NCERT10th MathematicsExercise 3.7Topic: 3 Pair of Linear Equations in Two Variables
Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables-Ex-3.6
EXERCISE 3.7
Explanation:
1) Here, let x and y be two variables.
2) Convert the variables suitably to get linear equations in two variables.
3) Then solve these equations to get the values of our variables.
Solution:
Q1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani, and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
1) Let Ani's age be x and Biju's age be y.
2) The age of Ani's father Dharam is 2x.
3) The age of Biju's sister Cathy is y/2.
4) As the ages of Cathy and Dharam differ by 30 years, as Dharam is older than
Cathy, we have,
2x - y/2 = 30
4x - y = 60 ---------------------- equation 1
5) Age difference between Ani and Biju is 3 years. Here we need to take 2 cases.
6) Case 1: a) Ani is older than Biju.
x - y = 3 ---------------------- equation 2.
7) Subtract equation 2 from equation 1, and we get,
4x - y = 60
x - y = 3
(-) (+) (-)
-----------------------
3x = 57
x = 57/3
x = 19 ---------------------- equation 3.
8) Put the value of x = 19 from equation 3 in equation 2, and we get,
x - y = 3
19 - y = 3
y = 19 - 3y = 16 ---------------------- equation 4
9) If Ani is older than Biju, Ani's age is 19 years and Biju's age is 16 years.
10) Case 2: b) Biju is older than Ani.
y - x = 3
- x + y = 3 ---------------------- equation 5.
11) Add equation 5 to equation 1, and we get,
4x - y = 60
- x + y = 3
-----------------------
3x = 63
x = 63/3
x = 21 ---------------------- equation 6.
12) Put the value of x = 21 from equation 6 in equation 5, and we get,
- x + y = 3
-21 + y = 3
y = 21 + 3y = 24 ---------------------- equation 4
13) If Biju is older than Ani, Ani's age is 21 years and Biju's age is 24 years.
14) If Ani is older than Biju, Ani's age is 19 years and Biju's age is 16 years.
Q2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
1) Let the first friend has Rs x and the other friend has Rs y.
2) If the first friend gets Rs 100 from the second friend, the new amount with them,
a) First friend has (x + 100)
b) Second friend has (y - 100), so, according to first condition, we have,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = - 200 - 100
x - 2y = - 300
x = 2y - 300 ------------ equation 1
3) If the second friend gets Rs 10 from the first friend, the new amount with them,
a) First friend has (x - 10)
b) Second friend has (y + 10), so, according to second condition, we have,
y + 10 = 6(x - 10)
y + 10 = 6x - 60
6x - y = 10 + 60
6x - y = 70 ------------ equation 2
4) Put the value of x = 2y - 300 from equation 1 in equation 2, we get,
6x - y = 70
6(2y - 300) - y = 70
12y - 1800 - y = 70
12y - y = 70 + 1800
11y = 1870
y = 1870/11
y = 170 -------------------------- equation 3.
5) Put the value of y = 170 from equation 3 in equation 2, we get,
6x - y = 70
6x - 170 = 70
6x = 70 + 170
6x = 240
x = 240/6
x = 40 -------------------------- equation 4.
6) The first friend has Rs 40 and the second friend has Rs 170.
Q3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
1) Let uniform speed be x km/h and the time taken to cover the distance be y hrs.
2) We know that distance = speed x time, so the total distance = xy km.
3) When the train travels 10 km/h faster, then the time will be 2 hrs less, so,
(x + 10)(y - 2) = xy
xy - 2x + 10y - 20 = xy
- 2x + 10y - 20 = 0
- 2x + 10y = 20 ------------ equation 1
4) When the train travels 10 km/h slower, then the time will be 3 hrs more, so,
(x - 10)(y + 3) = xy
xy + 3x - 10y - 30 = xy
3x - 10y - 30 = 0
3x - 10y = 30 ------------ equation 2
5) Add equation 1 to equation 2, and we get,
3x - 10y = 30
- 2x + 10y = 20--------------------------
x = 50
x = 50 ---------------------- equation 3.
6) Put the value of x = 50 from equation 3 in equation 2, we get,
3x - 10y = 30
3(50) - 10y = 30
150 - 10y = 30
10y = 150 - 30
10y = 120
y = 120/10
y = 12 -------------------------- equation 4.
7) The uniform speed is 50 km/h and the time taken to cover the distance is 12 hrs.
8) So, the total distance traveled by the train is 50 x 12 = 600 km.
Q4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
1) Let the number of rows be x and the number of students in a row is y.
2) Total number of students = Number of rows x Number of students in a row = xy.
3) When 3 students are more in a row, then the number of rows will be less by 1,
(x - 1)(y + 3) = xy
xy + 3x - y - 3 = xy
3x - y - 3 = 0
3x - y = 3 ------------ equation 1
4) When 3 students are less in a row, then the number of rows will be more by 2,
(x + 2)(y - 3) = xy
xy - 3x + 2y - 6 = xy
-3x + 2y - 6 = 0
-3x + 2y = 6 ------------ equation 2
5) Add equation 2 to equation 1, and we get,
3x - y = 3
- 3x + 2y = 6--------------------------
y = 9
y = 9 ---------------------- equation 3.
