Simultaneous equations

Solving quadratic equations

1. Simultaneous equations Elimination and substitution method

2. Objectives: solve simultaneous linear equations by elimination method Solve simultaneous linear equations by substitution method Elimination method of solving simultaneous equations Elimination method:- this is to get rid off one of the variables.

3. Conditions for elimination The coefficient of the variables to get rid off must be the same. Addition operation is applied when the variables are of different signs Subtraction is used when the variables are of the same signs. Example 1: solve simultaneously x +2y = 7 x – y = 4 Solution: to eliminate means to get rid of one of the variables

4. x +2y = 7 (1) x – y = 4 (2) x – x + 2y – (-Y) = 7- 4 2y + y = 3 3y = 3 y = 1 Substitute for y in any of the two equations x – y = 4 x – 1 = 4 x = 4 + 1 x = 5 To get rid of x subtract equation 2 from equation 1

5. other way x +2y = 7 (1) x – y = 4 (2) x +2y = 7 (1) 2x -2y = 8 (3) 3x = 15 x = 5 Substitute for x in any of the equation x – y = 4 5 – y = 4 5 -4 = y y = 1 To get rid of y, the coefficients of y in the two equations must be the same Multiply equation 2 by 2 Add equation 1 to equation 3

6. Solve for x and y in x + y = 12 3x- y = 20 Solution: x + y = 12 3x- y = 20 x + 3x + y + (-y) = 12 +20 4x = 32 x = 8 x + y = 12 8 + y = 12 y = 12 – 8 y = 4 (1) (2) To get rid of x, the coefficients of x in the two equations must be the same Multiply equation 1 by 3 3x + 3y = 36 (3) 3x - y = 20 (2) subtract equation 2 from equation 3 3y –(-y) = 36 - 20 4y = 16 y = 4 substitute for y in any of the equations x + y = 12 x + 4 = 12 x = 12 - 8 x = 4

7. Example 2 : solve for x and y in 2x + 3y = 7 x + 5y = 0 Solution : 2x + 3y = 7 (1) x + 5y = 0 (2) 2x + 10y = 0 (3) 2x + 3y = 7 (1) 10y – 3y = 0 -7 7y = - 7 y = -1 x + 5y = 0 x + 5(-1) = 0 x - 5 = 0 x = 5 To get rid of y, the coefficients of y in the two equations must be the same Multiply equation 1 by 5 and equation 2 by 3 10x + 15y = 35 (3) 3x + 15y = 0 (4) subtract equation 4 from equation 3 10x - 3x + 15y -15y = 35 - 0 7x = 35 x = 5 x + 5y = 0 5 + 5y = 0 y = -1

8. Activity 2 : Solve for x and y in 2x – 3y – 10 = 0 10x – 6y = 5 Solution : 2x – 3y = 10 (1) 10x – 6y = 5 (2) Multiply equation 1 by 5 10x – 15y = 50 (3) 10x – 6y = 5 (3) 10x -10x – 15y – (-6y) = 50 - 5 – 15y + 6y = 45 -9y = 45 y = -5 2x – 3(-5) = 10 2x + 15 = 10 2x =- 15 + 10 2x =- 5 x = - 5/2

9. Substitution method Objectives : to make one of the variables the subject in one of the equations to substitute the new equation correctly into the other equation To solve for the unknown

10. Example 1:solve for x and y in 2x – y =9/2 (!) X + 4y = 0 (2) Make y or x the subject Make x the subject in equation2 X = - 4y Sub. For x in equation (1) 2(-4y) - y = 9/2 -8y –y = 9/2 -9y = 9/2 y = 9/2 ÷ -9 y = 9/2 × 1/-9 y = - 1/2 Sub. For y in x = -4y x = -4 (-1/2) x = 2

11. Activity 1: solve for x and y in 4x – 3 = 3x + y =2y + 5x – 12 Solution: 4x – 3 = 3x + y 4x - 3x – y = 3 x – y =3 (1) 3x + y = 2y + 5x - 12 3x -2y + y - 5x = -12 -2x – y = -12 2x + y = 12 (2) From equation 1 x = 3 +y Sub. For x in eq.(2) 2(3 + y) + y = 12 6 + 2y + y = 12 6 + 3y = 12 3y = 12 - 6 3y = 6 3y 3 y =2 Recall x =3 + y x =3 + 2 x = 5

12. Activity 2: solve for p and q in 3p -5q -4 =5p +8 =2p + q + 7 Solution: 3p – 5q – 4 = 2p + q +7 3p – 2p – 5q –q = 7 + 4 P – 6q = 11 (1) 3p – 5q – 4 = 5p + 8 3p – 5p -5q = 8 + 4 -2p – 5q = 12 (2) From eq. (1) p = 11 + 6q Sub. For p in eq. (2) -2(11 + 6q) – 5q =12 -22 - 12q – 5q =12 -12q – 5q =12 + 22 -17q =-34 q = –2 Recall p =11 + 6q p =11 + 6(-2) p = 11 -12 p = -1

