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SIMULTANEOUS EQUATIONS
Problem of the Day!
There are 100 animals in a zoo, some which have 2 legs and some have 4 legs. If there are 262 animal legs altogether, how many 4 legged animals are there?
Answer:31 four legged and69 2 legged!
Please note: This Powerpoint has been uploaded to
the wiki: http://mazenod-10max.wikispaces.com
Solving Simultaneous Equations
Q. What does the word “simultaneous” mean?
A. “at the same time”
To solve simultaneous equations you need to have the same number of equations as you have variables (unknowns).
Here we will consider cases where we have two equations and two unknowns eg x + y = 3 and 2x – y = 5
Substitution MethodThis method is generally used in situations where one of the variables (say y) is the subject or can easily be made the subject of one (or both) of the equations.
Eg 1: y = 3x + 1 and y = 5 – 4x (y is the subject in both)
Eg 2: y = 2x - 1 and 2x + 3y = 5 (y is the subject of equation 1)
Eg 3: y + 2x = 1 and 2x + 3y = 5 (y can easily be made the subject in equation 1)
Eg 4: 3y = 2x - 1 and x + 3y = 5 (x can easily be made the subject of equation 2)
Substitution MethodIn this method we will make use of the mathematical truth:
“If things equal the same thing then they must equal each other”
ie If a = b and a = c then b = c
Example:
if y = x+1 and y = p
then x+1 = p
ie we can swap (substitute) x+1 for y in equation 2 or…
Substitution Method
Example
3 yx115 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
Step 2 Make one of the variables (we will use y) the subject of one of the equations (we will use equation 1)
From (1) xy 3 (1’ )
Step 3
Step 4
Substitute 3 – x for y
in equation (2)
xy 3
115 yx (2)
11)3(5 xx
Solve the equation
11515 xx11415 x
44 x1xStep 5
Substitute this value back into (1’ ) 2
13
y
(1’ )
Therefore the solution is (1, 2)
What does the solution (1,2) mean? It means that the point (1,2) is the
only solution that satisfies both equations.
x + y = 3 1 + 2 = 3 x + 5y = 11 1 + 5(2) = 11 Can you interpret this graphically?
Another example
934 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
Step 2 Equation (2) already has x the subject
12 yx
Step 3
Step 4
Substitute 2y + 1 for x in equation 1
(2)
93)12(4 yy
Solve the equation
9348 yy945 y55 y1yStep 5
Substitute this value back into (2 ) 3
1)1(2
x
Therefore the solution is (3, 1)
934 yx (1)
12 yx
A more difficult example
2321245 yxyx and
Solve the simultaneous equations:
To get started: Which variable in which equation is the easiest to make the subject?
I will choose to make x the subject in equation 2
)2.......(232 and)1.....(1245 yxyx
Substitute this for x in equation 1 124
2
325
yy
10247 y2y
Therefore the solution is (–4, –2)
2
32 yx
2481510 yy
Substitute y = -2 into (2’) 2
)2(32 x
4x
Making x the subject in equation 2 gives:
Solve for y
Elimination Method
This method is best used when you have equations in the form:
3 yx115 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
Elimination Method
Best used when you have equations in the form:
3 yx115 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
3 yx115 yx
Which variable has the same coefficient?
x
As the sign of x is the same, we subtract equation (2) from equation (1)
Step 2
(1)
(2)Step 3
840 y84 y
2yStep 4x is eliminated, solve for y
Step 5Substitute the value for y into equation (1) and solve for x
32 x1x
Therefore the solution is (1, 2)
What does the solution mean ?
We had two equations and two unknowns, x and y
3 yx 115 yxand
The solution (1, 2) gives the values of x and y that will make both sentences true.
Another example
832 yx143 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
Which variable has the same coefficient?
y
Step 2
As the sign of the 3y is different, we add equation (2) to equation (1)
(1)
(2)Step 3
603 x63 x2x
Step 4y is eliminated, solve for x
Step 5Substitute the value for x into equation (1) and solve for y
83)2(2 y123 y
Therefore the solution is (-2, 4)
832 yx143 yx
4y
A more difficult example
1032 yx265 yx
Label the equations (1) and (2)
(1)
(2)
Step 1
Which variable has the same coefficient? Neither
Step 2
We can make x have the same coefficient if we multiply equation (2) by 2
As the sign is the same, we subtract equation (2’) from equation (1)
(1)
(2’)Step 3
427 y6yStep 4
x is eliminated, solve for y
Step 5Substitute the value for y into equation (2) and solve for x
26)6(5 x2630 x
Therefore the solution is (-4, 6)
4x
1032 yx52102 yx
Try questions 1 to 4 on the worksheet
A more difficult example 2321245 yxyx and
Using elimination
Multiply (1) by 3 and (2) by 4
361215 yx (1’)
1245 yx (1)(2)232 yx
8128 yx (2’)
Subtract (2’) from (1’)
2807 x4x
Substitute into (1)
124)4(5 y20124 y
2y
Therefore the solution is (-4, -2)
2321245 yxyx andUsing substitution
From (2)
1245 yx (1)(2)232 yx
Substitute (2’) into (1)
1242
325
yy
10247 y2y
Therefore the solution is (–4, –2)
2
32 yx
2481510 yy
Substitute y = -2 into (2’)
2
)2(32 x
4x
(2’)
A more difficult example 2321245 yxyx and
Using elimination
Multiply (1) by 3 and (2) by 4
361215 yx (1’)
1245 yx (1)(2)232 yx
8128 yx (2’)
Subtract (2’) from (1’)
2807 x4x
Substitute into (1)
124)4(5 y20124 y
2y
Therefore the solution is (-4, -2)
2321245 yxyx andUsing substitution
From (2)
1245 yx (1)(2)232 yx
Substitute (2’) into (1)
1242
325
yy
10247 y2y
Therefore the solution is (–4, –2)
2
32 yx
2481510 yy
Substitute y = -2 into (2’)
2
)2(32 x
4x
(2’)
Step 3
Step 4
Substitute the new equation into the other equation in this case equation (2)
xy 3
115 yx (2)
11)3(5 xx
Solve the equation
11515 xx11415 x
44 x1xStep 5
Substitute this value back into (1’ ) 2
13
y
(1’ )
Therefore the solution is (1, 2)
Step 3
Step 4
Substitute equation (2) into equation (1)
(2)
93)12(4 yy
Solve the equation
9348 yy945 y55 y1yStep 5
Substitute this value back into (2 ) 3
1)1(2
x
Therefore the solution is (3, 1)
934 yx (1)
1 2 y x
Complete the questions on the worksheet
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