RD Sharma Solutions Chapter 3 Pair of Linear Equation in Two Variables Exercise 3.2 Class 10 Maths
Chapter Name  RD Sharma Chapter 3 Pair of Linear Equation in Two Variables 
Book Name  RD Sharma Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 3.2 Solutions
Solve the following systems of equations graphically:
1. x + y = 3
2x + 5y = 12
Solution
We have
x + y = 3
2x + 5y = 12
Now,
x + y = 3
When y = 0,we have
x = 3
When x = 0,we have
y = 3
Thus, we have the following table giving points on the line x + y= 3
x  0  3 
y  3  0 
Now,
2 + 5y = 12
⇒ y = (122x)/5
When x =1,we have
y = (12 – 2×1)/5 = 2
When x = −4, we have
y = (12 – 2× 4)/5 = 4
Thus, we have the following table giving points on the line 2x + 5y = 12
x  1  4 
y  2  4 
Graph of the equation x + y = 3 and 2x + 5y = 12;
Clearly, two lines intersect at P(1,2).
Hence, x = 1, y = 2 is the solution of the given system of equations.
2. x – 2y = 5
2x + 3y = 10
Solution
We have
x – 2y = 5
2x +3y = 10
Now,
x – 2y = 5
⇒ x = 5 +2y
When y = 0, we have
x = 5 + 2×0 = 5
When y =−2, we have
x = 5 + 2×(−2) = 1
Thus, we have the following table giving points on the line x – 2y = 5
x  5  1 
y  0  2 
Now,
2x + 3y = 10
⇒ 2x = 10 – 3y
⇒ x = (10 – 3y)/2
When y = 0, we have
x =10/2 = 5
When y = 0, we have
x = 10/2 = 5
When y = 2,we have
x = (10 – 3×2)/2 = 2
Thus, we have the following table giving points on the line 2x + 3y = 10
x  5  2 
y  0  2 
Graph of the equation x – 2y = 5 and 2x + 3y = 10:
Clearly, two lines intersect at (5,0).
Hence, x = 5, y = 0 is the solution of the given system of equations.
3. 3x + y + 1 = 0
2x – 3y + 8 = 0
Solution
We have,
3x + y + 1 = 0
2x – 3y + 8 = 0
Now,
3x + y + 1 = 0
⇒ y =  1 – 3x
When x = 0, we have
y = −1
When x = −1, we have
y = 1 – 3×(1) = 2
Thus, we have the following table giving points on the line 3x + y +1 = 0
x  1  0 
y  2  1 
Now,
2x – 3y + 8 = 0
⇒ 2x = 3y – 8
⇒ x = (3y – 8)/2
When y = 0, we have
x = (3×0 – 8)/2 =  4
When y = 2, we have
x = (3×2 – 8)/2 =  1
Thus, we have the following table giving points on the line 2x – 3y + 8 = 0
x  4  1 
y  0  2 
Graph of the equation are:
Clearly, two lines intersect at ( 1, 2).
Hence, x = −1, y = 2 is the solution of the given system of equations.
4. 2x + y – 3 = 0
2x – 3y – 7 = 0
Solution
We have
2x + y – 3 = 0
2x – 3y – 7 = 0
Now,
2x + y – 3 = 0
⇒ y = 3 – 2x
When x = 0,we have
y = 3
When x =1, we have
y =1
Thus, we have the following table giving points on the line 2x + y − 3 = 0
x  0  1 
y  3  1 
Now,
2x – 3y – 7 = 0
⇒ 3y = 2x – 7
⇒ y = (2×5 – 7)/3 = 1
When x = 2,we have
y = (2×2 – 7)/3 = 1
Thus, we have the following table giving points on the line 2x – 3y – 7 = 0
x  2  5 
y  1  1 
Graph of the given equation are
Clearly, two lines intersect at (2, −1).
Hence, x = 2, y= −1 is the solution of the given system of equations.
5. x + y = 6
x – y = 2
Solution
We have.
x + y = 6
x – y = 2
Now,
x + y = 6
⇒ y = 6 –x
When x = 2, we have
y = 4
When x = 3, we have
y = 3
Thus, we have the following table giving points on the line x + y = 6
x  2  3 
y  4  3 
Now,
x – y = 2
⇒ y = x – 2
When x = 0,we have
y = −2
When x = 2, we have
y = 0
Thus, we have the following table giving points on the line x y − = 6
x  0  2 
y  2  0 
Graph of the given equation are
Clearly, two lines intersect at (4,2).
Hence, x = 4, y = 2 is the solution of the given system of equations.
6. x – 2y = 6
3x – 6y = 0
Solution
We have.
x – 2y = 6
3x – 6y = 0
Now,
x – 2y = 6
⇒ x = 6 + 2y
When y = −2, we have
x = 6 + 2×−2 = 2
When y =−3, we have
x = 6 + 2×−3 = 0
Thus, we have the following table giving points on the line x – 2y = 6
X  2  0 
y  2  3 
Now,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y
When y = 0, we have
x = 0
When y =1, we have
x = 2
Thus, we have the following table giving points on the line 3x – 6y = 0
x  0  2 
y  0  1 
Graph of the given equation are
Clearly, two lines are parallel to each other. So, the two lines have no common point Hence, the given system of equations has no solution.
7. x + y = 4
2x – 3y = 3
Solution
We have.
x + y = 4
2x – 3y = 3
Now,
x + y = 4
⇒ x = 4 – y
When y = 0, we have
x = 4
When y = 2, we have
x = 2
Thus, we have the following table giving points on the line x + y = 4
x  4  2 
y  0  2 
Now,
2x – 3y = 3
⇒ 2x = 3y + 3
⇒ x = (3y + 3)/2
When y =1, we have
x = 3
When y = −1,we have
x = 0
Thus, we have the following table giving points on the line 2x – 3y = 3
x  3  0 
y  1  1 
Graph of the given equation are
Hence, x = 3, y = 1 is the solution of the given system of equations.
8. 2x + 3y = 4
x – y + 3 = 0
Solution
We have
2x + 3y = 4
x – y + 3 = 0
Now,
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x = (4 – 3y)/2
When y = 0, we have
x = (4 – 3×2)/2 = 1
When y = 2, we have
x = (4 – 3×2)/2 = 1
Thus, we have the following table giving points on the line 2x + 3y = 4
x  1  2 
y  2  0 
Now,
x – y+ 3 = 0
⇒ x = y – 3
When y = 3, we have
x = 0
When y = 4, we have
x =1
Thus, we have the following table giving points on the line x − y + 3 = 0
x  0  1 
y  3  4 
Graph of the given equation are
Clearly, two lines intersect at (−1, 2).