6) Put the value of y = 9 from equation 3 in equation 1, we get,
3x - y = 3
3x - 9 = 3
3x = 9 + 3
3x = 12
x = 12/3
x = 4 -------------------------- equation 4.
7) The number of rows is 4 and the number of students in a row is 9.
8) So, the number of students in the class = 9 x 4 = 36.
Q5. In a ∆ ABC, < C = 3 < B = 2 (< A + < B). Find the three angles.
1) Let the < A be x, and < B be y.
2) According to the first condition,
< C = 3y -------------------------- equation 1.
3) We know that the sum of the angles of a triangle is 180 degrees,
< A + < B + < C = 180
x + y + 3y = 180
x + 4y = 180 -------------------------- equation 2.
4) According to the given condition,
< C = 2 (< A + < B)
3y = 2(x + y)
3y = 2x + 2y
3y - 2y = 2x
y = 2x -------------------------- equation 3.
5) Put the value of y = 2x from equation 3 in equation 2, we get,
x + 4y = 180
x + 4(2x) = 180
x + 8x = 180
9x = 180
x = 20 -------------------------- equation 4.
6) Put the value of x = 20 from equation 4 in equation 3, we get,
y = 2x
y = 2(20)
y = 40 -------------------------- equation 5.
7) The angles are < A = 200, < B = 400, and < C = 1200.
Q6. 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 the y-axis.
1) First equation: 5x - y = 5, so, y = 5x - 5.
2) Take ant two points as follows:
3) Second equation: 3x - y = 3, so, y = 3x - 3.
4) Take ant two points as follows:
5) Now we will draw the graph:
Q7. Solve the following pair of linear equations:
(i) px + qy = p – q; qx – py = p + q(ii) ax + by = c; bx + ay = 1 + c(iii) (x/a) - (y/b) = 0; ax + by = a2 + b2(iv) (a – b)x + (a + b) y = a2 – 2ab – b2; (a + b)(x + y) = a2 + b2
(v) 152x – 378y = – 74; –378x + 152y = – 604
Solution:
(i) px + qy = p – q; qx – py = p + q
1) The given equations are
px + qy = p – q -------------------------- equation 1
qx – py = p + q -------------------------- equation 2
2) Multiply equation 1 by p, we get,
p(px + qy) = p(p – q)
p2x + pqy = p2 – pq -------------------------- equation 3
3) Multiply equation 2 by q, we get,
q(qx – py) = q(p + q)
q2x - pqy = pq + q2 -------------------------- equation 4
4) Add equation 4 to equation 3, we get,
----------------------------------------p2x + pqy = p2 – pqq2x - pqy = pq + q2
(p2 + q2) x = (p2 + q2)
x = (p2 + q2)/(p2 + q2)
5) Put the value of x = 1 from equation 5 in equation 1, and we get,x = 1 -------------------------- equation 5
px + qy = p – q
p(1) + qy = p – q
p + qy = p – q
qy = – q
y = – q/q
6) So, x = 1 and y = - 1.y = – 1 -------------------------- equation 6
(ii) ax + by = c; bx + ay = 1 + c
1) The given equations are
ax + by = c -------------------------- equation 1
bx + ay = 1 + c -------------------------- equation 2
2) Multiply equation 1 by a, we get,
a(ax + by) = ac
a2x + aby = ac -------------------------- equation 3
3) Multiply equation 2 by b, we get,
b(bx + ay) = b(c + 1)
b2x + aby = bc + b -------------------------- equation 4
4) Subtract equation 4 from equation 3, we get,
a2x + aby = ac
(-) (-) (-)b2x + aby = bc + b
----------------------------------------
(a2 - b2) x = (ac - bc - b)
5) Put the value of x = [c(a - b) - b]/(a2 - b2) from equation 5 in equation 1, we get,x = [c(a - b) - b]/(a2 - b2) -------------------------- equation 5
ax + by = c
a[c(a - b) - b]/(a2 - b2) + by = c
[ac(a - b) - ab]/(a2 - b2) + by = c
by = c - [ac(a - b) - ab]/(a2 - b2)
by = {c(a2 - b2) - [ac(a - b) - ab]}/(a2 - b2)
by = [ca2 - cb2 - a2c + abc + ab]/(a2 - b2)
by = [abc - cb2 + ab]/(a2 - b2)
by = [bc(a - b) + ab]/(a2 - b2)
by = b[c(a - b) + a]/(a2 - b2)
y = [c(a - b) + a]/(a2 - b2) -------------------------- equation 6
6) So, x = [c(a - b) - b]/(a2 - b2) and y = [c(a - b) + a]/(a2 - b2).
