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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 and so on.

We can represent the set of solutions graphically.

0

3

3 x

y

The coordinates of every point on the line satisfy the equation.

x + y = 3

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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:

This is the point (1, 2). At this point x = 1 and y = 2.

0 x

y

-1

1

y – x = 1

x + y = 3

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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.

Linear simultaneous equations can be solved algebraically using:

The substitution method.The elimination method, or

The solution to the equations can be illustrated graphically by finding the points where the two lines representing the equations intersect.

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.

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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:

Subtracting gives: 3x + 7y = 223x + 4y = 10–

3y = 12

The terms in x have been eliminated.

y = 4

Substituting y = 4 into the first equation gives:

3x + 28 = 22

x = –23x = –6

Solve the simultaneous equations3x + 7y = 22 and 3x + 4y = 10.

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The elimination method

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y = 2x – 3

2x + 3y = 23

Solve:

The substitution method

Two simultaneous equations can also be solved by substituting one equation into the other. For example:

Substitute 1 into 2 :2x + 3(2x – 3) = 23

2x + 6x – 9 = 23

8x – 9 = 23

8x = 32

x = 4

1

2

y = 2x – 3

Call these equations 1 and 2 .

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The substitution method

y = 2 × 4 – 3

y = 5

Substituting x = 4 into 1 gives

Check by substituting x = 4 and y = 5 into 2 :

So the solution is x = 4, y = 5.

LHS = 2 × 4 + 3 × 5

= 8 + 15

= 23

= RHS

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