Hence, x = −1, y = 2 is the solution of the given system of equations.
9. 2x – 3y + 13 = 0
3x – 2y + 12 = 0
Solution
We have,
2x – 3y + 13 = 0
3x – 2y + 12 = 0
Now,
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x = (3y – 13)/2
When y =1, we have
x = (3×1 – 13)/2 =  5
When y = 3, we have
x = (3×3 – 13)/2 = 2
Thus, we have the following table giving points on the line 2x – 3y + 13 = 0
x  5  2 
y  1  3 
Now,
3x – 2y + 12 = 0
⇒ 3x = 2y – 12
⇒ x = (2y – 12)/3
When y = 0,we have
x = (2× 0 – 12)/3 = 14
When y = 3, we have
x = (2×3 – 12)/3 = 2
Thus, we have the following table giving points on the line 3y – 2y + 12=0
x  4  2 
y  0  3 
Graph of the given equations are:
Clearly, two lines intersect at (2, 3)
Hence, x = −2, y = 3 is the solution of the given system of equations.
10. 2x + 3y + 5 = 0
3x + 2y – 12 = 0
Solution
We have,
2x + 3y + 5 = 0
3x + 2y – 12 = 0
Now,
2x + 3y + 5 = 0
⇒ 2x = 3y – 5
⇒ x = (3y – 5)/2
When y =1, we have
x = (3×1 – 5)/2 =  4
When y = −1, we have
x = [3 ×(1) – 5]/2 = 1
Thus, we have the following table giving points on the line 2x + 3y + 5 = 0
x  4  1 
y  1  1 
Now,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
⇒ x = (2y + 12)/3
When y = 0, we have
x = (2×0 + 12)/3 = 4
When y = 3, we have
x = (2×3 + 12)/3 = 6
Thus we have the following table giving points on the line 3x  2y – 12 = 0
x  4  6 
y  0  3 
Graph of the given equations are:
Clearly, two lines intersect at (2, −3).
Hence, x = − 2, y =  3 is the solution of the given system of equations.
Show graphically that each one of the following systems of equations has infinitely many solutions:
11. 2x +3y = 6
4x + 6y = 12
Solution
We have,
2x + 3y = 6
4x + 6y = 12
Now,
2x + 3y = 6
⇒ 2x = 6 – 3y
⇒ x = (6 – 3y)/2
When y = 0, we have
x = 3
When y = 2, we have
x = (6 – 3×2)/2 = 0
Thus, we have the following table giving points on the line 2x + 3y = 6
x  0  3 
y  2  0 
Now,
4x + 6y = 12
⇒ 4x = 12 – 6y
⇒ x = (12 – 6y)/4
When y = 0,we have
x = 3
When y = 2, we have
x =(12 – 6×2)/3 = 0
Thus, we have the following table giving points on the line 4x +6y = 12
x  0  3 
y  2  0 
Graph of the given equations:
Thus, the graphs of the two equations are coincident.
Hence, the system of equations has infinitely many solutions.
12. x – 2y = 5
3x – 6y = 15
Solution
We have,
x – 2y = 5
3x – 6y = 15
Now,
x – 2y = 5
⇒ x = 2y + 5
When y = −1, we have
x = 2(1) + 5 = 3
When y = 0, we have
x = 2 × 0 +5 = 5
Thus, we have the following table giving points on the line x − 2y = 5
x  3  5 
y  1  0 
Now,
3x – 6y = 15
⇒ 3x = 15 + 6y
⇒ x = (15 + 6y)/3
When y =−2, we have
x = [15 + 6(2)]/3 = 1
When y = −3, we have
x = [15 + 6(3)]/3 = 1
Thus, we have the following table giving points on the line 3x  6y = 15
x  1  1 
y  2  3 
Graph of the given equations:
13. 3x + y = 8
6x + 2y = 16
Solution
We have,
3x + y = 8
6x + 2y = 16
Now,
3x + y = 8
⇒ y = 8 – 3x
When x = 2, we have
y = 8, 3 ×2 = 2
When x = 3, we have
y = 8, 3 × 3 = −1
Thus we have the following table giving points on the line 3x + y = 8
x  2  3 
y  2  1 
Now,
6x + 2y = 16
⇒ 2y = 16 – 6x
⇒ y = (16 – 6x)/2
When x =1, we have
y = (16 – 6×1)/2 = 5
When x = 3, we have
y = (16 – 6 ×3)/2 = 1
Thus we have the following table giving points on the line 6x + 2y = 16
x  1  3 
y  5  1 
Graph of the given equations:
Thus, the graphs of the two equations are coincident.
Hence, the system of equations has infinitely many solutions.
14. x + 2y + 11 = 0
3x + 6y + 33 = 0
Solution
We have,
x + 2y + 11 = 0
3x + 6y + 33 = 0
Now,
x – 2y + 11 = 0
⇒ x = 2y – 11
When y = 5, we have
x = 2×5 − 11= −1
When x = 4, we have
x =2 × 4 −11 = −3
Thus we have the following table giving points on the line x – 2y +11 = 0
x  1  3 
y  5  4 
Now,
3x – 6y + 33 = 0
⇒ 3x = 6y – 33
⇒ x = (6y – 33)/3 = 1
When y = 6, we have
x = (6×6 – 33)/3 = 1
When y = 5, we have
x = (6×5 – 33)/2 = 1
Thus we have the following table giving points on the line 3x + 6y +33 = 0
x  1  1 
y  6  5 
Graph of the given equations:
Thus, the graphs of the two equations are coincident,
Hence, the system of equations has infinitely many solutions.
Show graphically that each one of the following systems of equations is inconsistent (i.e., has no solution)
15. 3x – 5y = 20
6x – 10y = 40
Solution
We have,
3x – 5y = 20
6x – 10y = 40
Now
⇒ 3x – 5y = 20
⇒ x = (5y + 20)/3
When y = −1, we have
x = (5×1 + 20)/3 = 5
When y =−4, we have
x = (5×4 + 20)/3 = 0
Thus we have the following table giving points on the line 3x – 5y =20
x  5  0 
y  1  4 
Now
6x – 10y = 40
⇒ 6x = 40 + 10y
⇒ x =(40 + 10y)/6
When y = 4, we have
x = (40 + 10 × 4)/6 = 0
When y =1, we have
x = (40 + 10×1)/6 = 5
Thus we have the following table giving points on the line 6x −y = − 40
x  0  5 
y  4  1 
Graph of the given equations:
Clearly, there is no common point between these two lines
Hence, given system of equations is inconsistent.
