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7.4 Graphical method of solving
Simultaneous linear equations:
Introduction :
We have seen in earlier sections(2.7) that a linear equation is represented a by a
line on the graph sheet and hence the name linear equation.
We have also learnt that we need
two equations to solve simultaneous liner equations. From these observations, it becomes logical that
why not represent the linear equations
graphically and then find a point
on the graph sheet which cuts the lines representing the given linear equations.
7.4 Problem 1: Solve 2x-y =3 and x+2y =6 graphically.
Solution:
Consider the equation 2x-y =3 and draw a graph for this
line on a graph sheet. Step 1 : Convert the given equation in
the form of y = (i.e. LHS will
have only y) Given
equation 2x-y =3 i.e. -y =3-2x(transposition). Multiply both sides by -1 to get y=2x-3 Step 2 : For few values of x (though 2 is enough)
get values of y and record them in a
table as shown below:
Step 3 : Plot the points represented by (x, y)
coordinates on a graph sheet. Step 4
: Join the points to get a straight line to represent the line y=2x-3(or
2x-y=3) Note : 1.
Though only two points are enough to draw a straight line, we have calculated
more values of (x, y) to indicate that there are many solutions to 2x-y =3. 2. It
is also clear from the graph that many coordinates like (0.5,-2) on the line 2x-y =3 satisfy the given equation. We need
another line crossing this line on the graph so that we get an intersection
point .The coordinates of this point will give unique solution to these 2
equations. Let us draw graph for x+2y = 6(Steps 5 to 7 given below) Step 5 : Transpose x to
get 2y = 6-x i.e. y = (6-x)/2 Step 6 : For few values of x, calculate corresponding
values of y and record them in a table as shown below(repetition of Step 2):
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Step 7 : Plot the points represented by (x, y) coordinates on a graph sheet and join them to get a
straight line. Step 8 : Let this and the earlier line meet at point
P( If the 2 lines don’t meet then extend the 2 lines in both
directions so that they meet) Step 9 : Find the coordinates
of P which is (2.4,1.8). |
Conclusion:
The solution the given equation is x =2.4 and y =1.8
Verfication:
Solve the above 2 equations using
elimination method:
2x-y = 3 ====è(1)
x+2y = 6 ====è(2)
Multiply (1) by 2: 4x-2y =6 ====è(3)
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Add (2) & (3) 5x+0 = 12
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i.e. x= 2.4
By Substituting 2.4 for x in (2).
We get 2.4+2y=6 and on
transposition we get 2y =3.6(=6-2.4) i.e. y=1.8
This is exactly the coordinates of
point P, which was obtained after drawing the graph
7.4 Problem 2 : Solve 2x+2y = 4 and x+y = 2
graphically
Solution:
Given equation 2x+2y =4 Step 1: (Express the equation as y=) :2y
= 4-2x (On transposition) i.e. y = 2-x
(On simplification) Step 2: For few values of x get values of y and record
them in a table as shown below:
Step 3: Draw a graph using these (x, y) coordinates to
represent the line 2x+2y=4 Step 4: Take 2nd equation x+y =2 i.e. y =2-x ( On transposition) Step 5: Since this equation is same as2x+2y=4, The (x,y) coordinate pairs can be obtained from step 2.
Step 6 : Since the
coordinates are same as in Step 2, this line is same as the one that was drawn in step
3(The two lines are collinear) Conclusion : There are infinite values of (x, y) which satisfy the given
set of equations. Hence there is no unique solution to the simultaneous
liner equations: 2x+2y = 4 and x+y = 2. |
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Verification: Refer 2.14 problem
3.
7.4 Problem 3: Solve 2x+2y = 4 and x+y = 3 graphically
Solution:
Given equation is 2x+2y =4 Step 1: (Express the given equation as y=)i.e. 2y =4-2x
(On transposition) i.e. y = 2-x
(On simplification)) Step 2: For few values of x calculate values of y and
record them in a table as below:
Step 3:
Draw a graph using these (x, y) coordinates to represent the line 2x+2y=4 Step 4: Consider the 2nd equation x+y =3 i.e. y =3-x (On
transposition) Step 5: For few values of x, calculate values of y and
record them in a table as below:
(repetition of Step 2)
Step 6:
Draw a graph using these (x, y) coordinates to represent the line x+y=3 (repetition
of step 3) Step 7 : Since these two lines are parallel, they never
meet Conclusion : There are no values of x and y
which satisfy both the given equations: 2x+2y = 4 and x+y =3
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Verification:
Refer 2.14 problem 4.
Observations: The graphs of
a1 x+ b1 y = c1 and
a2 x+b2 y = c2
Will be
1. Parallel if they have no
solution
2. Coincident (concurrent) if they
have infinite solutions
3. Intersecting if they have a
unique solution
7.4 Problem 4 :
The cost of manufacturing x
articles is Rs 50+3x. The selling price of x articles is 4x.
on a graph sheet draw two graphs, first for the cost
of manufacturing and second for selling price. From the graph find out the number of
articles to be manufactured and sold to break even(no profit and no loss)
Solution:
The equation for cost of
manufacturing articles is CP=3x+50.
The equation for selling price of
articles is SP=4x
Using the below mentioned scale,
draw two straight lines on a graph sheet representing the above two equations (CP
and SP)
x axis for number of articles : 1
cm = 10 articles
y axis for Rupees 1cm = 10Rs
Table for CP
x à |
0 |
10 |
20 |
CP à |
50 |
80 |
110 |
(x, y) |
(0,5) |
(1,8) |
(2,11) |
Table for SP
x à |
0 |
10 |
20 |
SP à |
0 |
40 |
80 |
(x, y) |
(0,0) |
(1,4) |
(2,18) |
We note that these two straight
lines meet at (5,20)
This means that the cP of manufacturing 50
articles = SP of 50 articles. Thus the break even happens at the manufacture of
50 articiles
Verification:
if x = 50 then CP = 3x+50=150+50=200
With x=50, SP = 4x=200
Thus for x=50, CP=SP=Rs 200 and
hence our solution is correct.
7.4 Summary of learning
No |
Points studied |
1 |
Simultaneous
linear equations can also be solved by drawing graph for each of the lines. |
2 |
The
intersection of the two lines gives the solution. |
3 |
If the
lines are parallel then there is no solution |
4 |
If the
lines are concurrent then there are infinite solutions |