STRAND: ALGEBRA Unit 3 Solving Equations

Transcription

1 CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet STRAND: ALGEBRA Unit Solving Equations TEXT Contents Section. Algebraic Fractions. Algebraic Fractions and Quadratic Equations. Algebraic of Simultaneous Equations One Linear and One Quadratic

2 CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Solving Equations. Algebraic Fractions Some algebraic fractions can be simplified if the numerator and denominator are factorised. Worked Eample Simplify Both the top and bottom of the fraction can be factorised to give ( ) ( ) ( ) ( ) The term appears as a factor on both the top and bottom of the fraction so it can be cancelled to give ( ) ( ) No further simplification is possible. Worked Eample Simplify ( ) First multiply together the fractions to give The bracket can be factorised using the difference of two squares to give ( ) ( ) ( ) ( ) Dividing numerator and denominator by common factors gives: ( ) No further simplification is possible.

3 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Worked Eample Simplify Recall from working with fractions that So a b c a d d b c. ( ) ( ) ( ) Worked Eample Simplify Algebraic fractions are added in the same way as ordinary fractions, by using the lowest ( ) common denominator. In this case it will be. ( ) ( ) No further simplification is possible. ( ) ( ) ( ) ( )

4 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Eercises. Simplify each of the following epressions. 6 (b) (c) ( ) ( ) ( ) ( ) (d) ( ) ( ) ( ) ( ) (e) 6 6 (f) ( ) ( ) (g) ( ) ( ) ( ) (h) ( ) ( ) ( ) ( ) ( ) (i) ( ) ( ) ( ) ( ) ( ) ( ). Simplify each of the following epressions. (b) (c) 6 (d) (e) (f) 6 (g) (h) (i) 9 (j) (k) 0 6 (l) 6 8 (m) 6 (n) 9 8 (o) 0 (p) 0 8 (q) 0 (r) 9 (s) 0 (t) (u) 9 6. Simplify each of the following epressions. (b) ( ) (c) 6 (d)

5 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet (e) (f) (g) 6 6 (h) (i) 6 8. Epress each of the following as a single fraction, simplifying if possible. (b) (c) (d) (e) 9 (f) 6 (g) (h) (i) 6 (j) 6 (k) (l) (m) 6 (n) (o). Simplify: (b) ab a b ab. Algebraic Fractions and Quadratic Equations This section deals with finding the solutions of equations such as 6 and uses techniques from the earlier sections. Worked Eample Find the solution of 6

6 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet The first stage is to combine the two fractions so that they have a common denominator. ( ) ( ) ( ) 6 ( ) Then dividing both sides by gives 6 ( ) ( ) Then multiplying both sides by and 6 gives Therefore and ( ) and dividing by 6 0 This factorises as so ( ) ( 6 ) 0 0 or 6 0 and or Worked Eample Solve the equation The first step is to combine the two fractions using the ccmmon denominator ( ). 6 That is,

7 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Then multiplying by and ( ) gives This quadratic can be solved by using the formula (or by completing the square) to give where a, b and c 6, giving b ± b ac a ± 9 6 ± 9 8 ± 9 9 or 9 Eercises Solve each of the following equations

8 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Algebraic of Simultaneous Equations One Linear and One Quadratic In Section. you solved simultaneous equations where both of the equations were linear. In this section we etend this to solving simultaneous equations where one equation is linear and the other is quadratic. This will normally give you a quadratic equation to solve. Worked Eample Solve the simultaneous equations y y () () Substituting y from equation () into equation () gives This equation simplifies to 0 so 0 6 We now solve this quadratic equation by factorisation. ( ) 0 so 0 or 0 therefore or These values of are now substituted into one of the original equations to find the corresponding values of y. It is usually easiest to use the linear equation for this. Substituting the first solution into equation () gives y 8. Substituting the second solution into equation () gives y. The solutions are, y 8 and, y.

9 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Note In Worked Eample, the first equation, y, can be represented by a straight line graph. The second equation, y, can be represented by a quadratic curve. When we solve this pair of simultaneous equations, we are finding the coordinates of the two points where the line and the curve intersect. This is shown on the graph, the intersection points being (, 8) and (, ). y 8 y y Worked Eample Solve the simultaneous equations Subtract equation () from equation (). y () y () - - y () y () This equation simplifies to 0 () () 0 0 This quadratic does not factorise, so we need to use the quadratic formula b b ac ± a 8

10 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet to find the two possible values of. Here a, b and c So ± ( ) ( ) ± 8 6 ± 88 6 so (to d.p.) or (to d.p.) Substituting the un-rounded values for into the linear equation () gives y (to d.p.) and y (to d.p.). The solutions are 90., y 69. and., y 0. (to d.p.). Worked Eample Solve the simultaneous equations Subtract equation () from equation (). This equation simplifies to y () y () y () y () ( ) 0 () () 0 so hence ± These values for are now substituted into the linear equation. Substituting the first solution,, into equation () gives y. Substituting the second solution,, into equation () gives y. The solutions are, y and, y. 9

11 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Worked Eample Solve algebraically the simultaneous equations y 0 () y () (b) Write down the coordinates of the points of intersection of the circle y 0 and the straight line y. We use the linear equation () to substitute for y into equation (), the equation for the circle. The square of equation () is y ( ) so y Substituting into equation () now gives 0 which simplifies to 6 0 which, dividing by, simplifies again to 0 We can now solve this equation for by factorising ( ) ( ) 0 so or When, equation () gives y When, equation () gives y Therefore the solutions of the two equations are, y and, y (b) The points of intersection of the circle y 0 and the straight line y are, and,. 0

12 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet Eercises. Solve the simultaneous equations y (b) y (c) y y y 0 y. Solve the following simultaneous equations, giving the values of and y correct to decimal places. y (b) t 8 (c) y y 6t y. Solve the simultaneous equations y (b) y 9 (c) y 8 8y 8 8y 0 y. Copy and complete the table below. 0 y 9 (b) Draw the graph of y 9 for the domain, using a scale of cm to unit on the horizontal ais and cm to unit on the vertical ais. (c) Draw the graph y 9. (d) On the same graph, draw the straight line y 6. (e) Write down the coordinates of the points where the line y 6 intersects the curve y 9. (f) (g) Use an algebraic method to solve the simultaneous equations y 9 y 6 Use the solutions found in part (f) to confirm your answer to part (e).. Solve the following pairs of simultaneous equations. Where appropriate, give your solutions correct to decimal places. y (b) y (c) y 8 y y y (d) y 68 (e) y (f) y 0 y y y

13 . CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet 6. Find the coordinates of the points where the line y intersects the circle y.. Draw the centrally-placed - and y-aes and scale them using the ranges, y. (b) Accurately construct the locus y 00. y R (c) On the same graph, draw the line y. (d) Write down the coordinates of the points P and Q (as labelled in the diagram) where the -ais meets the circle. PQ is a diameter of the circle. Q P (e) By solving the simultaneous equations y 00 y find the - and y-coordinates of the point, R, where the line meets the circle in the first quadrant. (f) (g) (h) (i) Calculate the area of Δ PQR. Show that the gradient of the line segment QR is 0. Find the gradient of the line segment PR. Show that PR is perpendicular to QR. N.B This illustrates the fact that the angle in a semicircle is a right angle. Information Both linear and quadratic equations have been used for over four thousand years. The early Chinese and Babylonians made use of equations to solve daily problems such as the sharing of an inheritance.