16. x – 2y = 6
3x – 6y = 0
Solution
We have
x – 2y = 6
3x – 6y = 0
Now,
x 2y = 6
⇒ x = 6 +2y
When y = 0,we have
x = 6 + 2×0 = 6
When y =−2, we have
x = 6+2 × (−2) = 2
Thus, we have the following table giving points on the line x − 2y = 6
x  6  2 
y  0  2 
Now,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 6y/3
⇒ x = 2y
When y = 0,we have
x = 2×0 = 0
When y =1,we have
x = 2× 1= 2
Thus, we have the following table giving points on the line 3x  6y = 0
x  0  2 
y  0  1 
Graph of the given equations:
We find the lines represented by equations x – 2y = 6 and 3x – 6y = 0 are parallel. So, the two lines have no common point.
Hence, the given system of equations is inconsistent.
17. 2y – x = 9
6y – 3x = 21
Solution
We have
2y – x = 9
6y – 3x = 21
Now,
2y –x = 9
⇒ 2y – 9 = x
⇒ x = 2y – 9
When y = 3,we have
x = 2× 3 – 9 = −3
When y = 4,we have
x = 2 × 4 – 9 = −1
Thus, we have the following table giving points on the line2 9 x x − =
x  3  1 
y  3  4 
Now,
6y – 3x = 21
⇒ 6y – 21 = 3x
⇒ 3x = 6y – 21
⇒ x = 3(2y – 7)/3
⇒ x = 2y – 7
When y = 2, we have
x = 2×2 – 7 = −3
When y = 3, we have
x = 2×3 − 7= −1
Thus, we have the following table giving points on the line 6y – 3x = 21.
x  3  1 
y  2  3 
Graph of the given equations:
We find the lines represented by equations 2y – x = 9 and 6y – 3x = 21 are parallel. So, the two lines have no common point.
Hence, the given system of equations is inconsistent.
18. 3x – 4y – 1 = 0
2x – (8/3)y + 5 = 0
Solution
We have
3x – 4y – 1 = 0
2x – (8/3)y + 5 = 0
Now,
3x – 4y – 1 = 0
⇒ 3x = 1 + 4y
⇒ x = (1 + 4y)/3
When y = 2, we have
x = (1+ 4×2)/3 = 3
When y = −1, we have
x = (1 + 4× 1)/3 = 1
Thus, we have the following table giving points on the line 3 x – 4y − 1= 0
x  1  3 
y  1  2 
Now,
2x – (8/ 3)y + 5 = 0
⇒ (6x – 8y + 15)/3 = 0
⇒ 6x – 8y + 15 = 0
⇒ 6x = 8y – 15
⇒ x = (8y – 15)/6
When y = 0, we have
x = (8×0 – 15)/6 = 2.5
When y = 3, we have
x = (8×3 – 15)/6 = 1.5
Thus, we have the following table giving points on the line 2x – (8/3)y + 5 = 0 .
x  2.5  1.5 
y  0  3 
Graph of the given equations:
We find the lines represented by equations 3x – 4y – 1 = 0 and 2x – (8/3)y + 5 = 0 are parallel. So, the two lines have no common point.
Hence, the given system of equations is inconsistent.
19. Determine graphically the vertices of the triangle, the equations of whose sides are given below:
2y – x = 8
5y – x = 14
(i) y – 2x = 1
y = x
y = 0
(ii) 3x + 3y = 10
Solution
We have
2y – x = 8
5y – x = 14
y – 2x = 1
Now,
2y –x = 8
⇒ 2y = 8 – x
⇒ x = 2y – 8
When y = 2,we have
x = 2× 2−8 = − 4
When y = 4, we have
x = 2×4 – 8 = 0
Thus, we have the following table giving points on the line 2y −x =8.
x  4  0 
y  2  4 
Now,
5y –x = 14
⇒ 5y – 14 =x
⇒ x = 5y – 14
When y = 2, we have
x = 5×2 – 14 = 1
When y = 3, we have
x = 5×3 −14 = 1
Thus, we have the following table giving points on the line 5y – x = 14.
x  4  1 
y  2  3 
We have
y – 2x = 1
⇒ y – 1 = 2x
⇒ x = (y – 1)/2
When y = 3, we have
x = (3 – 1)/2 = 1
When y = −1, we have
x = (1 – 1)/2 = 1
Thus, we have the following table giving points on the line y – 2x = 1.
x  1  1 
y  1  3 
Graph of the given equations:
From the graph of the lines represented by the given equations, we observe that the lines taken in pairs intersect each other at points A(4, 2), B(1, 3) and C(2, 5)
Hence, the vertices of the triangle are A (4,2), B(1, 3) and C(2, 5).
The given system of equations is
y = x
y = 0
3x + 3y = 10
We have,
y = x
When x =1, we have
y =1
When x = −2, we have
y =−2
Thus, we have the following table points on the line y = x
x  1  2 
y  7/3  4/3 
Graph of the given equation:
From the graph of the lines represented by the given equations, we observe that the lines taken in pairs intersect each other at points A(0, 0), B(10/3 , 0) and C(5/3, 5/3).
Hence, the required vertices of the triangle are A(0, 0), B(10/3 , 0) and C(5/3, 5/3)
20. Determine, graphically whether the system of equations x − 2y = 2, 4x − 2y = 5 is consistent or inconsistent.
Solution
We have
x 2y = 2
4x – 2y = 5
Now
x – 2y = 2
⇒ x = 2 + 2y
When y = 0,we have
x = 2 +2 ×0 = 2
When y = −1, we have
x = 2+2 ×(−1) = 0
Thus, we have the following table giving points on the line x – 2y= 2
x  2  0 
y  0  1 
Now,
4x – 2y = 5
⇒ 4x = 5+ 2y
⇒ x = (5 + 2y)/4
When y = 0, we have
x = (5 + 2×0)/4 = 5/4
When y =1, we have
x = (5 + 2×1)/4= 7/4
Thus, we have the following table giving points on the line 4x −2y = 5
x  5/4  7/4 
y  0  1 
Graph of the given equations:
Clearly, the two lines intersect at (1, 1/2).
Hence, the system of equations is consistent.
21. Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not:
(i) 2x − 3y = 6, x + y = l
(ii) 2y = 4x − 6, 2x = y + 3
Solution
We have
2x – 3y = 6
x + y = 1
Now
2x – 3y = 6
⇒ 2x = 6 + 3y
When y = 0, we have
x = (6 + 3y)/2
When y =−2, we have
x = (6 + 3×2)/2 = 0
Thus, we have the following table giving points on the line 2x – 3y = 6
x  3  0 
y  0  2 
Now,
x + y = 1
⇒ x =1 −y
When y =1, we have
x =1 – 1 = 0
When y = 0, we have
x = 1− 0 = 1
Thus, we have the following table giving points on the line x +y =1
x  0  1 
y  1  0 
Graph of the given equations:
We have,
2y = 4x – 6
2x = y + 3
Now,
2y = 4x – 6
⇒ 2y + 6 = 4x
⇒ 4x = 2y + 6
⇒ x = (2y+6)/4
When y = −1, we have
x = (2×1 +6)/4 = 1
When y = 5, we have
x = (2×5 + 6)/4 = 4
Thus, we have the following table giving points on the line 2y = 4x − 6
x  1  4 
y  1  5 
Now,
2x = y + 3
⇒ x = (y + 3)/2
When y =1, we have
x = (1 + 3)/2 = 2
When y = 3, we have
x = (3 +3)/2 = 3
Thus, we have the following table giving points on the line 2x = y +3
x  2  3 
y  1  3 
Graph of the given equations:
We find the graphs of the two equations are coincident,
∴ Hence, the system of equations has infinity many solutions
22. Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.
(i) 2x – 5y + 4 = 0
2x + y – 8 = 0
(ii) 3x + 2y = 12
5x – 2y = 4
(iii) 2x + y – 11 = 0
x – y – 1 = 0
(iv) x +2y – 7 = 0
2x – y – 4 = 0
(v) 3x + y – 5 =0
2x – y – 5 = 0
(vi) 2x – y – 5 = 0
x – y – 3 = 0
Solution
We have
2x – 5y + 4 = 0
2x + y – 8 = 0
Now,
2x – 5y + 4 = 0
⇒ 2x = 5y − 4
⇒ x = (5y – 4)/2
When y = 2,we have
x = (5×2 – 4)/2 = 3
When y = 4, we have
x = (5×4 – 4)/2 = 8
Thus, we have the following table giving points on the line 2x – 5y + 4 = 0
x  3  8 
y  2  4 
Now,
2x + y – 8 = 0
⇒ 2x = 8 – y
⇒ x = (8 – y)/2
When y = 4,we have
x = (8 – 4)/2 = 2
When y = 2,we have
x = (8 – 2)/2 = 3
Thus, we have the following table giving points on the line 2x −5y + 4 = 0
x  3  8 
y  2  4 
Graph of the given equations:
Clearly, two intersect at P(3, 2)
Hence, x =2, y = 3 is the solution of the given system of equations.
We also observe that the lines represented by 2x – 5y + 4 and 2x + y – 8 = 0 meet y – axis at A(0, 4/5) and B(0, 8) respectively.
We have,
3x + 2y = 12
5x – 2y = 4
Now,
3x + 2y = 12
⇒ 3x = 12 – 2y
⇒ x = (12 – 2y)/3
When y = 3, we have
x = (12 – 2×3)/3 = 2
When y =−3, we have
x = (12 – 2×3)/3 = 6
Thus, we have the following table giving points on the line 3x + 2y = 12
x  2  6 
y  3  3 
Now,
5x – 2y = 4
⇒ 5x = 4 +2y
⇒ x = (4 + 2y)/5
When y = 3,we have
x = (4 + 2×3)/5 = 2
When y =−7, we have
x = (4 + 2×7)/5 = 2
Thus, we have the following table giving points on the line 5x – 2y = 4
x  2  2 
y  3  7 
Graph of the given equation
Clearly, two intersect at p (2,3)
Hence, x =2, y = 3 is the solution of the given system of equations.
We also observe that the lines represented by 3x + 2y = 12 and 5x – 2y = 4 meet yaxis at A(0, 6) and B (0, 2) respectively.
We have,
2x + y – 11 = 0
x – y – 1= 0
Now,
2x + y – 11= 0
⇒ y = 11 – 2x
When x = 4, we have
y = 11 − 2×4 = 3
When x = 5, we have
y =11 − 2×5 = 1
Thus, we have the following table giving points on the line 2x + y −11 =0
x  4  5 
y  3  1 
Now,
x –y – 1 = 0
⇒ x – 1 = y
⇒ y =x − 1
When x = 2, we have
y = 2− 1= 1
When x = 3, we have
y = 3− 1= 2
Thus, we have the following table giving points on the line x −y −1 = 0
x  2  3 
y  1  2 
Graph of the given equation
We have,
2x + y – 11 = 0
x – y – 1 = 0
Now,
2x + y – 11 = 0
⇒ y = 11 – 2x
When x = 4, we have
y = 11−2 ×4 = 3
When x = 5, we have
y =11 − 2×5 = 1
Thus, we have the following table giving points on the line 2x + y − 11= 0
x  4  5 
y  3  1 
Now,
x – y – 1 = 0
⇒ x −1 = y
⇒ y = x −1
When x = 2, we have
y = 2 – 1 = 1
When x = 3, we have
y =3 − 1= 2
Thus, we have the following table giving points on the line x –y – 1= 0
x  2  3 
y  1  2 
Graph of the given equations:
Clearly, two intersect at P(4,3)
Hence, x = 4, y = 3 is the solution of the given system of equations.
We also observe that the lines represented by 2x + y – 11= 0 and x –y – 1 = 0 meet yaxis at, A(0, 11) and B (0, 1) respectively.
We have, x + 2y – 7 = 0
Now,
2x – y – 4 = 0
x + 2y – 7 = 0
x = 7 – 2y
When
y = 1, x = 5
y = 2, x = 3
x  5  3 
y  1  2 
Also,
2x – y – 4 = 0
y = 2x – 4
x  2  0 
y  0  4 
From the graph, the solution is A(3, 2).
Also, the coordinates of the points where the lines meet the yaxis are
B(0, 3.5) and C(0, − 4).
We have
3x + y – 5= 0
2x – y – 5 = 0
Now,
3x + y – 5 = 0
⇒ y = 5 – 3x
When x =1, we have
y = 5−3 × 1= 2
When x = 2, we have
y = 5 − 3×2 = − 1
Thus, we have the following table giving points on the line 3x +y −5 = 0
x  1  2 
y  2  1 
Now,
2x – y – 5 = 0
⇒ 2x – 5 = y
⇒ y = 2x − 5
When x = 0, we have
y =−5
When x = 2, we have
y = 2×2 −5 = −1
Thus, we have the following table giving points on the line 2x – y −5 = 0
x  0  2 
y  5  1 
Graph of the given equations:
Clearly, two intersect at P(2, −1).
