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Simultaneous equations

Sketching straight lines Solving simultaneous equations by straight line graphs Solving simultaneous equations by substitution Solving simultaneous equations by elimination Knowing two points which lie on a line find the equation of the line Using simultaneous equations to solve problems

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Simultaneous equations

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  1. Sketching straight lines • Solving simultaneous equations by straight line graphs • Solving simultaneous equations by substitution • Solving simultaneous equations by elimination • Knowing two points which lie on a line find the equation of the line • Using simultaneous equations to solve problems • Simultaneous equation by substitution • Problems from credit past papers Simultaneous equations

  2. Sketching straight lines Equations of the type y = x ; y = x + 3 ; y = 4x - 5; x + y = 2 are equations of straight lines . The general equation of a straight line is y = ax =b By finding the coordinates of some points which lie on the straight lines, plotting them and joining them up the graph of the straight line can be drawn.

  3. Table of values To draw the graph of a straight line we must find the coordinates of some points which lie on the line.We do this by forming a table of values . Give the x coordinate a value and find the corresponding y coordinate for several points Make a table of values for the equation y = 2x + 1 1 2 3 0 3 7 1 5 So (0,1) (1,3) (2,5) and (3,7) all lie on the line with equation y=2x + 1 Now plot the points on a grid and join them up

  4. . y Plot the points (0,1) (1,3) (2,5) and (3,7) on the grid 7 6 5 4 3 2 1 . Now join then up to give a straight line . . x 0 1 2 3 4 All the points on the line satisfy the equation y = 2x + 1

  5. Sketching lines by finding where the lines cross the x axis and the y axis A quicker method Straight lines cross the x axis when the value of y = o Straight lines cross the y axis when the value of x =0 Sketch the line 2x + 3y = 6 Line crosses x axis when y = 0 Line crosses y axis when x = 0 2x + 0 =6 0 + 3y = 6 3y =6 2x =6 y = 2 x =3 at ( 3,0) at ( 0,2) Ex 2 page 124

  6. Plot 0,2) and (3,0) and join them up with a straight line y 7 6 5 4 3 2 1 Now join then up to give a straight line . . x 0 1 2 3 4 All the points on the line satisfy the equation 2x + 3y = 6

  7. Solving simultaneous equations by straight line graphs Two lines either meet at a point intersect or they are parallel they never meet To find where two lines meet draw the graphs of both lines on the same grid and read off the point of intersection

  8. Find where the lines x + y = 5 and x – y =1 intersect get two points that lie on each line or better still three x + y = 5 0 1 5 x (0,5) and (5,0) and (1,4)lie on the line x + y = 5 y 5 4 0 x – y =1 0 4 1 x (0,-1) and (1,0) and ( 4,3) lie on the line y -1 3 0 Now plot the points and find where the two lines meet

  9. y 7 6 5 4 3 2 1 0 -1 . . x + y =1 . (3,2) X x + y =5 . . x . 1 2 3 4 5 The lines intersect at (3,2)

  10. Solving simultaneous equations algebraically Whoopee no more drawing graphs But you will have to learn a strategy to solve the equations Yikes!

  11. Elimination method Solve the simultaneous equations algebraically HOW DO I SOLVE TWO EQUATIONS WITH TWO UNKNOWNS ? x + y = 8 x - y = 4 2x = 12 ADD the two equations together This gives one equation with one unknown which is easy to solve x = 6 ADD How do I find y. It is gone.It has been eliminated ?

  12. SUBSTITUTE THE VALUE OF X INTO ONE OF YOUR EQUATIONS x = 6 x + y = 8 Now I have the solution 6 + y = 8 y = 2 Am I correct? Check by putting the values of x and y into , the other equation, x-y,and see if you get 4 6 – 2 = 4  Solution is x =6 and y = 2

  13. Solve the simultaneous equations x + 3y = 7 x - 3y = -5 Remember to check ADD 2x = 2 x = 1 Substitute x =1 into x + 3y = 7 1 + 3y = 7 3y =6 1 – 3 X 2 y = 2 = 1 -6 Solution is x = 1 and y = 2 = -5 

