Exercise 3.1
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a 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 that of one pen.
Answer:
(i) Let number of boys = x
Let number of girls = y
According to given conditions, we have
x + y = 10
And, x = 10 – y
putting y=0,5,10,we get,
X=10-0=10
X=10-5=5,
X=10-10=0
Number of girls is 4 more than number of boys ……..
Given, so,
Y=x+4
putting x=-4,0,4 we get,
Y=-4+4=0
Y=0+4
Y=4+4=8
We plot the points for both of the equations to find the solution.
ii.Let the cost of one pencil=Rs.X
and Let the cost of one pen=Rs.Y
According to the given conditions, we have:
=5x + 7y = 50
=5x=50-7y
=x=10-7/5y
Now
7 Pencils and 5 pens together cost Rs. 46
7x+5y =46
5y =46-7x
Y= 9.2 -1.4x
Putting x =0, 2, 4 we get,
2. On comparing the ratios a1/a2,b1/b2and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x − 4y + 8 = 0 7x + 6y – 9 = 0
(ii)9x + 3y + 12 = 0 18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0 2x – y + 9 = 0
Ans:
(i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0
Comparing equation 5x − 4y + 8 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
We get,
= a1 = 5, b1= -4, c1 = 0
= a2 = 7, b2 = 6, c2 = -9
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
Comparing equation 9x + 3y + 12 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Hence, lines are coincident.
(iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0
Comparing equation 6x − 3y + 10 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Hence, lines are parallel to each other
3. On comparing the ratios a1/a2,b1/b2and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5, 2x − 3y = 8 (ii) 2x − 3y = 7, 4x − 6y = 9 (iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14 (iv) 5x − 3y = 11, −10x + 6y = −22
Ans:
(i) 3x + 2y = 5, 2x − 3y = 7
Comparing equation 3x + 2y = 5 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(ii) 2x − 3y = 8, 4x − 6y = 9
Comparing equation 2x − 3y = 8 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore, these linear equations are parallel to each other and have no possible solution.in
And,pair of linear euations is inconsistent
(iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14
Comparing equation 3x/2 + 5y/3 = 7, with a1x + b1y + c1 = 0and 9x − 10y = 14 with a2x + b2y + c2 = 0,
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 5x − 3y = 11, −10x + 6y = −22
Comparing equation 5x − 3y = 11 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore these pair of lines have infinite number of solutions
And, pair of linear equations is consistent
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x − 3y = 16 (iii) 2x + y = 6, 4x − 2y = 4 (iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
(i) x + y = 5, 2x + 2y = 10
Therefore these pair of lines have infinite number of solutions and pair of linear equation is consistent.
X + Y = 5
X = 5-Y
Putting Y= 1,2,3 we get,
And 2X + 2Y =10
X =10-2Y/2
Putting Y = 1,2,3 we get
(ii) x – y = 8, 3x − 3y = 16
Therefore, these linear equations are parallel to each other and have no possible solution.in
And,pair of linear euations is inconsistent
(iii) 2x + y = 6, 4x − 2y = 4
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden graphically.
Let the length of the garden be x m and its breadth be y m
Then according to the question,
We have:
x = y + 4
x –y =4……..(i)
And x + y =6…….(ii)
For graphical representation:
From equation (i) x – y = 4
We get
From equation (ii) x + y = 36
We get
Plotting the points of each table of solutions, we obtain the graph of two lines interesting at (20, 16).
Therefore, the solution is x= 20 and y= 16
Thus the length of the garden is 20m and its breadth is 16m.
6. 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
(i) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;
(a1/a2) ≠ (b1/b2)
Thus, another equation could be 2x – 7y + 9 = 0, such that;
(a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7
Clearly, you can see another equation satisfies the condition.
(ii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;
(a1/a2) = (b1/b2) ≠ (c1/c2)
Thus, another equation could be 6x + 9y + 9 = 0, such that;
(a1/a2) = 2/6 = 1/3
(b1/b2) = 3/9= 1/3
(c1/c2) = -8/9
Clearly, you can see another equation satisfies the condition.
(iii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;
(a1/a2) = (b1/b2) = (c1/c2)
Thus, another equation could be 4x + 6y – 16 = 0, such that;
(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2
Clearly, you can see another equation satisfies the condition.
7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Ans: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.
For, x – y + 1 = 0 or x = -1+y
For, 3x + 2y – 12 = 0 or x = (12-2y)/3
Exercise 3.2
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
(s/3) + (t/2) = 6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2 x+√3 y = 0
√3 x-√8 y = 0
(vi) (3x/2) – (5y/3) = -2
(x/3) + (y/2) = (13/6)
Soln:
(i) Given,
x + y = 14 and x – y = 4 are the two equations.
From the 1st equation, we get,
x = 14 – y
Now, substitute the value of x to the second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the exact value of x.
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5
(ii) Given,
s – t = 3 and (s/3) + (t/2) = 6 are the two equations.