Hence, x = 2, y = −1 is the solution of the given system of equations.
We also observe that the lines represented by 3x + y – 5 = 0 and 2x –y − 5= 0 meet yaxis at A(0,5) and B(0, −5 ) respectively.
We have,
2x – y – 5 = 0
x – y – 3 = 0
Now,
2x – y – 5 = 0
⇒ 2x – 5 = y
⇒ y =2x − 5
When x =1, we have
y = 2× 1−5 = − 3
When x = 2, we have
y =2 × 2− 5= −1
Thus, we have the following table giving points on the line 2x −y – 5 = 0
x  1  2 
y  3  1 
Now,
x –y – 3 = 0
⇒ x − 3 = y
⇒ y = x − 3
When x = 3, we have
y = 3 − 3= 0
When x = 4, we have
y = 4− 3= 1
Thus, we have the following table giving points on the line x − y− 3= 0
X  3  4 
Y  0  1 
Graph of the given equations:
Clearly, two intersect at P(2, −1).
Hence, x = 2, y= −1 is the solution of the given system of equations.
We also observe that the lines represented by 2x – y – 5 = 0 and x – y −3 =0 meet yaxis at A(0, 5) and B(0, −3) respectively.
23. Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are:
(i) y = x
y = 2x
y + x = 6
(ii) y =x
3y = x
x +y = 8
Solution
The system of the given equations is,
y = x
y = 2x
y + x = 6
Now,
y = x
When x = 0, we have
y = 0
When x =−1, we have
y =−1
Thus, we have the following table:
x  0  1 
y  0  2 
We have
y = 2x
When x = 0, we have
y = 2×0 = 0
When x =−1, we have
y = 2× −1 = − 2
Thus, we have the following table:
x  0  1 
y  0  2 
We have
y +x = 6
⇒ y = 6 – x
When x = 2, we have
y = 6−2 = 4
When x = 4, we have
y = 6−4 = 2
Thus, we have the following table:
x  2  4 
y  4  2 
Graph of the given system of equations:
From the graph of the three equations, we find that the three lines taken in pairs intersect each other at points A(0,0) , B(2, 4) and C(3, 3).
Hence, the vertices of the required triangle are (0,0) , (2,4) and (3,3).
The system of the given equations is,
y = x
3y = x
x + y = 8
Now,
y = x
⇒ x = y
When y = 0,we have
x = 0
When y = −3, we have
x = −3
Thus, we have the following table.
x  0  3 
y  0  3 
We have
3y =x
⇒ x = 3y
When y = 0, we have
x = 3× 0= 0
When y = −1, we have
y = 3× −1 = −3
Thus, we have the following table:
x  0  3 
y  0  1 
We have
x + y = 8
⇒ x = 8 − y
When y = 4, we have
x =8 − 4= 4
When y = 5, we have
x =8 − 5= 3
Thus, we have the following table:
x  4  5 
y  4  3 
Graph of the given system of equations:
From the graph of the three equations, we find that the three lines taken in pairs intersect each other at points A(0,0) , B(4,4) and C(6, 2).
Hence, the vertices of the required triangle are A(0,0) , B(4,4) and C(6, 2).
24. Solve the following system of linear equations graphically and shade the region between the two lines and xaxis:
(i) 2x + 3y = 12
x – y = 1
(ii) 3x + 2y – 4 = 0
2x – 3y – 7 = 0
(iii) 3x + 2y – 11 = 0
2x – 3y + 10 = 0
Solution
The system of given equations is
2x + 3y = 12
x – y = 1
Now,
2x + 3y = 12
⇒ 2x =12 − 3y
⇒ x = (12 – 3×2)/2 = 3
When y = 2, we have
x = (12 – 3×2)/2 = 3
When y = 4,we have
x = (12 – 3×4)/2 = 0
Thus, we have the following table:
x  0  3 
y  4  2 
We have,
x – y = 1
⇒ x =1 +y
When y = 0, we have
x =1
When y =1,we have
x = 1+ 1= 2
Thus, we have the following table:
x  1  2 
y  0  1 
Graph of the given system of equations:
Clearly, the two lines intersect at P(3, 2) .
Hence, x =3, y= 2 is the solution of the given system of equations. The system of the given equations is,
3x + 2y – 4 = 0
2x – 3y – 7 = 0
Now,
3x + 2y – 4 = 0
⇒ 3x = 4 −2y
⇒ x = (4 – 2y)/3
When y = 5, we have
x = 4 – (2×5)/3 =  2
When y = 8, we have
x = (4 – 2×8)/3 =  4
Thus, we have the following table:
x  2  4 
y  5  8 
We have,
2x – 3y – 7 = 0
⇒ 2x = 3y + 7
⇒ x = (3y + 7)/2
When y =1, we have
x = (3×1 + 7)/2 = 5
When y = −1, we have
x = (3 × 1 + 7)/2 = 2
Thus, we have the following table:
x  5  2 
y  1  1 
Graph of the given system of equations:
Clearly, the two lines intersect at P(2, 1).
Hence, x =2 y = 1 is the solution of the given system of equations.
The system of the given equations is,
3x + 2y – 11 = 0
Now,
3x + 2y  11 = 0
⇒ 3x = 11 – 2y
⇒ x = (11 – 2y)/3
When y =1, we have
x = (11 – 2×1)/3 = 3
When y = 4, we have
x = (11 – 2×4)/3 = 1
Thus, we have the following table:
x  3  1 
y  1  4 
We have,
2x – 3y + 10 = 0
⇒ 2x = 3y − 10
⇒ x = (3y – 10)/2
When y = 0, we have
x = (3×0 – 10 )/2 = 5
When y = 2, we have
x = (3×2 – 10)/2 = 2
Thus, we have the following table:
x  5  2 
y  0  2 
Graph of the given system of equations:
Clearly, the two lines intersect at P(1,4).
Hence, x = 1, y = 4 is the solution of the given system of equations.
25. Draw the graphs of the following equations on the same graph paper:
2x + 3y = 12
x – y = 1
Solution
The system of the given equations is
2x + 3y = 12
x – y = 1
Now,
2x + 3y = 12
⇒ 2x =12 − 3y
⇒ x =(12 3y)/2
When y = 0, we have
x = (12 – 3×0)/2 = 6
When y = 2, we have
x = (12 – 3×2)/2 = 3
Thus, we have the following table:
x  6  3 
y  0  2 
We have
x –y = 1
⇒ x = 1 + y
When y = 0, we have
x =1
When y = −1, we have
x = 1 −1 = 0
Thus, we have the following table:
x  1  0 
y  0  1 
Graph of the given system of equations:
Clearly, the two lines intersect at A(3, 2).
