Решения NCERT для класса 10. Глава 3. Пара линейных уравнений с двумя переменными. Упражнение 3.2.
Вопрос 1. Составьте пары линейных уравнений в следующих задачах и найдите их решения графически.
(i) 10 учеников X класса приняли участие в викторине по математике. Если девочек на 4 больше, чем мальчиков, найдите количество мальчиков и девочек, принявших участие в викторине.
Решение:
Let’s take,
Number of girls = x
Number of boys = y
According to the given conditions,
x + y = 10 -(1)
x – y = 4 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
x + y = 10, So, we can use the following table to draw the graph:
x y 0 10 10 0 For equation (2)
x – y = 4, So, we can use the following table to draw the graph:
x y 0 -4 4 0 The graph will be as follows for Equation (1) and (2):
Now, from the graph, we can conclude the given lines intersect each other at point (7, 3).
Hence, the number of girls are 7 and number of boys are 3 in a class.
(ii) 5 карандашей и 7 ручек вместе стоят 50 ₹, тогда как 7 карандашей и 5 ручек вместе стоят 46 ₹. Найдите стоимость одного карандаша и одной ручки.
Решение:
Let’s take,
Cost for one pencil = x
Cost for one pencil = y
According to the given conditions,
5x + 7y = 50 -(1)
7x + 5y = 46 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
5x + 7y = 50, So, we can use the following table to draw the graph:
x y 3 5 10 0 For equation (2)
7x + 5y = 46, So, we can use the following table to draw the graph:
x y 3 5 8 -2 The graph will be as follows for Equation (1) and (2):
Now, from the graph, we can conclude the given lines intersect each other at point (3, 5).
Hence, the cost of a pencil is ₹ 3 and cost of a pen is ₹ 5.
Вопрос 2. О сравнении соотношений 
, а также 
, выяснить, пересекаются ли прямые, представляющие следующие пары линейных уравнений, в точке, параллельны или совпадают:
(i) 5х – 4у + 8 = 0; 7х + 6у – 9 = 0
Решение:
In the given equations,
a1 = 5
a2 = 7
b1 = -4
b2 = 6
c1 = 8
c2 = -9
Now, here
a1/a2 = 5/7
b1/b2 = -4/6 = -2/3
c1/c2 = 8/-9
As, here
Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.
(ii) 9x + 3y + 12 = 0; 18х + 6у + 24 = 0
Решение:
In the given equations,
a1 = 9
a2 = 18
b1 = 3
b2 = 6
c1 = 12
c2 = 24
Now, here
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2
c1/c2 = 12/24 = 1/2
As, here
Hence, the given pairs of equations have infinite many solutions and the lines are coincident.
(iii) 6х – 3у + 10 = 0; 2х – у + 9 = 0
Решение:
In the given equations,
a1 = 6
a2 = 2
b1 = -3
b2 = -1
c1 = 10
c2 = 9
Now, here
a1/a2 = 6/2 = 3
b1/b2 = -3/-1 = 3
c1/c2 = 10/9
As, here
Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.
Вопрос 3. О сравнении соотношений , а также , выясните, являются ли следующие пары линейных уравнений совместными или противоречивыми.
(i) 3x + 2y = 5; 2х – 3у = 7
Решение:
In the given equations,
a1 = 3
a2 = 2
b1 = 2
b2 = -3
c1 = -5
c2 = -7
Now, here
a1/a2 = 3/2
b1/b2 = 2/-3
c1/c2 = -5/-7
As, here
Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.
Pair of linear equations are CONSISTENT.
(ii) 2x – 3y = 8; 4х - 6у = 9
Решение:
In the given equations,
a1 = 2
a2 = 4
b1 = -3
b2 = -6
c1 = -8
c2 = -9
Now, here
a1/a2 = 2/4 = 1/2
b1/b2 = -3/-6 = 1/2
c1/c2 = -8/-9 = 8/9
As, here
Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.
Pair of linear equations are INCONSISTENT.
(iii) 
= 7; 9х – 10у = 14
Решение:
In the given equations,
a1 = 3/2
a2 = 9
b1 = 5/3
b2 = -10
c1 = -7
c2 = -14
Now, here
c1/c2 = -7/-14 = 1/2
As, here
Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.
Pair of linear equations are CONSISTENT.
(iv) 5х – 3у = 11; – 10х + 6у = –22
Решение:
In the given equations,
a1 = 5
a2 = -10
b1 = -3
b2 = 6
c1 = -11
c2 = 22
Now, here
a1/a2 = 5/-10 = -1/2
b1/b2 = -3/6 = -1/2
c1/c2 = -11/22 = -1/2
As, here
Hence, the given pairs of equations have infinite many solutions and the lines are coincident.
Pair of linear equations are CONSISTENT.
(в) 
+ 2у = 8; 2х + 3у = 12
Решение:
In the given equations,
a1 = 4/3
a2 = 2
b1 = 2
b2 = 3
c1 = -8
c2 = -12
Now, here
b1/b2 = 2/3
c1/c2 = -8/-12 = 2/3
As, here
Hence, the given pairs of equations have infinite many solutions and the lines are coincident.
Pair of linear equations are CONSISTENT.