  14. Solve the simultaneous equations If I add I get another equation in x and y 3x + y = 8 x + y = 4 3x + 2=12 no use 3x + y = 8 -x -y = -4 Multiply one of the equations by ( -1) ADD 2x = 4 x = 2 It changes the sign of everything SUBSTITUTE Put x=2 into x + y =4 x + y = 4 2 + y = 4 3X2 + 2 = 8  y = 2 check Solution x = 2 and y = 2

  15. Using simultaneous equations to solve problems example

  16. Slightly more difficult But not for you IF you have learned the work so far Solve the simultaneous equations Adding gives 5x + 5y = 55 no use 3x + 2y = 29 2x + 3y = 26 Multiplying by a negative does not eliminate any of the letters

  17. Could I multiply and then add? That ‘s good let’s try it but be careful Remember when we add numbers together which are the negatives of each other they are eliminated

  18. Solve the following simultaneous equations algebraically 3x + 2y = 292x + 3y = 26 Multiply the equations by two suitable numbers so that the coeficients of x or y are the negatives of each other We choose y to be eliminated X 3 3x + 2y = 292x + 3y = 26 X -2 Giving two new equivalent equations 9x +6y = 87 -4x + -6y = -52 Now add

  19. 9x +6y = 87 -4x + -6y = -52 Remember practice makes perfect 5x = 35 ADD x = 7 check substitute x = 7 into 9x+6y = 87 -4 X 7 - 6X 4 = -28 –24 = -52  9 X 7 + 6y = 87 63 +6y = 87 6y = 24 y = 4 Ex 7B p 139 Solution x = 7 and y = 4

  20. Using simultaneous equation to solve problems

  21. Problem Solving With Simultaneous Equations Example : The problem For the cinema 2 adults’ tickets and 5 children’s tickets cost £26 4 adults’ tickets and 2 children tickets cost £28 Find the cost of each kind of ticket. Introduce Letters Let the cost of adult ticket = £A Let the cost of children’s ticket = £C Write the equations

  22. Solve the Problem using simultaneous equations Conclusion Adult Ticket costs £5.50 Children's tickets cost £3

  23. strategy • Read the problem • Introduce two letters • Write two equations that describe the information • Solve the problem using simultaneous equations • check

  24. Substitution method This an excellent method of solving simultaneous equations when the equations are given to you in a certain form Solve the equations y = 2x and y = x + 10 Example 1 We take the equation y = x + 10 And substitute 2x for y 2x = x + 10 Take out y and replace it with 2x x = 5 Sub x=5 into y=2x y = 10 Now solve the equation Solution x = 5 y = 10

  25. Example 2 Solve the equations y = 2x –8 and y – x = 1 Sub y =2x-8 in the equation y - x = 1 2x –8 -x = 1 x – 8 = 1 x = 9 y = 18-8 Sub x = 9 in the equation y = 2x -8 y = 10 Solution x = 9 and y =10 Check 10 –9 =1

  26. Problems with simultaneous equations from past papers • 1. The tickets for a sports club cost £2 for members and £3 for non- members • The total ticket money collected was £580 • x tickets were sold to members and y tickets were sold to non-members. • Use this information to write down an equation involving x and y 2x + 3y =580

  27. b) 250 people bought raffle tickets for the disco. Write down another equation involving x and y. x + y = 250 c ) How many tickets were sold to members ? We now have two equations in x and y so let’s solve them simultaneously 2x + 3y = 580 x + y = 250

  28. 2x + 3y = 580 x + y = 250 X 1 X (-2) 2x + 3y = 580 -2x + (-2 )y = -500 ADD y = 80 Sub y = 80 into x + y = 250 check x + 80 =250 2 x170 + 3x80 =340 + 240 =580 x = 170 170 tickets were sold to members