We also observe that the lines represented by the equations 2x + 3y = 12 and x – y =  1 meet yaxis at B(0,  1) and C (0, 4).
Hence, the vertices of the required triangle are A (3,2) , B(0, 1 ) and C (0, 4).
26. Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x axis and shade the triangular area. Calculate the area bounded by these lines and xaxis.
Solution
The given system of equations is
x –y + 1= 0
3x + 2y – 12 = 0
Now,
x –y + 1 = 0
⇒ x = y − 1
When y = 3, we have
x = 3− 1= 2
When y = −1, we have
x = −1 −1 = −2
Thus, we have the following table:
x  2  2 
y  3  1 
We have
3x + 2y – 12 = 0
⇒ 3x = 12− 2y
⇒ x = (12 – 2y)/3
When y = 6, we have
x = (12 – 2×6)/3 = 0
When y = 3,we have
x = (12 – 2×3)/3 = 2
Thus, we have the following table:
x  0  2 
y  6  3 
Graph of the given system of equations:
Clearly, the two lines intersect at A(2,3)
We also observe that the lines represented by the equations
x  y + 1 = 0 and 3x + 2y − 12 = 0 meet xaxis at B(1, 0) and C (4,0) respectively.
Thus, x = 2, y =3 is the solution of the given system of equations.
Draw AD perpendicular from A on xaxis.
Clearly, we have
AD = y −coordinate of point A(2,3)
⇒ AD = 3and, BC = 4 – ( 1) = 4 + 1 = 5
27. Solve graphically the system of linear equations:
4x – 3y + 4 = 0
4x + 3y – 20 = 0
Find the area bounded by these lines and xaxis.
Solution
The given system of equation is
4x – 3y + 4 = 0
4x + 3y – 20 = 0
Now,
4x – 3y + 4 = 0
⇒ 4x = 3y − 4
⇒ x = (3y – 4)/4
When y = 0,we have
x = (3×0 – 4)/4 =  1
When y = 4, we have
x = (3×4 – 4)/ 4 = 2
Thus, we have the following table:
x  2  1 
y  4  0 
We have
4x + 3y – 20 = 0
⇒ 4x = 20 – 3y
⇒ x = (20 – 3y)/4
When y = 0,we have
x = (20 – 3×0)/4 = 5
When y = 4,we have
x = (20 – 3×4)/ 4 = 2
Thus, we have the following table:
x  5  2 
y  0  4 
Graph of the given system of equation:
Clearly, the two lines intersect at A(2,4 .)Hence x = 2, y = 4 is the solution of the given system of equations.
We also observe that the lines represented by the equations
4x – 3y + 4 = 0 and 4x + 3y 20= 0 meet xaxis at B (1, 0) and C(5,0) respectively.
Thus, x = 2, y = 4 is the solution of the given system of equations.
Draw AD perpendicular from A on xaxis.
Clearly, we have
AD = y −coordinate of point A(2, 4)
⇒ AD = 4 and, BC = 5 – (−1) = 5 +1 = 6
∴ Area of the shaded region = Area of Î”ABC
⇒ Area of the shaded region = ½ (Base × Height)
= ½(BC ×AD)
= ½ × 6 ×4
= 6 ×2
= 12 sq. units
∴ Area of shaded region =12sq. units
28. Solve the following system of linear equations graphically:
3x + y – 11 = 0
x –y – 1 = 0
Shade the region bounded by these lines and y axis. Also, find the area of the region bounded by these lines and yaxis.
Solution
The given system of equation is
3x + y – 11 = 0
x –y – 1 = 0
Now,
3x + y – 11 = 0
⇒ y = 11 – 3x
When x = 0, we have
y =11 − 3×0 = 11
When x = 3 we have
y = 11− 3× 3= 2
Thus, we have the following table:
x  0  3 
y  11  2 
We have
x – y  1 = 0
⇒ x − 1= y
⇒ y = x − 1
When x = 0, we have
y =0 −1 = − 1
When x = 3, we have
y =3 − 1= 2
Thus, we have the following table:
x  0  3 
y  1  2 
Graph of the given system of equations:
Clearly, the two lines intersect at A(3, 2). Hence x = 3, y = 2 is the solution of the given system of equations.
We so observe that the lines represented by the equations 3x + y 11 = 0 and x  y – 1 = 0 meet yaxis at B (0, 11) and C(0,  1) respectively.
Thus, x = 3, y = 2 is the solution of the given system of equations.
Draw AD perpendicular from A on yaxis.
Clearly, we have
AD = x − coordinate of point A(3,2)
⇒ AD = 3and, BC = 11− (−1) =11 +1 = 12
∴ Area of the shaded region = Area of Î”ABC
⇒ Area of the shaded region = ½(base × height)
= ½ × (BC ×AD)
= ½ ×12 × 3
= 6×3
=18sq. units
∴Area of the shaded region=18sq. units
29. Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system:
(i) 2x – y = 2
4x – y = 8
(ii) 2x – y = 2
4x – y = 8
(iii) x + 2y = 5
2x – 3y = 4
(iv) 2x + 3y = 8
x – 2y = 3
Solution
The given system of equation is
2x – y = 2
4x –y = 8
Now,
2x + y = 2
⇒ 2x = y + 2
⇒ x = (y + 2)/2
When y = 0, we have
x = (0 + 2)/2 = 1
When y = 2, we have
x = (2 + 2)/2 = 2
Thus, we have the following table:
x  1  2 
y  0  2 
We have,
4x  y = 8
⇒ 4x = y + 8
⇒ x = (y + 8)/4
When y = 0,we have
x = (0+8)/4 = 2
When y = −4 we have
x = (4 +8)/4 = 1
Thus, we have the following table:
x  2  1 
y  0  4 
Graph of the given system of equations:
Clearly, the two lines intersect at A(2,2).
Hence x = 2, y = 2 is the solution of the given system of equations.
We so observe that the lines represented by the equations 2x + y = 6 and x 2y =  2 meet xaxis at B(3,0) and C (−2,0 )respectively.