Вопрос 4. Какие из следующих пар линейных уравнений являются совместными/несовместными? Если непротиворечиво, получить решение графически:
(i) х + у = 5, 2х + 2у = 10
Решение:
In the given equations,
a1 = 1
a2 = 2
b1 = 1
b2 = 2
c1 = -5
c2 = -10
Now, here
a1/a2 = 1/2
b1/b2 = 1/2
c1/c2 = -5/-10 = 1/2
As, here
Hence, the given pairs of equations have infinite many solutions and the lines are coincident.
Pair of linear equations are CONSISTENT.
x + y = 5 -(1)
2x + 2y = 10 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
x + y = 5, So, we can use the following table to draw the graph:
x y 0 5 5 0 For equation (2)
2x + 2y = 10, So, we can use the following table to draw the graph:
x y 0 5 5 0 The graph will be as follows for Equation (1) and (2):
(ii) х – у = 8, 3х – 3у = 16
Решение:
In the given equations,
a1 = 1
a2 = 3
b1 = -1
b2 = -3
c1 = -8
c2 = -16
Now, here
a1/a2 = 1/3
b1/b2 = 1/3
c1/c2 = -8/-16 = 1/2
As, here
Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.
Pair of linear equations are INCONSISTENT.
(iii) 2х + у – 6 = 0, 4х – 2у – 4 = 0
Решение:
In the given equations,
a1 = 2
a2 = 4
b1 = 1
b2 = -2
c1 = -6
c2 = -4
Now, here
a1/a2 = 2/4 = 1/2
b1/b2 = 1/-2
c1/c2 = -6/-4 = 3/2
As, here
Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.
Pair of linear equations are CONSISTENT.
2x + y – 6 = 0 -(1)
4x – 2y – 4 = 0 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
2x + y – 6 = 0, So, we can use the following table to draw the graph:
x y 0 6 3 0 For equation (2)
4x – 2y – 4 = 0, So, we can use the following table to draw the graph:
x y 0 -2 1 0 The graph will be as follows for Equation (1) and (2):
Now, from the graph, we can conclude the given lines intersect each other at point (2, 2).
(iv) 2х – 2у – 2 = 0, 4х – 4у – 5 = 0
Решение:
In the given equations,
a1 = 2
a2 = 4
b1 = -2
b2 = -4
c1 = -2
c2 = -5
Now, here
a1/a2 = 2/4 = 1/2
b1/b2 = -2/-4 = 1/2
c1/c2 = -2/-5 = 2/5
As, here
Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.
Pair of linear equations are INCONSISTENT.
Вопрос 5. Половина периметра прямоугольного сада, длина которого на 4 м больше ширины, равна 36 м. Найдите размеры сада.
Решение:
Let’s take,
length = x
breadth = y
Half the perimeter of a rectangular garden =
= x + y
According to the given conditions,
x = y + 4 -(1)
x + y = 36 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
x = y + 4, So, we can use the following table to draw the graph:
x y 0 -4 4 0 For equation (2)
x + y = 36, So, we can use the following table to draw the graph:
x y 0 36 36 0 The graph will be as follows for Equation (1) and (2):
Now, from the graph, we can conclude the given lines intersect each other at point (20, 16).
Hence, length is 20 m and breath is 16 m of rectangle.
Вопрос 6. Имея линейное уравнение 2x + 3y – 8 = 0, напишите другое линейное уравнение с двумя переменными так, чтобы геометрическое представление образованной таким образом пары было:
(i) пересекающиеся линии
Решение:
Linear equation in two variables such that pair so formed is intersecting lines, so it should satisfy the given conditions
By rearranging, we get
Hence, the required equation should not be in ratio of 2/3
Hence, another equation can be 2x – 9y + 9 = 0
where the ratio is 2/-9
and,
(ii) параллельные линии
Решение:
Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions
By rearranging, we get
Hence, the required equation a2/b2 should be in ratio of 2/3 and b2/c2 should not be equal to 3/-8
Hence, another equation can be 4x + 6y + 9 = 0
where the ratio a2/b2 is 2/3
and,
(iii) совпадающие линии
Решение:
Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions
By rearranging, we get
Hence, the required equation a2/b2 should be in ratio of 2/3 and b2/c2 should be equal to 3/-8
Hence, another equation can be 4x + 6y -16 = 0
where the ratio a2/b2 is 2/3
and, b2/c2 = 3/-8
Вопрос 7. Нарисуйте графики уравнений x – y + 1 = 0 и 3x + 2y – 12 = 0. Определите координаты вершин треугольника, образованного этими прямыми и осью x, и заштрихуйте треугольную область.
Решение:
x – y + 1 = 0 -(1)
3x + 2y – 12 = 0 -(2)
So, to construct a graph, we need to find at least two solutions of the given equation.
For equation (1)
x – y + 1 = 0, So, we can use the following table to draw the graph:
x y 0 -1 1 0 For equation (2)
3x + 2y – 12 = 0, So, we can use the following table to draw the graph:
x y 0 6 4 0 The graph will be as follows for Equation (1) and (2):
Now, from the graph, we can conclude the given lines intersect each other at point (2, 3), and x-axis at (−1, 0) and (4, 0).
Hence, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).