From the 1st equation, we get,
s = 3 + t ________________(1)
Now, substitute the value of s to the second equation to get,
(3+t)/3 + (t/2) = 6
⇒ (2(3+t) + 3t )/6 = 6
⇒ (6+2t+3t)/6 = 6
⇒ (6+5t) = 36
⇒5t = 30
⇒t = 6
Now, substitute the value of t to the equation (1)
s = 3 + 6 = 9
Therefore, s = 9 and t = 6
(iii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From the 1st equation, we get,
x = (3+y)/3
Now, substitute the value of x to the second equation to get,
9(3+y)/3 – 3y = 9
⇒9 +3y -3y = 9
⇒ 9 = 9
Therefore, y has infinite values, and since x = (3+y) /3, x also has infinite values.
(iv) Given,
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 are the two equations.
From the 1st equation, we get,
x = (1.3- 0.3y)/0.2 _________________(1)
Now, substitute the value of x to the second equation to get,
0.4(1.3-0.3y)/0.2 + 0.5y = 2.3
⇒2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3
Now, substitute the value of y in equation (1), and we get,
x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2
Therefore, x = 2 and y = 3
Soln:
(v) Given,
√2 x + √3 y = 0 and √3 x – √8 y = 0
are the two equations.
From the 1st equation, we get,
x = – (√3/√2)y __________________(1)
Putting the value of x in the given second equation to get,
√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0
⇒ y = 0
Now, substitute the value of y in equation (1), and we get,
x = 0
Therefore, x = 0 and y = 0
(vi) Given,
(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.
From 1st equation, we get,
(3/2)x = -2 + (5y/3)
⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)
Putting the value of x in the second equation, we get,
((-12+10y)/9)/3 + y/2 = 13/6
⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6
Now, substitute the value of y in equation (1), and we get,
(3x/2) – 5(3)/3 = -2
⇒ (3x/2) – 5 = -2
⇒ x = 2
Therefore, x = 2 and y = 3
2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Soln:
2x + 3y = 11…………………………..(I)
2x – 4y = -24………………………… (II)
From equation (II), we get
x = (11-3y)/2 ………………….(III)
Substituting the value of x to equation (II), we get
2(11-3y)/2 – 4y = 24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(IV)
Putting the value of y in equation (III), we get
x = (11-3×5)/2 = -4/2 = -2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.
3. Form the pair of linear equations for the following problems and find their solution by the substitution method.
(i) The difference between two numbers is 26, and one number is three times the other. Find them.
Soln:
Let the two numbers be x and y, respectively, such that y > x.
According to the question,
y = 3x ……………… (1)
y – x = 26 …………..(2)
Substituting the value of (1) to (2), we get
3x – x = 26
x = 13 ……………. (3)
Substituting (3) in (1), we get y = 39
Hence, the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds, the smaller by 18 degrees. Find them.
Soln:
Let the larger angle be xo and the smaller angle be yo.
We know that the sum of two supplementary pairs of angles is always 180o.
According to the question,
x + y = 180o……………. (1)
x – y = 18o ……………..(2)
From (1), we get x = 180o – y …………. (3)
Substituting (3) in (2), we get
180o – y – y =18o
162o = 2y
y = 81o ………….. (4)
Using the value of y in (3), we get
x = 180o – 81o
= 99o
Hence, the angles are 99o and 81o.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.
Soln:
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (I)
3x + 5y = 1750 ………………. (II)
From (I), we get
y = (3800-7x)/6………………..(III)
Substituting (III) to (II), we get,
3x+5(3800-7x)/6 =1750
⇒3x+ 9500/3 – 35x/6 = 1750
⇒3x- 35x/6 = 1750 – 9500/3
⇒(18x-35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500
x = 500 ……………………….. (IV)
Substituting the value of x to (III), we get
y = (3800-7 ×500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500, and the cost of a ball is Rs 50.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge be Rs x and the per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)
From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x to (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (4)
Putting the value of y in (3), we get
x = 105 – 10 × 10 = 5
Hence, the fixed charge is Rs 5 and the per km charge = Rs 10
Charge for 25 km = x + 25y = 5 + 250 = Rs 255
(v) A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the fraction.
Solution:
Let the fraction be x/y.
According to the question,
(x+2) /(y+2) = 9/11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (1)
(x+3) /(y+3) = 5/6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get x = (-4+9y)/11 …………….. (3)
Substituting the value of x to (2), we get
6(-4+9y)/11 -5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y to (3), we get
x = (-4+9×9 )/11 = 7
Hence, the fraction is 7/9.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solutions:
Let the age of Jacob and his son be x and y, respectively.
According to the question,
(x + 5) = 3(y + 5)
x – 3y = 10 …………………………………….. (1)
(x – 5) = 7(y – 5)
x – 7y = -30 ………………………………………. (2)
From (1), we get x = 3y + 10 ……………………. (3)
Substituting the value of x to (2), we get
3y + 10 – 7y = -30
-4y = -40
y = 10 ………………… (4)
Substituting the value of y to (3), we get
x = 3 x 10 + 10 = 40
Hence, the present age of Jacob and his son is 40 years and 10 years, respectively.