13. Class work: solve for x and y in X + 3y = 6 2x – y = 5 2. 3x – 5y -3 = 0 2y – 6x + 5=0

14. Simultaneous equation with fractions Objectives : To change the fractional equation to simple linear equation To solve for the unknowns

15. Solve for x and y in Solution : step 1 Find the lcm of the denominators of each of the equation Step 2: multiply each equation by the corresponding lcm (1) (2) from eq. 1 x = 2 - y Sub. For x in 2x + 3y = -1 2(2-y) + 3y = -1 4 – 2y + 3y = -1 4 + y = -1 y = -1 - 4 y = -5

16. Recall x = 2 – y x = 2 – (-5) x = 2 + 5 x = 7 Activity 1: Solve for x and y in Solution (1) (2) Multiply eq (1) by2 10x – 8y = 4 (3) 10x – 9y = 2 (2) 10x-10x -8y –(-9y ) = 2 -8y + 9y = 2 y = 2 10x – 9(2) = 2 10x – 18 = 2 10x = 20 x = 2

17. Activity 2 : Solve for x and y in Solution (1) (2) from eq. 2 y = -4 -2x Sub. for y in 8x + 9y = 24 8x + 9(-4-2x) = 24 8x -36 – 18x= 24 8x – 18x= 24 + 36 -10x= 60 x= -6 y = -4 -2x y = -4 -2(-6) y = 8

18. Activity 3 : Solve for x and y in Solution: (1) (2) From eq(1) y = 1 – 2x Sub. for y in 3x – y =9 3x – (1 – 2x) =9 3x – 1 + 2x =9 5x = 9 + 1 5x = 10 x = 2 y = 1 - 2x y = 1 – 2(2) y = 1 – 4 y = – 3

19. Solve the following equations simultaneously

20. Simultaneous equation with quadratic equation Objectives: to make the variable the subject To be able to substitute correctly To be able to solve for the unknown from quadratic equations

21. Example 1: solve for x and y in X + 3y= 2 (1) x 2 + 2y =3 (2) Solution: make x the Subject in eq.(1) x = 2 - 3y Sub. for x in eq.(2) (2- 3y) 2 + 2y = 3 4 – 6y – 6y + 9y 2 + 2y = 3 4 – 12y + 2y +9y 2 - 3 = 0 9y 2 -10y + 1 = 0 9y 2 – 9y – y + 1= 0 9y(y – 1) –1(y – 1)= 0 (9y –1)(y – 1)= 0 y = 1/9 or 1 Sub. For y in x = 2- 3y x =2 – 3(1) x =2 – 3 x = –1 x =2 – 3(1/9) x =2 –1/3 x = 5/3

22. Activity 1 solve for x and y in 3x + y = 1 2x 2 – y 2 = -2 Solution: y = 1 – 3x 2x 2 – (1- 3x) 2 = - 2 2x 2 – (1 – 3x – 3x + 9x 2 )= -2 2x 2 – 1 + 6x - 9x 2 = - 2 9x 2 -2x 2 -6x + 1 -2 = 0 7x 2 -6x -1 =0 7x 2 -7x + x -1 = 0 7x(x -1) + 1(x-1) = 0 (7x +1) (x - 1) =0 X = -1/7 or 1 Recall y = 1 – 3x y = 1 – 3(-1/7) 1 + 3/7 = 7 + 3 =10/7 7 y = 1 – 3x 1 – 3(1) 1 – 3 = -2

23. Activity2: Solve for x and y in 2x + y = 4 and x 2 + xy = -12 Solution: 2x + y = 4 y = 4 – 2x Sub. for y in x 2 + xy = -12 x 2 + x(4 - 2x) = -12 x 2 + 4x -2x 2 =-12 -x 2 + 4x =-12 -x 2 + 4x + 12 = 0 -x 2 – 2x + 6x + 12 = 0 -x (x + 2) + 6(x +2) = 0 (6 - x)(x + 2) = 0 x = -2 or 6 y = 4 – 2x y = 4 – 2(-2) y = 4 + 4 = 8 y = 4 – 2(6) y = 4 – 12 y = -8

24. Activity3: Solve for x and y in x + y = 14 and 2xy + 1 = 21 Solution: x =14 – y Sub. for x in 2xy + 1 = 21 2y(14 - y) + 1 = 21 28y -2y 2 + 1 = 21 2y 2 – 28y + 20 = 0 y 2 - 14y + 10 = 0