(2) Each Solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
For Example: 2x + 3y = 5 has
(1, 1) as its Solution. So,(1, 1) will lie on the line 2x + 3y = 5.
3) The general form for a pair of linear equations in two variables x and y is
a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, where a1 , b1 , c1 , a2 , b2 , c2 are all real numbers and
a12 + b12 ≠ 0, a22 + b22 ≠ 0.
For Example: 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0 forms a pair of linear equations.
(4) A pair of linear equations which has no Solution is called an inconsistent pair of linear equations. In this case, the lines may be
𝐏𝐚𝐫𝐚𝐥𝐥𝐞𝐥 𝐚𝟏/𝐚𝟐 = 𝐛𝟏/𝐛𝟐 ≠ 𝐜𝟏/𝐜𝟐
For Example: x + 2y – 4 = 0 and 2x + 4y – 12 = 0 are parallel lines.
Exercise 4.1:
Question 1:
By comparing the ratios 
- 5x- 4y + 8 = 0
7x+6y-9 = 0 - 9x+3y +12 = 0
18x+6y + 24 = 0 - 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution :
- 5x – 4y + 8 = 0
7x + 6y – 9 = 0 - 9x + 3y + 12 = 0
18x + 6y + 24 = 0 - 6x – 3y + 10 = 0
2x – y + 9 = 0
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Question 2:
Check whether the following equations are consistent or inconsistent. Solve them graphically.
Question 2(a):
3x + 2y = 5
2x-3y = 7
Solution :
Question 2(b):
2x – 3y = 8
4x – 6y = 9
Solution :
Question 2(c):
9x – 10y = 12
Solution :
Question 2(d):
5x – 3y = 11
-10x + 6y = -22
Solution :
5x – 3y = 11 ⇒ 5x – 3y – 11 = 0
-10x + 6y = -22 ⇒ -10x + 6y + 22 = 0
So the equation is consistent. Therefore, they are coincident lines. So, the pair of linear equations are dependent and have infinitely many solutions.
For the equation 5x – 3y – 11 = 0
Question 2(e):
4/3 x + 2y = 8
2x + 3y = 12
Solution :
So the equation is consistent. Therefore, they are coincident lines. So, the pair of linear equations are dependent and have infinitely many solutions.
Question 2(f):
x + y = 5
2x + 2y = 10
Solution :
x + y – 5=0
2x + 2y – 10=0
So the equations are inconsistent. Therefore, they are coincident lines. So, the pair of linear equations are dependent and have infinitely many solutions.
For the equation x + y – 5=0
Question 2(g):
x – y = 8
3x – 3y = 16
Solution :
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Question 2(h):
2x + y – 6 = 0
4x – 2y – 4 = 0
Solution :
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Question 2(i):
2x – 2y – 2 = 0
4x – 4y – 5 = 0
Solution :
Question 3:
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are 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 friend to find how many pants and skirts Neha bought.
Solution :
Let the number of pants purchased by Neha be ‘x’ and
no. of skirts purchased be ‘y’.
Twice the number of pants = 2x.
From the given condition,
y = 2x – 2 (i)
i.e (Number of skirts are 2 less than twice the number of pants.)
y = 4x – 4 (ii)
i.e (Number of skirts are 4 less than four times the number of pants)
subtracting (i) from (ii), we get
2x = 2
x = 1
So, number of pants purchased = 1 and number of skirts = 0.
Question 4:
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boy s then, find the number of boys and the number of girls who took part in the quiz.
Solution :
Let the number of boys be ‘x’
and that of girls be ‘y’.
Total no. of students = 10 (given) —- (1)
According to given condition,
y = x + 4 —– (2)
[No of girls is 4 more(add 4) then no.of boys]
But, x + y = 10 [From (1)] —— (3)
x + x + 4 = 10 [Substituting (2) in (3)]
2x + 4 = 10
2x = 6
x = 3 —–(4)
y = 3 + 4 //Substituting (4) in (1)
y = 7
Hence, there are 3 boys and 7 girls in the class.
Question 5:
5 pencils and 7 pens together cost D50 whereas 7 pencils and 5 pens together cost D46. Find the cost of one pencil and that of one pen.
Solution :
Let the cost of one pen be ‘y’;
Cost of one pencil be ‘x’;
So, according to given conditions,
5x + 7y = 50 (i)
7x + 5y = 46 (ii)
Solving (i) and (ii),
Multiplying (i) by 7and multiplying (ii) by 5
y=5
Now, substituting y = 5 in equation (i)
x = 3
Hence, the cost of one pencil is Rs. 3 and the cost of one pen is Rs. 5.
Question 6:
Half the perimeter of a rectangular garden is 36m. If the length is 4m more than its width, is 36m. Find the dimensions of the garden.
Solution :
Let the length and width of the rectangular garden be x and y respectively.
Perimeter = 2(l + b)
Half the perimeter = (l + b)
⇒ l + b = 36 — (1)
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Question 7:
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines.
Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Solution
Question 8:
The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq units. Find the length and breadth of the rectangle.
Solution :
Let the present area, length and breadth of the rectangle be ‘z’, ‘x’ and ‘y’ respectively.
