
remove a letter
Solve for the
remaining letter
Substitute into
the first equation
to find y
2
+ 3y = 35
the solution is
3
+ 2y = 7
the solution is
…solve simultaneous equations algebraically by elimination
Sometimes it is necessary to multiply one or both of the
equations before adding or subtracting.
Examples
Place the letters in
the same order
Multiply as required
3) 3x + 4y = 26
6x – y = 7
4(6x – y) = 4 x 7
4) 6x + 2y = 38
2x – 3y = 20
3(2x – 3y) = 3 x 20
Add/subtract to
remove a letter
Solve for the
remaining letter
Substitute into
the first equation
to find y
3
+ 4y = 26
the solution is
6x + 2
the solution is
= 38
…solve simultaneous equations algebraically by elimination
Example
Place the letters in
the same order
Multiply as required
5) 3x + 2y = 7
4x + 3y = 9
3(3x + 2y) = 3 x 7
Add/subtract to
remove a letter
Substitute into
the first equation
to find y
3
+ 2y = 7
the solution is
2(4x + 3y) = 2 x 9
…use context to create simultaneous equations
Examples
1) A jug and two glasses hold 1·6 litres altogether.
Two jugs and three glasses hold 2·9 litres altogether.
How much does each hold?
Choose appropriate
letters
Multiply as required
2(j + 2g) = 2 x 1.6
If answering a question set in
a particular context you must
write your final answer in
context.
Add/subtract to
remove a letter
Substitute into
the first equation
to find y
Some problems can be solved
using simultaneous equations
by turning the problem into a
set of equations.
j+2×
= 1·6
So a Jug holds 1 litre and
a glass holds 0·3 litres