The system of the given equations is
2x + y = 6
x – 2y =  2
Now,
2x + y = 6
⇒ x = (6 – y)/2
When y = 0,we have
x = (6 – 0)/2 = 3
When y = 2,we have
x = (6 – 2)/2 = 2
Thus, we have the following table:
x  3  2 
y  0  2 
We have,
x – 2y = 2
⇒ x = 2y – 2
When y = 0, we have
x = 2×0 – 2 = 2
When y =1,we have
x =2 × 1−2 = 0
Thus, we have the following table:
x  2  0 
y  0  1 
Graph of the given system of equations:
Clearly the two lines intersect at A(3, 4 .)
Hence x = 3, y = 4 is the solution of the given system of equations.
We so observe that the lines represented by the equations 2x – y = 2 and 4x – y = 8 meet xaxis at B(1, 0) and C (2,0) respectively
The system of the given equations is
x + 2y = 5
2x – 3y =  4
Now,
x + 2y = 5
⇒ x = 5 −2y
When y = 2, we have
x = 5−2×2 = 1
When y = 3,we have
x = 5− 2×3 = −1
Thus, we have the following table:
x  1  1 
y  2  3 
We have,
2x – 3y =  4
⇒ 2x = 3y − 4
⇒ x =(3y – 4)/2
When y = 0,we have
x = (3×0 – 4)/2 =  2
When y = 2, we have
x = (3×2 – 4)/2 = 1
Thus, we have the following table:
x  2  1 
Y  0  2 
Graph of the given system of equations:
The given system of equation is
2x + 3y = 8
x – 2y =  3
Now,
2x + 3y = 8
⇒ 2x =8 −3y
⇒ x = (8 – 3y)/2
When y = 2, we have
x = (8 – 3×4)/2 = 1
When y = 4,we have
x = (8 – 3×4)/2 =  2
Thus, we have the following table:
x  1  2 
y  2  4 
We have,
x – 2y = 3
⇒ x = 2y – 3
When y = 0,we have
x = 2×0 – 3 =  1
When y =1,we have
x =2 ×1 −3 = − 1
Thus, we have the following table:
x  3  1 
y  0  1 
Graph of the given system of equations:
Clearly, the two lines intersect at A(1,2).
Hence x = 1, y = 2 is the solution of the given system of equations.
We also observe that the lines represented by the equations 2x + 3y = 8 and x – 2y =  3 meet xaxis at B(4,0) and C (−3, 0 )respectively.
30. Draw the graphs of the following equations:
2x – 3y + 6= 0
2x + 3y – 18 = 0
y – 2 = 0
Find the vertices of the triangle so obtained. Also, find the area of the triangle.
Solution
The given system of equation is
2x – 3y + 6 = 0
2x – 3y – 18 = 0
y – 2 = 0
Now,
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
⇒ x = (3y – 6)/2
When y = 0,we have
x = (3×0 – 6)/2 =  3
When y = 2,we have
x = (3×2 – 6 )/2 = 0
Thus, we have the following table:
x  3  0 
y  0  2 
We have,
2x + 3y – 18 = 0
⇒ 2x = 18 – 3y
⇒ x = (18 – 3y)/2
When y = 2,we have
x = (18 – 3×2)/2 = 6
When y = 6, we have
x = (18 – 3 × 6)/2 = 0
Thus, we have the following table:
x  6  0 
y  2  6 
We have
y – 2 = 0
⇒ y = − 2
Graph of the given system of equations:
From the graph of the three equations, we find that the three lines taken in pairs intersect each other at points A(3, 4), B(0, 2) and C (6,2).
Hence, the vertices of the required triangle are (3, 4), (0, 2) and (6,2).
From graph, we have
AD = 4 – 2 = 2
BC = 6 – 0 = 6
Area of Î”ABC = ½ (Base ×Height)
= ½ ×BC ×AD
= ½ ×6 ×2
= 6 sq. units
∴Area of Î”ABC = 6 sq units.
31. Solve the following system of equations graphically:
2x – 3y + 6 = 0
2x + 3y – 18 = 0
Also, find the area of the region bounded by these two lines and yaxis.
Solution
The given system of equation is
2x – 3y + 6 = 0
2x + 3y – 18 = 0
Now,
2x – 3y + 6 = 0
⇒ 2x +6 = 3y
⇒ 3y = 2x + 6
⇒ y = (2x + 6)/3
When x = 0, we have
y = (2×0 + 6)/3 = 2
When x = −3,we have
y = (2×  3 + 6)/3 = 0
Thus, we have the following table:
x  0  3 
y  2  6 
Graph of the given system of equations:
Clearly, the two lines intersect at A(3, 4 ).
Hence, x = 3, y = 4 is the solution of the given system of equations.
We also observe that the lines represented by the equations 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0 meet yaxis at B(0, 2) and C (0,6 ) respectively.
Thus, x = 3, y = 4 is the solution of the given system of equations.
Draw AD perpendicular from A on yaxis.
Clearly, we have,
AD = x −coordinate of point A(3,4)
⇒AD = 3and, BC = 6 – 2 = 4
Area of the shaded region = Area of Î”ABC
Area of the shaded region =1/2 × (Base ×Height)
= ½ (BC ×AD)
= ½ ×4×3
= 2 ×3 =6 sq.units
∴Area of the region bounded by these two lines and yaxis is 6 sq. units.
32. Solve the following system of linear equations graphically:
4x – 5y – 20 = 0
3x + 5y – 15= 0
Determine the vertices of the triangle formed by the lines representing the above equation and the yaxis.
Solution
The given system of equation is
4x – 5y – 20 = 0
3x + 5y – 15 = 0
Now,
4x – 5y – 20 = 0
⇒ 4x = 5y + 20
⇒ x = (5y + 20)/4 = 5
When y = 0, we have
x = (5×0+20)/4 = 5
When y =−4,we have
x = (5×4 + 20)/4 = 0
Thus, we have the following table:
x  5  0 
y  0  4 
We have,
3x + 5y – 15 = 0
⇒ 3x = 15 – 5y
⇒ x = (15 – 5y)/3
When y = 0,we have
x = (15 – 5×0)/3 = 5
When y = 3,we have
x = (15 – 5 ×3)/3 = 0
Thus, we have the following table:
x  5  0 
y  0  3 
Graph of the given system of equations:
Clearly, the two lines intersect at A(5,0). Hence, (x – 5), (y −0) is the solution of the given system of equations.
We also find that the two lines represented by the equations 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0 meet yaxis at B(0,  4) and C (0,3) respectively,
∴The vertices of the required triangle are (5,0) , (0, 4) and (0,3).
33. Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and yaxis. Calculate the area of the triangle so formed.