  29. 2. Alloys are made by mixing metals.Two different alloys are made using iron and lead. To make the first alloy, 3 cubic cms of iron and 4 cubic cms of lead are used. This alloy weighs 65 grams a ) Let x grams be the weight of 1 cm3 f iron and y grams be the weight of 1 cm3 of lead Write down an equation in x and y which satisfies the above equation. 3x + 4 y = 65 To make the second alloy, 5 cm3 of iron and 7cm3 of lead are used . This alloy weighs 112 grams. b ) Write down a second equation in x and y which satisfies this condition 5x + 7y = 112

  30. C ) Find the weight of 1 cm3 of iron and the weight of 1cm3 of lead. X (-5) 3x + 4 y = 65 5x + 7y = 112 X 3 Check 5x7 + 7x11 = 35 + 77= 112 -15x + -20 y = -325 15x + 21y = 336 ADD y = 11 y = 11 into 3x + 4y = 65 SUB 3x + 44 = 65 1cm3 of iron weighs 7gms and 1cm3 of lead weighs 11 gms 3x = 21 x = 7

  31. 3. A rectangular window has length , lcm and breadth, b cm . A security grid is made to fit this window. The grid as 5 horizontal wires and 8 vertical wires a ) The perimeter of the window is 260 cm. Use this information to write down an equation involving l and b . 2 l + 2 b = 260 b) In total, 770 cm of wire is used. Write down another equation involving l and b 5 l + 8 b = 770

  32. Find the length and breadth of the window 2 l + 2 b = 260 5 l + 8 b = 770 X (4) X (-1) 8 l + 8 b = 1040 -5 l + (-8) b = -770 ADD 3 l = 270 l = 90 5x90 +8x40 = 450 + 320=770 l = 90into 2 l + 2 b = 260 SUB 180 + 2b = 260 Length is 90cm and the breadth is 40 cm 2b = 80 b = 40

  33. 4. A number tower is built from bricks as shown in fig 1.The number on the brick above is always equal to the sum of the two numbers below. fig1 9 12 -3 8 4 -7 3 5 -1 -6

  34. Find the number on the shaded brick in fig 2 34 Fig 2 16 18 5 11 7 34 -2 7 4 -11

  35. In fig 3, two of the numbers on the bricks are represented by p and q Show that p + 3q = 10 -3 p +2q -5 q-8 p+q q -5 -3 Fig 3 p q -5 2 Adding the numbers in the second row to equal the top row p + 2q –5 + q -8 = -3 P + 3q – 13 = -3 simplifying P + 3q = 10

  36. Use fig 4 to write down a second equation in p and q 14 2q+1 8-p fig 4 2q-2 3 5-p 2q -2 5 -p Adding the numbers in the second row to equal the top row 2q+1+8-p =14 2q +9 – p = 14 simplifying 2q – p = 5 d) Find the values of p and q

  37. 3q + P = 10 2q – p = 5 Change the terms around q under q and p under p Remember to check 3q + P = 10 2q - p = 5 ADD Check 3x3 +1 = 10 5q = 15 q = 3 SUB q = 3 into 2q –p =5 p = 1 and q = 3 6 –p = 5 p=1

  38. 5. A sequence of numbers is 1 ,5 ,12 ,22 ,………Numbers from this sequence can be illustrated in the following way using dots • First number ( N = 1) • • • • • Second number (N=2) • • • • • • • • • • • • • • Third number (N=3) • • • • • • • • • • • • • • • • • • • • Fourth number (N=4)

  39. a )Write the fifth number in the pattern Form a table of values When N=2 D = 5 When N= 1 D = 1 b) The number of dots needed to illustrate the nth number in this sequence is given by the formula D =aN² - bN Find the values of a and b

  40. N = 1 D = 1 When N=2 D = 5 D =aN² - bN Form two equations by substituting the values of n and D into the equation D =aN² - bN N = 1 D = 1 1 = a - b 5 = 4a -2b N=2 D = 5 Now solve simultaneously

  41. X (-2) 1 = a - b X1 5 = 4a -2b -2 = -2a + 2b 5 = 4a - 2b Check 4x3/2-2x1/2 =12/2-2/2 =10/2=5 ADD 3 = 2a a = 3/2 a = 3/3 b= 1/3 1 = 3/2 -b SUB b = 1/2

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