Simultaneous Linear Equation II – SS1 Mathematics Lesson Note (First Term)

Simultaneous equations involving one linear and one quadratic

One of the equations is in linear form while the other is in quadratic form.

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

Examples:

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

3x + y = 10 & 2×2 +y2 = 19

Solution

3x + y = 10 ———– eq 1

2×2 + y2 = 19 ——— eq 2

Make y the subject in eq 1 (linear equation)

y = 10 – 3x ———- eq 3

Substitute eq 3 into eq 2

2×2 + (10-3x) 2   = 19

2×2+ (10 – 3x) (10 – 3x) = 19

2×2 + 100 – 30x – 30x + 9×2 = 19

2×2 + 9×2 – 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

(when x = 3, y = 1

x = 27   , y =  59

      11               11

2. Solve the equations simultaneously:

 3x + 4y = 11   &xy = 2

Solution

3x + 4y = 11      ——– eq 1

xy = 2        ——– eq 2

Make y the subject in eq 1

4y  = 11 – 3x

y =   11 – 3x   …………   eq3

            4

substitute eqn 3 into eq 2    

x ( 11- 3x )  =  2

         4

x (11-3x) = 2×4

11x – 3×2 = 8

-3×2   + 11x – 8 = 0

-3×2   + 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 = ¾.

Word  problems leading to linear and quadratic equations

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)