(iii) (x/a) - (y/b) = 0; ax + by = a2 + b2
1) The given equations are
(x/a) - (y/b) = 0 -------------------------- equation 1
ax + by = a2 + b2 -------------------------- equation 2
2) So, from equation 1, we have
(x/a) - (y/b) = 0
(x/a) = (y/b)
x = (ay/b) -------------------------- equation 3
3) Put the value of x = (ay/b) from equation 3 in equation 2, we get,
ax + by = a2 + b2
a(ay/b) + by = a2 + b2
(a2y/b) + by = a2 + b2
(a2y + b2y)/b = a2 + b2
(a2 + b2)y/b = a2 + b2
(a2 + b2)y = b(a2 + b2)
y = b(a2 + b2)/(a2 + b2)
y = b -------------------------- equation 4
4) Put the value of y = b from equation 4 in equation 3, we get,
x = (ay/b)
x = (ab/b)
x = a
5) So, x = a and y = b.
(iv) (a – b)x + (a + b) y = a2 – 2ab – b2; (a + b)(x + y) = a2 + b2
1) The given equations are
(a – b)x + (a + b) y = a2 – 2ab – b2 -------------------------- equation 1
(a + b)(x + y) = a2 + b2 -------------------------- equation 2
2) So, from equation 2, we have,
(a + b)(x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 -------------------------- equation 3
3) Subtract equation 1 from equation 3, we get,
(a + b)x + (a + b)y = a2 + b2
(a – b)x + (a + b)y = a2 – 2ab – b2
(-) (-) (-)
---------------------------------------------------------
---------------------------------------------------------
2bx = 2b2 + 2ab
2bx = 2b(b + a)
x = (a + b) -------------------------- equation 4
4) Put the value of x = (a + b) from equation 4 in equation 1, we get,
(a – b)x + (a + b) y = a2 – 2ab – b2
(a – b)(a + b) + (a + b) y = a2 – 2ab – b2
(a2 – b2) + (a + b) y = (a2 – b2) – 2ab
(a + b) y = – 2ab
y = – 2ab/(a + b) -------------------------- equation 5.
5) So, x = (a + b) and y = – 2ab/(a + b).
(v) 152x – 378y = – 74; –378x + 152y = – 604
1) The given equations are
152x – 378y = – 74 -------------------------- equation 1
–378x + 152y = – 604 -------------------------- equation 2
2) Add equation 2 to equation 1, we get,
----------------------------------------------152x – 378y = – 74–378x + 152y = – 604
– 226 x – 226 y = – 678
x + y = (-678)/(-226)
x + y = 3 -------------------------- equation 3
3) Subtract equation 2 from equation 1, we get,
152x – 378y = – 74
–378x + 152y = – 604
(+) (-) (+)
---------------------------------------------------------
---------------------------------------------------------
530x – 530y = 530
x – y = 530/530
x – y = 1
x = y + 1 -------------------------- equation 4
4) Put the value of x = y + 1 from equation 4 in equation 3, we get,
x + y = 3
(y + 1) + y = 3
2y + 1 = 3
2y = 3 - 1
y = 2/2
y = 1 -------------------------- equation 5.
5) Put the value of y = 1 from equation 5 in equation 4, we get,
x = y + 1
x = 1 + 1
x = 2 -------------------------- equation 6.
6) So, x = 2 and y = 1.
Q8. ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.
1) We know that the sum of the opposite angles of a cyclic quadrilateral is 1800.
2) So according to the diagram, we have,
(-7x + 5) + (3y - 5) = 180 -------------------------- equation 1.
(-4x) + (4y + 20) = 180 -------------------------- equation 2.
3) From equation 1 we get,
(-7x + 5) + (3y - 5) = 180
-7x + 3y = 180
7x - 3y = -180 -------------------------- equation 3.4) From equation 2 we get,
(-4x) + (4y + 20) = 180
(-x) + (y + 5) = 45
-x + y = 45 - 5
-x + y = 40
y = x + 40 -------------------------- equation 4.
5) Put the value of y = x + 40 from equation 4 in equation 3, we get,
7x - 3y = -180
7x - 3(x + 40) = -180
7x - 3x - 120 = -180
4x = 120 -180
4x = - 60
x = - 60/4
x = - 15 -------------------------- equation 5.
6) Put the value of x = - 15 from equation 5 in equation 4, we get,
y = x + 40
y = - 15 + 40
y = 25 -------------------------- equation 6.
7) So, here, x = - 15 and y = 25, therefore we have,
< A = (4y + 20) = (4(25) + 20) = (100 + 20) = 120, so , < A =120.
< B = (3y - 5) = (3(25) - 5) = (75 - 5) = 70, so, < B = 70.
< C = (- 4x) = (- 4(- 15)) = 60, so, < C = 60.
< D = (- 7x + 5) = (- 7(- 15) + 5) = (105 + 5) = 110, < D = 110.
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