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SIMULTANEOUS EQUATIONS

In lower forms, we have learnt the way to solve a system of two simultaneous linear equations

in two unknowns. In this chapter, we are going to learn ways to solve a system of one linear

equation and one quadratic equation in two unknowns and have a look at some of its applications.

1. Method of Substitution

Similar to a system of two simultaneous linear equations in two unknowns, we have an

analogous method of substitution to solve a system of one linear equation and one quadratic

equation in two unknowns. We illustrate this method by means of some examples.

Example 1.1.

Solve the following system of equations.

( )

( )2

..........

.......... 2! 2

y x

y x x

= = + +

Solution.

Substituting "# into "2#, we get

( ) ( )

2

2

! 2

$ 2 %

2 %

x x x

x x

x x

= + ++ + =

+ + =

and so x = −2 or x = −. In order to find the corresponding values for y, we substitute these values of

x into "# to get y = −2 when x = −2 and y = − when x = −. &quivalently, we may write

" x, y# = "−2, −2# or "−, −#.

Example 1.2.


( )

( )2

.......... 2 ' (

.......... 22 2

y x

y x x

= − = + −

Solution.

)rom "#, we have

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( )' (

.......... $2 2

y x= − .

Substituting "$# into "2#, we get

( ) ( )

2

2

2

' (2 2

2 2

' ( 2 ! 2!

2 ' %

2 ' $ %

x x x

x x x

x x

x x

− = + −

− = + −

− − =

+ − =

and so '

2 x = − or x = $. Substituting these values of x into "$#, we get " x, y# = ( )' !$

2 !,− − or "$, $#.

Example 1..


( )

( )2 2

.......... 2

.......... 2$ 2 !$

y x

x y x y

= + + = − +

Solution.


( ) ( )

( ) ( )

22

2 2

2

2 $ 2 2 !$

! ! $ 2 ! !$ %

2 $ $' %2 + ' %

x x x x

x x x x x

x x x x

+ + = − + ++ + + − + + − =

+ − =− + =

and so +

2 x = or x = −'. Substituting these values of x into "#, we get " x, y# = ( )+

2 2, or "−', −$#.

Exe!"ise

. Solve the following system of equations.

2 2

2 2 %!

x y x y− + = + =

2. Solve the following system of equations.

2 $ $2

! $ !

x y

x y

+ =

+ =

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2. #!aphi"al Method

We may solve the system of equations graphically. We are going to show this by redoing the

above examples.

Example 2.1.


( )

( )2

..........

.......... 2! 2

y x

y x x

= = + +

Solution.

he graph of equations "# and "2# is shown in )igure .

)igure - he graph of2

! 2 y x x= + + and y = x.

ote that the coordinates of the points of intersection of the two lines are the required solutionsof the system. )rom )igure , the points of intersection are "−, −# and "−2, −2#.

Example 2.2.


( )

( )2

.......... 2 ' (

.......... 22 2

y x

y x x

= − = + −

Solution.

he graph of equations "# and "2# are shown in )igure 2.

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)igure 2- he graph of2

2 2 y x x= + − and 2 y = ' x / (.

)rom )igure 2, the points of intersection occurs at about x = $ and x = −2.' and hence the

points of intersection is "$, $# and "−2.', −%.+'#.

Example 2..


( )

( )2 2

.......... 2

.......... 2$ 2 !$

y x

x y x y

= + + = − +

ote that "2# can be written as

( )

( )

2 2

2 2

2 2

2

2

$ 2 !$

$ $$ 2 !$

2 2

$ 0' .........."$#

2 !

x y x y

x x y y

x y

+ = − +

− + + + + − − = ÷ ÷

− + + = ÷

1ny point lying on "$# is at a distance of 0'

! from the point ( )$

2, − . ence "$# is a circle

centered at ( )$

2, − with radius 0'

!.

Solution.

he graph of equations "# and "2# are shown in )igure $.

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)igure $- he graph of2 2

$ 2 !$ x y x y+ = − + and 2 y x= + .

)rom )igure $, the answers are "$.', '.'# and "−', −$#.

. Appli"ations in $!a"ti"al $!oblems

he system of one linear and one quadratic equation is quite useful when solving practical

problems. We will illustrate its use through some examples.

Example .1.

he area and perimeter of a rectangle are respectively 223 cm and 3% cm. )ind its length and

width.

Solution.

4et x cm and y cm be its length and width respectively. hen

( )( )( )

23 .......... 2 3% .......... 2

xy x y

= + =.

Since the width of a rectangle cannot be 5ero, by "#,

( )23

.......... $ x y

= .

Substituting "$# into "2#, we get

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( ) ( )

2

2

232 3%

23 $%

$% 23 %

2 0 %

y y

y y

y y

y y

+ = ÷

+ =

− + =− − =

So y = 2 cm or y = 0 cm. ote that the second answer is re6ected since y = 0 gives 23

02 x = = ,

which contradicts to the assumption that x is the length and y is the width. ow when y = 2,23

20 x = = . ence the length of the rectangle is 0 cm and the width of it is 2 cm.

Example .2.

7uring free fall, the distance fallen " s, in metres# of an ob6ect and the time during which the ob6ect

has been falling "t , in seconds# have the following relation

2

' s t = .

1t a certain instant, $ s = '3 / t . )ind the distance fallen at that instant.

Solution.

o find the distance fallen, it is equivalent to solve for s in the system

( )

( )

2 .......... '

.......... 2$ '3

s t

s t

=

= −.


