Transcript Boardworks Algebra 2 powerpoint

Boardworks Maths
Maths
Algebra 2
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Linear simultaneous equations
An equation with two unknowns has an infinite number of
solution pairs. For example:
x+y=3
is true when
x=1
and
y=2
x = –4 and
y=7
x = 6.4 and
y = –3.4
We can represent the set of
solutions graphically.
y
3
The coordinates of every point
on the line satisfy the equation.
0
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and so on.
x+y=3
3
x
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Linear simultaneous equations
Similarly, an infinite number of solution pairs exist for the
equation
y–x=1
Again, we can represent the set of
solutions graphically.
There is one pair of values that
satisfies both these equations
simultaneously.
This pair of values corresponds to
the point where the lines x + y = 3
and y – x = 1 intersect:
y
y–x=1
1
-1 0
x
x+y=3
This is the point (1, 2). At this point x = 1 and y = 2.
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Linear simultaneous equations
Two linear equations with two unknowns, such as x and y, can
be solved simultaneously to give a single pair of solutions.
When will a pair of linear simultaneous
equations have no solutions?
In the case where the lines corresponding to the equations are
parallel, they will never intersect and so there are no solutions.
Linear simultaneous equations can be solved algebraically
using:
The elimination method, or
The substitution method.
The solution to the equations can be illustrated graphically by
finding the points where the two lines representing the
equations intersect.
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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:
Solve the simultaneous equations
3x + 7y = 22 and 3x + 4y = 10.
Subtracting gives:
3x + 7y = 22
– 3x + 4y = 10
3y = 12
y=4
The terms in x have
been eliminated.
Substituting y = 4 into the first equation gives:
3x + 28 = 22
3x = –6
x = –2
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The elimination method
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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
Solve:
1
2
Call these equations 1 and 2 .
Substitute 1 into 2 :
2x + 3(2x – 3) = 23
2x + 6x – 9 = 23
8x – 9 = 23
8x = 32
x=4
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The substitution method
Substituting x = 4 into 1 gives
y=2×4–3
y=5
Check by substituting x = 4 and y = 5 into 2 :
LHS = 2 × 4 + 3 × 5
= 8 + 15
= 23
= RHS
So the solution is x = 4, y = 5.
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