Topics Covered: Solving Linear Equations Evaluation and Transposition of formulae

Transcription

1 Algebra Workshop: Algebra Topics Covered: Solving Linear Equations Evaluation and Transposition of formulae Solving Linear Equations When we talk about linear equations we are referring to equations that are in terms of one variable. Definition: A linear equation is an equation that can be written in the form ax + b 0 where a and b are real numbers. Linear equations have one root. To find the solution of the linear equation, get the variable involved on one side and all constants on the other side of the equals sign using inverse operations. The solution can be checked b substituting the value back into the original equation and making sure that the right hand side equals the left hand side. Example : Find the value of x in: x 0 - Solution: x x 8 Check: LHS: x RHS the solution x 8 is correct

2 Algebra Example : Find the value of x in the following equation: x + 7 Solution: x + 7 x 7 x x x Check: LHS: x + () + 7 RHS the solution x is correct Example 3: Find the value of x in the following equation: x 36 x x 36 x x x 7 x x 7x -4 7x -4 x Check: LHS: 3 x x 36 RHS: LHS RHS and so x -6 is the correct solution

3 Algebra Example 4: Find the solution to the equation (x ) + x (x + 3) (x + 3) Solution: (x ) + x (x + 3) (x + 3) x + x 0x + x 6 3x 8x x 3x - x x Check: LHS: (x ) + x RHS: (x + 3) (x + 3) Therefore LHS RHS and so x is the correct solution. 3

4 Algebra Questions (Solving linear equations): Solve the following linear equations:. + 3x x x + 7x + 3x 4. (x 3) + 4x x + 3(x ) 3x. ( x + 7) + 9 (Solutions on page 9) Evaluation and Transposition of formulae A formula, in mathematics, gives a relationship between different quantities. When there is more then one, we use the word formulae. When working with formulae, it ma be necessar to single out one of the quantities involved in terms of all the others. This procedure is often referred to as transpose the formula and make that quantit the subject of the equation. The procedures used to transpose a formula to make a certain quantit the subject is the same as those used to re-arrange linear equations. Example : Make x the subject of the formula 3(x + 7) Solution: 3(x + 7) Dividing both sides b 3: x Subtracting 7 from both sides: Therefore x x 3 4

5 Algebra Example : Make g the subject of the formula t π g Solution: t π g Squaring both sides t Multipling both sides b g gt 4π g 4π Dividing both sides b t 4π g t 4π Therefore g t Example 3: Make the subject of the formula a b + c x Solution: a b + c x Subtracting b from both sides a b c x Dividing b c a b c x Squaring both sides a b c x Subtracting x a b x c Multipling through b - a b x c Taking the square root of both sides: ± x a b c Therefore ± x a b c

6 Algebra Example 4: Make x the subject of the formula (x + ) x + Solution: (x + ) x + x appears on both sides so we firstl need to rearrange the formula to get all x terms on (x + ) x + x + x + x x one side. Factorising the LHS x( ) Divide both sides b x Therefore x Example : Rearrange the formula + x + x to make the subject. Solution: + x + x Multipl both sides b + x + ( + x) x( + x) Rearrange to get all terms on one side Factorise the LHS + + x x + x 6 x x x (6 x) x x Divide both sides b 6 x x x 6 x Therefore x x 6 x 6

7 Algebra Questions (Transposition of formulae): In each of the following, transpose the given formula to make the smbol in brackets the subject of the formula.. (w + h) (h);. m k a( x) (x); 3. a + x (x); 4. a(3b ) b + (b);. a 7b 3 + b (b); 6. n L r p (r); (Solutions on page 0) In some cases ou ma be given particular values to evaluate a formula. For example, the area of a circle can be calculated b using the formula: A πr where A is the area and r is the radius. If ou are given that the radius of a circle is 4cm, ou can calculate the area b substituting r 4 into the above formula, i.e. A πr π(4) 6π 0.7cm Example: If v u + at, find t when v 3, u and a 3. Solution: v u + at Substitute in the given values: 3 + 3t Rearrange for t: 3 3t 30 3t t 0 7

8 Algebra Questions (Evaluation of formulae):. If s ½(u + v)t, find s if u.6, v 3. and t.. If V πr h, find r when V 7 and h If t a + (n )d, find n when a.6, d and t If f. If F (Solutions on page ) vu, find v when f 0 and u. v + u m( v - u), find u when F 8, m 0, v 4 and t. t 8

9 Algebra Solutions (Solving linear equations):. + 3x 3 3x x x + 7x + 3x 9x + 0x + 0x 9x 7 x. -7 6x x x x (x 3) + 4x x + 3(x ) x + 4x x + 3x - 6 9x 6 x 6 9x x x 0 0 x 4 x. 3x ( x + 7) + 9 6x ( x + 7 ) + 8 ( x + 7 ) 6x + 8 x + 3 6x + 90 x 6x x x 9

10 Algebra Solutions (Transposition of formulae):. (w + h) 3. w + h w h a + w h x ( x) a( x) + x a ax + ax x a + x(a ) a +. m k a( x) m k a( x) m x k a m x k a x m k a 4. a(3b ) b + 3ab a b + 3ab b + a b(3a ) + a b + a 3a x a + - a. 7b a 3 + b a(3 + b) 7b 3a + ab 7b ab + 7b 3a b(a + 7) 3a 3a b a n n n L 4L r 4L r p p r 4n L p r p 0

11 Algebra Solutions (Evaluation of formulae):. s ½(u + v)t ½( )(.) ½(.8)(.) 7.. V πr h 7 πr (4) r 7 4π r 7 4π places) r.0 (rounded to decimal 3. t a + (n )d (n ).6.6 (n ) F 8 0 (n ) 4 n n m( v - u) t 0( 4 -u) 40 0(4 u) 4 u u 4. f 0 vu v + u v v + 0(v + ) v 0v + 00 v 00 v 0v 00 v v 00