Download Simultaneous equations

Simultaneous equations
1. The method of elimination
2. The method of substitution
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The elimination method – Example 1
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 – Example 1 continued
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 – Example 1 continued
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 – Example 2
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 – Example 2 continued
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 – Practice 1
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The elimination method – Example 3
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 - Example 3 continued
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 = 19
21 – 2 = 19
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The elimination method - Example 4
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 – Practice 2
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The substitution method – Example 1
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 – Example 1 continued
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 – Example 2
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 – Example 2 continued
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 – Example 2 continued
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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