SIMULTANEOUS EQUATION (One linear, One quadratic)

SUBJECT: MATHEMATICS

CLASS: SS 3

TERM: FIRST TERM

WEEK 5                                                                                 DATE:_____________________

TOPIC:  SIMULTANEOUS EQUATION (One linear, One quadratic)

Examples

Solve simultaneously for x and y (i.e. the points of their intersection)

3x + y = 10

2x² +y² = 19

Note: One linear, One quadratic is only possible analytically using substitution method.

Solution.

3x + y = 10 ———– eq 1

2x²+ y² = 19 ——— eq 2

Make y the subject in eq 1 (linear equation)

y = 10 – 3x ———- eq 3

Substitute eq 3 into eq 2

2x² + (10-3x)²   = 19

2x²₊ (10 – 3x) (10 – 3x) = 19

2x² + 100 – 30x – 30x + 9x² = 19

2x² + 9x² – 30x – 30x + 100 – 19 = 0

11x² – 60x + 81 = 0

11x² – 33x – 27x + 81= 0

11x (x-3) – 27 (x – 3) = 0

(11x – 27) (x – 3) = 0

11x – 27 = 0  or x-3 = 0

11x = 27 or  x = 3

\ x = 27/11  or 3

Substitute the values of x into eq 3.

When x = 3

y = 10 – 3(x)

y = 10  – 3(3)

y = 10 – 9 = 1

When x =27/11

y = 10 – 3(27/11)

y = 10 –  51/11

y =110 – 51

11

y = 59/11

\ w hen x = 3, y = 1

x = 27   ,   y = 59

11              11

EXAMPLE

Solve the equations simultaneously: 3x + 4y = 11 and  xy = 2

                       solution

3x + 4y = 11 ——– eq 1

x y = 2        ——– eq 2

Make y the subject in eq 1

4y  = 11 – 3x

y = 11 – 3x  …………eq3

4

substitute eq 3 into eq 2

x y = 2

x ( 11- 3x )=  2

4

x (11-3x) = 2×4

11x – 3x² = 8

-3x²   + 11x – 8 = 0

-3x²   + 3x + 8x – 8 = 0

-3x (x-1) +8 (x-1) = 0

(-3x + 8) (x-1) = 0

-3x + 8 = 0  or  x – 1 = 0

3x = 8  or  x = 1

x = 8/3 or 1

Substitute the values of x into eq 3

y = 11- 3x

4

when x = 1

y = 11 – 3(1)=11-3 = 8

4               4         2

y =  4

when x = 8/3

y = 11 – 3(8/3)

4

y = 33 – 24  =  9      =    3

12         12          4

\ x = 1, y = 2

x = 8/3, y = 3/4.

Evaluation

Solve for x and y

1. 3x ² – 4y = -1                           II.  4x² + 9y² = 20

2x – y = 1                                        2x – 9y =-2

Reading Assignment NGM for SS IIEx 7d, 1 b, e, 2 b, c.

Simultaneous Equation (Further Examples)

Solve simultaneously for x and y.

3x – y = 3 ——– eq 1

9x²   – y ² = 45 ——— eq 2

Solution

From eq 2

(3x)²  – y ²  = 45

(3x-y) (3x+y) = 45 ———- eq 3

Substitute eq 1 into eq 3

3 (3x + y) = 45

3x + y = 15 ……………..eq4

Solve eq 1 and eq 4 simultaneously.

3x – y = 3 ——— eq 1

3x + y = 15 ——– eq 4

eq 1 + eq 4

6x  =  18

x  =  18/ 6

x = 3

Substitute x = 3 into eq 4.

3x + y = 15

3 (3) + y = 15

9 + y = 15

y = 15 – 9

y = 6

\ x = 3, y = 6

Evaluation

1. a. Given that : 4x² – y² = 15 b. Given that : 3x² +5xy –y² = 3

2x – y = 5                                                                 x  –  y  = 4

Solve for x and y.                                                    solve for x and y.

WORD PROBLEM LEADING TO ONE LINEAR-ONE QUADRATIC EQUATION

  Example:

The product of two numbers is 12. The sum of the larger number and twice the smaller number is 11. Find the two numbers.

Solution

         Let    x  = the larger number

                   y  = the smaller number

Product,  x y  =  12    …………….eq1

From the last statement,

x + 2y  =  11  ………….. eq2

From eq2,   x  =  11 – 2y   ……………eq3

Sub. Into  eq1

y(11 – 2y) = 12

11y – 2y2  = 12

2y2 -11y + 12 = 0

2y2 – 8y – 3y + 12 = 0

2y(y-4) – 3(y-4) = 0

(2y-3)(y-4)  =0

2y-3 =0 or  y-4 =0

2y = 3 or   y = 4

y= 3/2 or 4

when y = 3/2                                             when  y=4

x = 11 – 2y                                      x = 11- 2y

x = 11 – 2(3/2)                                  x = 11 – 2(4)

x = 11 – 3                                           x = 11 – 8

x = 8                                                  x = 3

Therefore, (8 , 3/2)(3 , 4)

2.Find a two digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.

Solution

Let the two digit number be  ab, where a is the tens digit and b is the unit digit

From the first statement,

2a  + 3 = 3b

2a – 3b = -3 ………….eq1

From the second statement,

4(10a + b) – 99 = 10b + a

40a + 4b – 99 = 10b + a

40a – a + 4b – 10b = 99

39a – 6b = 99

Dividing through by 3

13a – 2b = 33 ………….eq2

Solving both equations simultaneously,

a = 3 , b = 3

Hence, the two digit number is 33

Reading Assignment

1. New General Mathematics for SS 2 by J B Chamnon & Co   page 73 – 78
2. Past SSCE Questions
3. Exam Focus (Mathematics).

Weekend Assignment

In each of the following pairs of equations solve simultaneously,

1. xy = -12 ; x – y = 7 a. (3 , -4)(4 ,-3)         b. (-2 ,4)(-3, -4) c. (-4, 5)(-2 , 3) d(3 ,-3)(4,-4)
2. x – 5y = 5 ; x² – 25y² = 55 a (-8, 0)(3/5 , 0) b(0, 0)(-8 , 3/5) c (8 , 3/5) d (0, 8)(0, 3/5)
3. y = x²and y = x + 6 (a).(0,6) (3,9) (b)(-3,0) (2,4)  (c)(-2,4) (3,9) (d).(-2, 3), (-3,2)
4. x – y = -3/2 ; 4x² + 2xy – y² = 11/4 : a(-1, 1/2)(1, 5/2).b.(3, 2/5)(1, 1/2)c.(3/2 , -1)(4,2)  d.(-1 , -1/2)(-1 , 5/2)
5. m² + n² = 25 ; 2m + n – 5 = 0 : a. (0,5)(4, -3) b.(5,0)(-3,4)c.(4,0)(-3,5) d(-5,3)(0,4)

Theory

1    a. Find the coordinate of the points where the line 2x – y = 5 meets the curve 3x² – xy -4 =10

1. Solve the simultaneous equations: x + 2y = 5, 7(x² +1) = y(x + 3y)
2. A woman is q years old while her son is p years old. The sum of their ages is equal to twice the difference of

their ages. The product of their ages is 675.

Write down the equations connecting their ages and solve the equations in order to find the ages of the woman

and her son. (WAEC)