module :ma0001np foundation mathematics lecture week 9

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Module :MA0001NP Foundation Mathematics Lecture Week 9

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Page 1: Module :MA0001NP Foundation Mathematics Lecture Week 9

Module :MA0001NPFoundation Mathematics

Lecture Week 9

Page 2: Module :MA0001NP Foundation Mathematics Lecture Week 9

Transposing formulae&

Simultaneous equations

Page 3: Module :MA0001NP Foundation Mathematics Lecture Week 9

Transposing formulae…What if we have the temperature in °C and what to convert it to °F ?We need to transpose the formula to make F the subject…C = (F – 32)

C = F – 32So, C + 32 = F

Here the subject of the formula is changed from C to F

95

59

59

Page 4: Module :MA0001NP Foundation Mathematics Lecture Week 9

Transposition of Formulae

• Changing the subject of a formula• e.g Formulae to find perimeter of rectangle • P = 2l + 2w• Rearrange the formula to make l the subject• = l + w

• So - w = l

2P

2P

Remember:

Whatever we do to one side we must do to the other side to maintain the equality

Page 5: Module :MA0001NP Foundation Mathematics Lecture Week 9

Transposition of Formulae• Rearrange the formula to make y the subject 3x + 2y = 7 2y = 7 – 3x y = 7 – 3x 2

Subtract 3x from both sides

Divide both sides by 2

Page 6: Module :MA0001NP Foundation Mathematics Lecture Week 9

Transposition of Formulae• Transpose the formula to make the subject

a. C = 2∏r for r b. Y – z = 3(x+2) for x c. T = 2∏√(l/g) for l d. x²+y² = 2 for y e. A² = B² + 2SR for B

Page 7: Module :MA0001NP Foundation Mathematics Lecture Week 9

x32

As the y-values are the same, the right-hand sides of the equations must also be the same.

3 xy52 xy

Two Lines At the point of intersection, we notice that the x-

values on both lines are the same and the y-values are the same.

523 xx

Substituting into one of the original equations, we can find y:3 xy

332 y

311y

The point of intersection is 31132 ,

x 323 xy

52 xy

Simultaneous Equations & Intersections

Page 8: Module :MA0001NP Foundation Mathematics Lecture Week 9

1 quadratic equation and 1 linear equatione.g.

xy 23 )(2

2xy )(1

xx 232 This is a quadratic equation, so we get zero on one side and try to factorise:

0322 xx

031 ))(( xx 31 xx or

To find the y-values, we use the linear equation, which in this example is equation (2)

11231 yyx )(

93233 yyx )(

The points of intersection are (1, 1) and (-3, 9)

Since the y-values are equal we can eliminate y by equating the right hand sides of the equations:

Simultaneous Equations & Intersections

Page 9: Module :MA0001NP Foundation Mathematics Lecture Week 9

13 xy )(2

32 xy )(1e.g.

Sometimes we need to rearrange the linear equation before eliminating y

Rearranging (2) gives 13 xy )2( a

Eliminating y: 1332 xx0432 xx

0)4)(1( xx

1x 4xor

Substituting in (2a): 21 yx134 yx

Simultaneous Equations & Intersections

Page 10: Module :MA0001NP Foundation Mathematics Lecture Week 9

Solving the equations simultaneously will not give any real solutions.

Special Cases

e.g. 1 Consider the following equations:

1 xy )(222 xy )(1

The line and the curve don’t meet.

042 acbThe discriminant

Simultaneous Equations & Intersections

Page 11: Module :MA0001NP Foundation Mathematics Lecture Week 9

e.g. 2

14 xy )(232 xy )(1

Eliminate y: 1432 xx

The discriminant, 0)4)(1(444 22 acb0442 xx

0)2)(2( xx

(twice)2 x

The quadratic equation has equal roots.

The line is a tangent to the curve.

72 yx

0442 xxSolving

Simultaneous Equations & Intersections