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Class 9 SELINA Solutions Maths Chapter 27 - Graphical Solution (Solution of Simultaneous Linear Equations, Graphically)

Graphical Solution (Solution of Simultaneous Linear Equations, Graphically) Exercise Ex. 27(A)

Solution 1

(i)

The graph x= 5 in the following figure is a straight line AB which is parallel to y axis at a distance of 5 units from it.

 

 

 

 

 

 

(ii)

x+5=0 Þx = -5

The graph x= -5 in the following figure is a straight line AB which is parallel to y axis at a distance of 5 units from it in the negative x direction.

 

 

 

 

 

(iii)

The graph y = 7 in the following figure is a straight line AB which is parallel to x axis at a distance of 7 units from it.

 

 

 

 

 

 

 

(iv)

y+7=0 rightwards double arrow y = -7

The graph y = -7 in the following figure is a straight line AB which is parallel to x axis at a distance of 7 units from it in the negative y direction.

 

 

 

 

 

 

 

(v)

2x + 3y = 0

rightwards double arrow 3y=-2x

 

 

 

 

 

 

 

 

(vi)

3x + 2y = 6

rightwards double arrow 2y=6-3x

 

 

 

 

 

 

 

 

 

 

 

 

(vii)

x-5y+4=0

rightwards double arrow 5y=4+x

 

 

 

 

 

 

 

 

 

(viii)

5x + y + 5 = 0

rightwards double arrowy = -5x - 5

 

 

 

 

 

 

 

 

 

 

 

Solution 2

 

 

 

 

 

From the figure it is clear that, the graph meets the coordinate
axes at (3, 0) and (0, 5)

 

 

 

 

 

 

From the figure it is clear that, the graph meets the coordinate
axes at (-9, 0) and (0, 6)

 

 

 

Solution 3

The straight line cuts the co-ordinate axis at A(0, 12) and B(-9, 0).

Solution 4

 

 

 

 

 

Solution 5

Solution 6

Solution 7

 

 

 

 

 

 

 

 

 

 

Solution 8

 

 

 

 

 

Solution 9

Solution 10

(i)

(ii)

Graphical Solution (Solution of Simultaneous Linear Equations, Graphically) Exercise Ex. 27(B)

Solution 1

(i)

 

 

 

 

 

 

(ii)

 

 

 



(iii)

 

 

 

 

 

(iv)

 

 

 

 

 

 

Solution 2

Solution 3

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 4

Solution 5

(i)

 

 

 

 

 

 

 

 

 

(ii)

(iii)



A p p l y i n g space P y t h a g o r a s space T h e o r e m comma space
t h e space d i s tan c e space f r o m space t h e space o r i g i n space equals square root of open parentheses minus 2 minus 0 close parentheses squared plus open parentheses minus 1 minus 0 close parentheses squared end root
equals square root of 2 squared plus 1 squared end root
equals square root of 4 plus 1 end root
equals square root of 5
equals 2.2 space c m space left parenthesis a p p r o x right parenthesis




Solution 6

(ii)

Solution 7

Solution 8

(i)

(ii)

Solution 9

 

(i)

No. of articles to be manufactured and sold are 50 when there is no loss and no profit.

C.P = S.P = Rs.200

(ii)

(a)

On article 30,

C.P = Rs.140 and S.P. = 120

Therefore Loss = 140 - 120 = Rs.20

(b)

On article 60,

C.P.=Rs.230 and S.P.= Rs.240

Therefore Profit = 240 - 230 = Rs.10

Solution 10

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 11

Solution 12

 

 

 

 

 

Therefore the solution of the given system of equations is (2,1).

 

 

 

 

Solution 13

Solution 14

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