Therefore, z = x ×y (∵ A = length × breadth)
For the first condition,
z = z – 80 (∵ Area is reduced by 80 sq. units)–(1)
x = x – 5 (∵ Length is reduced by 5 units)—-(2)
y = y + 2 (∵ Breadth is increased by 2 units) —(3)
Therefore,
z = x × y —(4)
Substituting new values i.e. (1),(2),(3) in (4)
(z – 80) = (x – 5)(y + 2) —(5)
For the second condition,
z = z + 50 (∵ Area is increased by 50 sq. units)–(1)
x = x+10 (∵ Length is increased by 10 units)—-(2)
y = y – 5 (∵ Breadth is reduced by 5 units) —(3)
Therefore,
Substituting new values i.e (1),(2),(3) in (4)
(z + 50) = (x + 10)(y – 5) —(5)
Question 9
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Solution :
Let the number of students be = x and
Number of benches be = y
If three students sit on each bench, one student will be left.
So, x = 3y – 1
x – 3y – 1= 0 ……(1)
If four students sit on each bench, one bench will be left. So,
x = 4(y – 1)
x – 4y + 4 = 0 …….(2)
For the equation x – 3y – 1= 0
Exercise 4.2:
Question 1:
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly income.
Solution :
Question 2:
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Solution :
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Question 3:
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Solution :ABCD
Let the larger of the two supplementary angles be ‘y’ and the smaller angle be ‘x’.
Therefore, x + y = 180 —(1)
(∵ Sum of supplementary angles is 180˚)
Also, y=x + 18 (∵ Given) ———-(2)
Substituting (2) in (1),we get,
x + x + 18 = 180
⇒ 2x + 18 = 180
⇒ 2x = 180 – 18
⇒ 2x = 162
⇒ x = 81 ———–(3)
Substitute (3) in (2),
y = 81+18
⇒ y = 99
Question 4:
The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is D220. For a journey of 15 the charge paid is D310.
- What are the fixed charges and charge per km?
- How much does a person have to pay for travelling a distance of 25 km?
Solution :
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Question 5:
A fraction becomes equal to 4/5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes equal to 1/2 . What is the fraction?
Solution :
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Question 6:
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution :
Let P and Q be the cars starting from A and B respectively. Let their speeds be x km/hr and y km/hr respectively.
Case – I: When the cars P and Q move in the same direction,
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Distance covered by car P in 5 hours = 5x km
Distance covered by car Q in 5 hours = 5y km
∴ AM = 5x km and BM = 5y km
∴ AM – BM = AB ⇒ 5x – 5y = 100
⇒ 5(x – y) = 100
⇒ x – y = 20 —(1)
Case – II: When the car P and Q move in the opposite direction,
Distance covered by P in 1 hour = x km
Distance covered by Q in 1 hour = y km
∴ AN = x km and BN = y km
∴ AN + BN = AB
⇒ x + y = 100 —-(2)
Adding (1) and (2) we get,
2x = 120
⇒ x = 60
Substituting x = 60 in (1) we get y = 40.
Therefore, the speeds of the cars P and Q are 60 km/hr and 40 km/hr respectively.
Question 7:
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.
Solution :
Let the measure of the smaller angle be x° and that of the larger angle be y°.
The larger angle is 3° less than twice the measure of the smaller angle, so
y°= 2x – 3° (1)
Given, the two angles are complementary,
x + y = 180°
⇒ x + (2x – 3°) = 180°
⇒ x + 2x – 3°= 180°
⇒ 3x = 183°
⇒ x = 61°
Substituting the value of x in equation (1)
y = 2(61) – 3
⇒ y = 119°.
Hence, the measures of the two angles are 61° and 119°.
Question 8:
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?
Solution :
Let the number of pages in the first part be x and in the second part be y.
Total pages number of pages is 1382,
Hence, x + y = 1382. (1)
The second part of the book has 64 pages more than the first part,
y = x + 64 —– (2)
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Substituting the value of y in equation (1)
x + x + 64 = 1382
2x = 1382 – 64
2x = 1318
X = 659
Substituting the value of x in equation (2),
y = 659 + 64
y = 723
The pages in the first and second part are 659 and 723 respectively.
Question 9:
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100ml of a 68% solution.
Solution :
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Question 10:
Suppose you have ₹12000 to save. You have to save some amount at 10% and the rest at 15%. How much should be saved at each rate to yield 12% on the total amount saved?
Solution :
Hence the amou to be invested at 10% is Rs. 7200 and the amount to be invested at 15% is Rs. 4800.
Exercise 4.3
Question 1.
2x + 4y = 5xy
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water = x kmph
and the speed of the stream = y kmph
then speed in downstream = x + y
Speed in upstream = x – y
and time =
By problem,
Take
the given equations reduce to
30a + 44b = 10 ……… (1)
40a + 55b = 13 ……… (2)
Answer:
Answer:
Let the time taken by 1 woman to complete the work = x days
and time taken by 1 man to complete the work = y days
∴ Work done by 1 woman in 1 day =
Work done by 1 man in 1 day =
By problem,