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Example 1. Solve \x = 4.
x = 3.5 0.7 x 5 = 3.5 or 3|
5 x = 5 x 4 Multiply both sides by 5
3jc = 20 3 20
-x = β Divide both sides by 3
x =6.1 20 -r- 3 = 6.7 (to 1 dp) or 6Β§
Example 2. Solve
β + 2 = - The two operations are add 2 and divide by 3
2 β 2 = - β 2 Subtract 2 from both sides of the equation
3 x - = β 1- x 3 Multiply both sides by 3.
Example 3. Solve
3x + 17 = β6 The two operations are add 17 and multiply by 3
LINEAR EQUATION 277
3x T 17 β 17 = β6 β 17 Subtracting 17 from both sides of the equation 3x = -23 3 β23
-x = βDividing both sides of equation by 3 a: = -7.7 β23 = 3 = -7.7 (to 1 dp) or - 7Β§
Example 4. Solve
3x β 5
Solution. The three operations on x, in order to form this equation, are multiply by 3, subtract 5, and divide by 7. In solving the equation this order is reversed, and of course the appropriate inverse operations are applied. Write
(3x 5) _ ^ Inserting brackets shows that all of 3a: β 5 is divided by 7
7(3x - 5)
=4x7 Multiplying both sides of the equation by 7
(3x - 5) = 28
3a: β 5 = 28 The brackets are no longer needed
3a:β5 + 5 = 28+ 5 Adding 5 to both sides of the equation
3a: = 33
-x = β Dividing both sides of the equation by 3
x = 11
Example 5. Solve 6x - 3 = 2(x + 5).
6a: β 3 = 2a: + 10 Expanding the brackets
6x β 3 + 3 = 2a: + 10 + 3 Adding 3 to both sides of the equation
6x = 2a: + 13
6x β 2x = 2a: + 13 β 2a: Subtracting 2a: from both sides to get a:βs on the
left side of equation
4x = 13
4x _ 13 T ~ ~4
x = 3.25
6x β 2x = 4x and 2x β 2x = 0 Dividing both sides by 4
Cross multiplying is a useful way of eliminating fractions in equations, and applies to equations that have one fraction on each side of the equals sign. Cross multiplying is based upon the fact that if | = |, then ad = be.
3.x x β 2 T β 4
3x x 4 = 5(x β 2)
12x = 5x β 10 12x β 5x = 5x β 10 β 5x Ix = -10 Ix -10
Inserting brackets and cross multiplying Expanding the brackets Subtracting 5x from both sides
Dividing both sides by 7 β 10^7 = -1.4 (toi dp) or -1
References: Balancing an Equation, Brackets, Degree, Equations, Inverse Operation, Operations.
A linear graph is a straight-line graph.
Reference: Gradient-Intercept Form.
The liter (abbreviation 1) is a unit of volume and is used for expressing the volume of liquids and capacity in general. For example, in countries where the metric system is used, milk is often sold in 1-liter quantities. There are 1000 milliliters (abbreviation ml) in 1 liter. The size of 1 liter corresponds to a cube of sides 10 cm. This means
that 1 liter = 1000 cubic centimeters (cm3). Other useful equivalent relationships are as follows:
1 milliliter = 1 cubic centimeter
1000 liters = 1 cubic meter
A volume of 1000 liters of water has a mass of 1 tonne, provided the water is at 4Β° C (for the distinction between the unit tonne and the U.S. ton, see the entry Gram). Rainfall of 1 millimeter is equivalent to 1 liter of water per square meter.
References: CGS Units, Metric Units, SI Units.
The plural of locus is loci. A locus is a path made up of a set of points, and the position of each point in the path obeys a certain rale. This concept is demonstrated in the following example (see figure a). Amanda is training her horse, George, for a show. She stands in one place, at the point A, holding the end of a rope, and the other end of the rope is tied to George (G). As George trots around her, Amanda turns so that she is always facing him, but keeps the rope tight and the same length. Different positions of George make up the set of points, and the path of George is the locus of G. In this example the locus of G is a circle. The rale for this locus is that each point G is the same distance from the fixed point A. This locus is made up of an infinite number of positions of George.
Like the circle above, many lines and curves in mathematics can be defined in terms of a locus. Some of them are described here.
The line that is the bisector of an angle can be defined as the locus of points that are equidistant from the βarmsβ of the angle. In figure b, three of the points of the locus, each at an equal distance from either arm, are marked on the angle bisector.
Example. Jo has a dog, Spot, who is occasionally tied up in the garden with the end of his leash free to pass along a length of wire AB which is fixed to the ground at each end. The length of the leash is 2 meters and the length of the wire is 6 meters. Draw the region in which Spot is free to move, assuming the leash does not get tangled up with the wire.
280 LOWEST COMMON MULTIPLE
Solution. This is a locus problem where the set of points covers a region, rather than represents a line. First we draw the locus of Spot at his furthest positions from the wire AB (see figure c). The locus of the furthermost position of Spot is made up of two semicircles each of radius 2 meters, and two straight and parallel line segments each of length 6 meters. The region over which Spot is free to move is shaded.