Basic Engineering Mathematics, Fifth Edition

110 Basic Engineering Mathematics

Ka=

x^2

v( 1 −x)

, determinex, the degree of

ionization, given thatv=10dm^3.

6. A rectangular building is 15m long by 11m
   wide. A concrete path of constant width is
   laid all the way around the building. If the
   area of the path is 60.0m^2 ,calculate its width
   correct to the nearest millimetre.


7. The total surface area of a closed cylindrical
   container is 20.0m^3. Calculate the radius of
   the cylinder if its height is 2.80m.


8. The bending momentMat a point in a beam
   is given byM=

3 x( 20 −x)

2

,wherexmetres

is the distance from the point of support.

Determine the value ofxwhen the bending

moment is 50Nm.

9. A tennis court measures 24m by 11m. In the
   layoutofanumberofcourtsanareaofground
   must be allowed for at the ends and at the
   sides of each court. If a border of constant
   width is allowed aroundeach court and the
   total areaofthecourt anditsborderis950m^2 ,
   find the width of the borders.


10. Two resistors, when connected in series, have
    a total resistance of 40 ohms. When con-
    nected in parallel their total resistance is 8.4
    ohms. If one of the resistors has a resistance
    ofRx, ohms,
    (a) show thatR^2 x− 40 Rx+ 336 =0and
    (b) calculate the resistance of each.

14.6 Solution of linear and quadratic

equations simultaneously

Sometimes a linear equation and a quadratic equation

need to be solved simultaneously. An algebraic method

of solutionis shown in Problem 27; a graphical solution

is shown in Chapter 19, page 160.

Problem 27. Determine the values ofxandy

which simultaneously satisfy the equations

y= 5 x− 4 − 2 x^2 andy= 6 x− 7

For a simultaneous solution the values ofymust be

equal, hence the RHS of each equation is equated.

Thus, 5 x− 4 − 2 x^2 = 6 x− 7

Rearranging gives^5 x−^4 −^2 x^2 −^6 x+^7 =^0

i.e. −x+ 3 − 2 x^2 = 0

or 2 x^2 +x− 3 = 0

Factorizing gives ( 2 x+ 3 )(x− 1 )= 0

i.e. x=−

3

2

orx= 1

In the equationy= 6 x−7,

whenx=−

3

2

, y=^6

(

−

3

2

)

− 7 =− 16

and whenx=1, y= 6 − 7 =− 1

(Checking the result iny= 5 x− 4 − 2 x^2 :

whenx=−

3

2

, y= 5

(

−

3

2

)

− 4 − 2

(

−

3

2

) 2

=−

15

2

− 4 −

9

2

=− 16 ,as above,

and whenx=1, y=^5 −^4 −^2 =−^1 ,as above.)

Hence, the simultaneous solutions occur when

x=−

3

2

,y=− 16 and whenx= 1 ,y=− 1.

Now try the following Practice Exercise

PracticeExercise 58 Solving linear and

quadratic equations simultaneously

(answers on page 346)

Determine the solutions of the following simulta-

neous equations.

1. y=x^2 +x+12.y= 15 x^2 + 21 x− 11
   y= 4 −xy= 2 x− 1


2. 2x^2 +y= 4 + 5 x
   x+y= 4