Solution
5x – y = 5
⇒ y = 5x – 5
Three solutions of this equation can be written in a table as follows:
x  0  1  2 
y  5  0  5 
3x – y = 3
⇒ y = 3x − 3
x  0  1  2 
y  3  0  3 
The graphical representation of the two lines will be as follows:
It can be observed that the required triangle is Î”ABC.
The coordinates of its vertices are A(1,0), B(0, 3), C (0 −5 ).
Concept insight: In order to find the coordinates of the vertices of the triangle so formed. Find the points where the two lines intersects the yaxis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with yaxis is taken and not with xaxis, this is became the question says to find the triangle formed by the two lines and the yaxis.
34. Form the pair of linear equations in the following problems, and find their solution graphically:
(i) 10 students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and a pen.
(iii)Champa went to a ‘sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased.” Help her friends to find how many pants and skirts Champa bought.
Solution
(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x – y = 4
x + y = 10
⇒ x =10 – y
Three solutions of this equation can be written in a table as follows:
x  4  5  6 
y  6  5  4 
x – y = 4
⇒ x = 4 + y
Three solutions of this equation can be written in a table as follows:
x  5  4  3 
y  1  0  1 
The graphical representation is as follows:
From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So. x = 7and y = 3.
Thus, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of one pencil and one pen be Rs x and Rs y respectively. According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46
5x + 7y = 50
⇒ x = (50 – 7y)/5
Three solutions of this equation can be written in a table as follows:
x  3  10  4 
y  5  0  10 
7x + 5y = 46
⇒ x = (46 – 5y)/7
Three solutions of this equation can be written in a table as follows:
x  8  3  2 
y  2  5  12 
The graphical representation is as follows:
From the graph. It can be observed that the two lines intersect each other at the point (3, 5).
So. x = 3 and y = 5.
Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.
(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:
y = 2x  2 ...(1)
y = 4x  4 ...(2)
Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.
They are given in table
x  2  0 
y = 2x2  2  2 
x  0  1 
y 4x4  4  0 
Plot the point and draw the lines passing through them to represent the equation, as shown in fig.
The t lines intersect at the point(1,0). So. x −1,y = 0 is the required solution of the pair of linear equations, i.e, the number of pants she purchased island she did not buy any skirt
Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it on the graph paper, mark the points accurately and neatly using a sharp pencil. Also take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.
35. Solve the following system of equations graphically:
Shade the region between the lines and the yaxis
(i) 3x – 4y = 7
5x + 2y = 3
(ii) 4x – y = 4
3x + 2y = 14
Solution
The given system of equations is
3x – 4y = 7
5x + 2y = 3
Now,
3x – 4y = 7
⇒ 3x – 7 = 4y
⇒ 4y = 3x −7
⇒ y = (3x – 7)/4
When x =1,we have
y = (3×1 – 7)/4 = 1
When x = −3,we have
y = (3×3 – 7)/4 = 4
Thus, we have the following table:
x  1  3 
y  1  4 
We have,
5x + 2y = 3
⇒ 2y = 3 – 5x
⇒ y = (3 – 5x)/2
When x =1,we have
y = (3 – 5×1)/2 =  1
When x = 3, we have
y = (3 – 5×3)/2 =  6
Thus, we have the following table:
x  1  3 
y  1  6 
Graph of the given system of equations:
Clearly, the two lines intersect at A(1, −1). Hence, x = 1, y = − 1, is the solution of the given system of equations.
We also observe that the required shaded region is Î”ABC
The given system of equations is
4x – y = 4
3x + 2y = 14
Now,
4x – y = 4
⇒ 4x – 4 = y
⇒ y = 4x – 4
When x = 0, we have
y = 4 × 0 – 4 = − 4
When x =−1, we have
y = 4 × 0 −4= − 4
Thus, we have the following table:
x  0  1 
y  4  8 
We have,
3x + 2y = 14
⇒ 2y = 14 – 3x
⇒ y =(14 – 3x)/2
When x = 0, we have
y = (14 – 3×0)/2 = 7
When x = 4,we have
y = (14 – 3×4)/2 = 1
Thus, we have the following table:
x  0  4 
y  7  1 
Graph of the given system of equations:
Clearly, the two lines intersect at A(2,4). Hence, x = 2, y = 4 is the solution of the given system of equations.
We also observe Î”ABCis the required shaded region.
36. Represent the following pair of equations graphically and write the coordinates of points where the lines intersects yaxis
x+ 3y = 6
2x – 3y = 12
Solution
The given system of equations is
x + 3y = 6
2x – 3y = 12
Now,
x + 3y = 6
⇒ 3y = 6 – x
⇒ y = (6 – x)/3
When x = 0, we have
y = (6 0)/3 = 2
When x = 3, we have
y = (6 3)/3 = 1
Thus we have the following table :
X  0  3 
Y  2  1 
we have,
2x + 3y = 12
⇒ 2x – 12 = 3y
⇒ y = (2x – 12)/3
When x = 0, we have
y = (2×0 – 12)/3 = 4
When x = 6, we have
y = (2×6 – 12)/3 = 0
Thus, we have the following table :
X  0  6 
Y  4  0 
Graph of the given system of equations:
We observe that the liens represented by the equations x + 3y – 6 and 2x – 3y – 12 meet y – axis at B(0,2) and C(0,  4) respectively.
Hence, the required co ordinates are (0,2) and (0, 4).
37. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) Parallel lines (iii) coincident lines
Solution
(i)For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0 , to be intersecting, we must have a_{1}/a_{2} ≠ b_{1}/b_{2}
So, the other linear equation can be 5x + 6y – 16 = 0
As a_{1}/a_{2} = 2/5 , b_{1}/b_{2} = 3/6 = 1/2, c_{1}/c_{2} = 8/16 = 1/2
(ii) For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0, to be parallel we must have a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
So, the other linear equation can be 6x + 9y + 24 = 0 ,
As a_{1}/a_{2} = 2/6 = 1/3 , b_{1}/b_{2} = 3/9 = 1/3, c_{1}/c_{2} = 8/24 = 1/3
(iii) For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0, to be coincident, we must have a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
So, the other linear equation can be 6x + 9y + 24 = 0,
As a_{1}/a_{2} = 2/8 = 1/4, b_{1}/b_{2} = 3/12 = 1/4, c_{1}/c_{2} = 8/32 = ¼
Concept insight : In orders to answer such type of problems, just remember the conditions for two lines to be intersecting parallel, and coincident .
This problem will have multiple answers as their can be marry equations satisfying the required conditions.