stroud worked examples and exercises are in the text programme f5 linear equations and simultaneous...

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Worked examples and exercises are in the text

PROGRAMME F5

LINEAR EQUATIONS and

SIMULTANEOUS LINEAR EQUATIONS

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Linear equations

Simultaneous linear equations with two unknowns

Simultaneous linear equations with three unknowns

Programme F5: Linear equations and simultaneous linear equations

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Linear equations

Solution of simple equations


A linear equation in a single variable (unknown) x is an equation of the form below (on the left), where the a, b, c, d are expressions not involving x (J.A.B.). A linear equation in one unknown is also referred to as a simple equation.

The solution of a simple equation consists essentially of moving the x occurrences to one side:

giving and hence

ax b cx d ax cx d b

d bxa c

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Linear equations




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Solution by substitution

Solution by equating coefficients


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Solution by substitution


A single linear equation in two variables has an infinite number of solutions. For two such equations there may be just one pair of x- and y-values that satisfy both simultaneously.

The substitution solution method = (1) solve one of the equations for one of the variables, and then (2) replace that variable in the other equation by the result of (1).

( ) 5 2 145( ) 3 4 24 from ( ): 5 2 14 2 14 5 72

a x yxb x y a x y y x y

5in ( ) 3 4 7 24 4

2

in ( ) 5(4) 2 14 3

xb x x

a y y

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Solution by equating coefficients


Example:

Multiply (a) by 3 (the coefficient of y in (b)) and multiply (b) by 2 (the coefficient of y in (a))

( ) 3 2 16( ) 4 3 10 a x yb x y

( ) 3 9 6 48( ) 2 8 6 20 add together to give 17 68 4

Substitute in ( ) to give 3(4) 2 16 2

a x yb x y x x

a y y

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Linear equations




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With three unknowns and three equations the method of solution is just an extension of the work with two unknowns.

By equating the coefficients of one of the variables (OR by using the substitution method – J.A.B.) that variable can be eliminated from one pair of equations, giving you one equation in the remaining two variables. The same thing can be done with a different pair. You now have two equations in two unknowns. These can be solved in the usual manner, and the value of the first variable above evaluated by substitution.

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Simultaneous linear equations with three unknowns (x,y,z), contd

(added by John Barnden)

METHOD in more detail:

There will now be THREE EQUATIONS, not two, each involving ALL THREE VARIABLES.

Take any two of the equations. Eliminate one of the variables, say z , using either the substitution method or the equating-coefficients method. So you now have created a new equation, involving only x and y.

Do the same with a different pair of the original equations. So you now have another new equation, involving only x and y.

Solve your two new equations for x and y, using the method for two variables.

Now use any one of the original equations to find the value of z.

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EXAMPLE (from textbook p.193, to be worked through in class):

3x + 2y – z = 19

4x – y + 2z = 4

2x + 4y – 5z = 32

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EXERCISE:

How would you proceed if

(a) only two of the equations used all three of x, y, z, the other equation using only one of them

(b) only one of the equations used all three of x, y, z, the other two equations using only x and y

(c) one of the equations used all three of x, y, z, another uses just x and y, and the third uses just y and z.

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Simultaneous linear equations

Pre-simplification


Sometimes, the given equations need to be simplified before the method of solution can be carried out. For example, to solve:

Simplification yields:

2( 2 ) 3(3 ) 384(3 2 ) 3( 5 ) 8

x y x yx y x y

11 389 7 8

x yx y

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Back to Simultaneous linear equations in TWO unknowns

STRAIGHT LINE VIEW (ADDED by John Barnden)STRAIGHT LINE VIEW (ADDED by John Barnden)

A linear equation in two variables can be viewed as defining a straight line.Suppose we have

Dx + Ey = F

Solve for y, to get:

y = – (D/E)x + F/E

But when you graph this you get a straight line …..

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Equation of a straight line (REMINDER)

Programme 8: Differentiation applications 1

(NB: from a different part of the book)

The basic equation of a straight line is:

where:

y mx c

gradient

intercept on the -axis

dym

dxc y

When m is negative, the line slopes downwards. (ADDED by JAB)

m is also called the slope.

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STRAIGHT LINE VIEW, contdSTRAIGHT LINE VIEW, contd

So if we have two linear equations in x and y:

Dx + Ey = FGx + Hy = J

(where neither E nor is H is zero) we get two straight lines:

y = – (D/E)x + F/Ey = – (G/H)x + J/H

And their intersection is at the x and y values that solve the original equations (Why?).

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If we have two (non-vertical) straight lines:

y = mx + cy = nx + d

We can find the x value at the intersection by solving those simultaneous equations, most easily by eliminating y:

mx + c = nx + d

So:

x = (d-c) / / (m-n) [provided m and n are unequal]

Now we can also find the y value, from either of the equations.

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We had the following earlier:Dx + Ey = FGx + Hy = J

From which we got:y = – (D/E)x + F/Ey = – (G/H)x + J/H

So “m” is –D/E and “n” is –G/H. So we can solve if and only if

D/E G/H, i.e.DH – EG 0

DH – EG is called the determinant of the original pair of equations. So you can test whether or not the equations can be solved by seeing whether or not the determinant is non-zero. (See textbook Prog 4 for optional extra material on determinants.)

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Simultaneous equations need not be linear!


Could for instance have a linear equation and an equation involving squares:

Dx + Ey = F

Ax2 + Bxy + C y = J

Here we could use the substitution method. Solve the first equation for y and then use the answer in the second equation. What sort of equation do you now have? (EXERCISE)

stroud worked examples and exercises are in the text programme f5 linear equations and simultaneous...

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