( )

( ) ( )

2

2

$ ' '3

' '3 %

$ + ' 0 %

t t

t t

t t

= −

+ − =

+ − =

So 0

't = sec or +

$t = − sec "re6ected as %t ≥ #. *ut 0

't = into "#, we get ( )

20 3!

' '' s m= = .

Example ..

1 rectangular garden of area

2

33 m is surrounded by a rectangular path of width m. If the area of the path is 2%! m , find the dimension of the garden.

Solution.

4et the dimension of the rectangle be x m by y m. hen we have xy = 33.

8n the other hand, " x 9 2#" y 9 2# / 33 = %!.

ence we have to solve the following system

( ) ( )

( )

( )

33 ..........

2 2 33 %! .......... 2

xy

x y

= + + − =

.

)rom "2#, we have

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( )

2 2 ! 33 %!

2 2 +3 .......... $

xy x y

xy x y

+ + + − =+ + =

.

:y substituting "# into "$#, we have

( )

33 2 2 +3

'%

'% .......... !

x y

x y

y x

+ + =+ =

= −.

:y substituting "!# into "#, we have

( )

( ) ( )

2

'% 33

'% 33 %

22 20 %

x x

x x

x x

− =− + =

− − =

So x = 22 m or x = 20 m. ow when x = 22, y = 20 and when x = 20, y = 22. So both values of x

gives the same dimension. ence the dimension of the garden is 22 m by 20 m.

Exe!"ise

. 1n isosceles triangle has perimeter +2 cm and base angle 3%°. )ind the length of its base and

its area "in surd form#.

2. 1 plate of area 2$3( cmπ is cut and bent to form two spheres. he sum of the diameters of the

two spheres is 2+ cm. )ind the volume of the smaller sphere.

$. 7uring free fall, an ob6ect is dropped freely from certain height. 1t any instant, the distance

fallen " s, in metre# and the instantaneous speed "v, in metre per second# is given by the

following relation

2 2%v s= .

When the ob6ect reaches the ground, its distance fallen is 2'; more than its speed. )ind the

height of the point when the ob6ect is released.

!. 4et C be a unit circle on the <artesian coordinate plane centered at the origin. hen the equationof C is 2 2 x y+ = . 4et A be the point "!, %#. )ind the coordinates of the point"s# on the circle

such that the point"s# is=are at a distance of ! from A.

'. 1 two>digit number is given. If the sum of the digits is ! and the sum of square of the digits is

$2 larger than the number, find the number.

3. 1 hollow cone of radius r cm and height h cm is cut straightly along its slant height and

unfolded to a sector. &xpress the area A of the sector in terms ofπ,

r and h. )urthermore, giventhat the square of h is larger than ! times of r by !( and A is 2220 cmπ , find the volume of the

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cone in terms of π in surd form.

%. Solution to Sele"ted Exe!"ise

?ethod of Substitution

. " x, y# = "%, 2# or ( )0 3

' ',− − .

2. " x, y# = "+, 3# or ( )20 22!

' !', .

1pplications in *ractical *roblems

. 4et x cm be half the length of the base and y cm be the length of the slant. So the perimeter of

the isosceles triangle is 2 x 9 2 y = +2. 8n the other hand,

2cos3%

x

y

= =o

. So we need to solve

the system of equations given below.

2 2 +2

2

x y

x

y

+ = =

he answer is " x, y# = "2, 2!# and so the height is 2 22! 2 !$2 2 $ cm− = = . So the area is

( )2

2 2 2 $ 2 !! $ cm× × ÷ = .

2. 4et r and s be the radii of the two spheres. hen 2"r 9 s# = 2+. 8n the other hand, since the

surface areas for the spheres are 2! r π and 2! sπ , we have

2 2! ! $3(r sπ π + = .

ence we need to solve the system of equations below.

( )2 2

2 2+

! ! $3(

r s

r sπ π π

+ =

+ =

So we know that the radius of the smaller sphere is 3 cm and so the volume of the smaller

sphere is$ $!

$3 200 cmπ π = .

$. When the ob6ect reaches the ground, s = "92';#v = 2.2'v. ence we need to solve for s in the

following system of equations.

2 2%

2.2'

v s

s v

=

=

So we found that 2.2' !' %.2' s m= × = or !%'

! m .

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!. 4et Q = " x, y# be the coordinates of the required point. Since Q lies on C , we have

( )2 2 .......... x y+ = .

In addition, QA = !. So we have, after simplification, 2 2 0 % x y x+ − = . Substituting "# into this,

we have / 0 x = % i.e.

0 x = and so we have 3$ $ +

0 0 y = ± = ± . ence the required points are

( )$ +

0 0, and ( )$ +

0 0, − .

'. 4et x be the tens digit and y be the unit digit. hen the number is % x 9 y. So we have x 9 y = !

and ( )2 2 $2 % x y x y+ = + + . :y solving, we get x = 3 or x = 2.'. :ut the later answer is

re6ected as x should be an integer which satisfies % ( x≤ ≤ . So y = 0 and so the number is 30.

3. A = the curved surface area of the cone = 2 2rl r r hπ π = + .

ext, 2 ! !(h r = + and A = 220π . So by solving "note that ( )22 ! !( +r r r + + = + #, we have r

= 2 cm or r = −( cm. :ut the later answer is re6ected as r > %. So ! 2 !( 2+h cm= × + =

and so the volume of the cone is $!0 2+ cmπ .

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