Q2 (i): 5 x-4 y+8=0 7 x+6 y-9=0

For a pair of linear equations a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0: If a_1/a_2 ≠ b_1/b_2, the lines intersect at a point. If a_1/a_2 = b_1/b_2 = c_1/c_2, the lines are coincident. If a_1/a_2 = b_1/b_2 ≠ c_1/c_2, the lines are parallel. (i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 Here,…

Q2 (ii): 9 x+3 y+12=0 18 x+6 y+24=0

For a pair of linear equations a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0: If a_1/a_2 ≠ b_1/b_2, the lines intersect at a point. If a_1/a_2 = b_1/b_2 = c_1/c_2, the lines are coincident. If a_1/a_2 = b_1/b_2 ≠ c_1/c_2, the lines are parallel. (i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 Here,…

Q2 (iii): 6 x-3 y+10=0 2 x-y+9=0

For a pair of linear equations a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0: If a_1/a_2 ≠ b_1/b_2, the lines intersect at a point. If a_1/a_2 = b_1/b_2 = c_1/c_2, the lines are coincident. If a_1/a_2 = b_1/b_2 ≠ c_1/c_2, the lines are parallel. (i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 Here,…

Q4 (i): x+y=5, 2 x+2 y=10

(i) x+y=5 and 2x+2y=10 Comparing ratios: a_1=1, b_1=1, c_1=-5 and a_2=2, b_2=2, c_2=-10. a_1/a_2 = 1/2 b_1/b_2 = 1/2 c_1/c_2 = -5/-10 = 1/2 Since a_1/a_2 = b_1/b_2 = c_1/c_2, the pair of equations is consistent with infinitely many solutions (coincident lines). Graphical Solution: Both equations…

Q4 (ii): x-y=8, 3 x-3 y=16

(i) x+y=5 and 2x+2y=10 Comparing ratios: a_1=1, b_1=1, c_1=-5 and a_2=2, b_2=2, c_2=-10. a_1/a_2 = 1/2 b_1/b_2 = 1/2 c_1/c_2 = -5/-10 = 1/2 Since a_1/a_2 = b_1/b_2 = c_1/c_2, the pair of equations is consistent with infinitely many solutions (coincident lines). Graphical Solution: Both equations…

Q4 (iii): 2 x+y-6=0, 4 x-2 y-4=0

(i) x+y=5 and 2x+2y=10 Comparing ratios: a_1=1, b_1=1, c_1=-5 and a_2=2, b_2=2, c_2=-10. a_1/a_2 = 1/2 b_1/b_2 = 1/2 c_1/c_2 = -5/-10 = 1/2 Since a_1/a_2 = b_1/b_2 = c_1/c_2, the pair of equations is consistent with infinitely many solutions (coincident lines). Graphical Solution: Both equations…

Q4 (iv): 2 x-2 y-2=0, 4 x-4 y-5=0

(i) x+y=5 and 2x+2y=10 Comparing ratios: a_1=1, b_1=1, c_1=-5 and a_2=2, b_2=2, c_2=-10. a_1/a_2 = 1/2 b_1/b_2 = 1/2 c_1/c_2 = -5/-10 = 1/2 Since a_1/a_2 = b_1/b_2 = c_1/c_2, the pair of equations is consistent with infinitely many solutions (coincident lines). Graphical Solution: Both equations…

Q6 (i): intersecting lines

Given Equation: 2x + 3y - 8 = 0 Here, a_1=2, b_1=3, c_1=-8. (i) Intersecting lines For intersecting lines, the condition is a_1/a_2 ≠ b_1/b_2. We can choose any values for a_2 and b_2 that do not satisfy 2/a_2 = 3/b_2. Let's choose a_2 = 3 and b_2 = 2. Then 2/3 ≠ 3/2. We can choose any value for…

Q6 (ii): parallel lines

Given Equation: 2x + 3y - 8 = 0 Here, a_1=2, b_1=3, c_1=-8. (i) Intersecting lines For intersecting lines, the condition is a_1/a_2 ≠ b_1/b_2. We can choose any values for a_2 and b_2 that do not satisfy 2/a_2 = 3/b_2. Let's choose a_2 = 3 and b_2 = 2. Then 2/3 ≠ 3/2. We can choose any value for…

Q6 (iii): coincident lines

Given Equation: 2x + 3y - 8 = 0 Here, a_1=2, b_1=3, c_1=-8. (i) Intersecting lines For intersecting lines, the condition is a_1/a_2 ≠ b_1/b_2. We can choose any values for a_2 and b_2 that do not satisfy 2/a_2 = 3/b_2. Let's choose a_2 = 3 and b_2 = 2. Then 2/3 ≠ 3/2. We can choose any value for…

Pair of Linear Equations in Two Variables Class 10 Mathematics NCERT Solutions

Q1Exercise 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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Solution

(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.

Let: The number of boys be

x

. The number of girls be

y

.

Equations: According to the problem, the total number of students is 10.

x + y = 10 \quad ...(1)

The number of girls is 4 more than the number of boys.

y = x + 4 \quad ...(2)

Graphical Solution: To solve graphically, we find two points for each equation to draw the lines.

For equation (1):

x + y = 10

or

y = 10 - x

$x$	0	10
$y$	10	0
Points are A(0, 10) and B(10, 0).

For equation (2):

y = x + 4

$x$	0	1
$y$	4	5
Points are P(0, 4) and Q(1, 5).

Now, we plot these points on a graph paper. The line for equation (1) passes through A and B. The line for equation (2) passes through P and Q. We observe that the two lines intersect at the point (3, 7).

Verification: Substituting

x = 3

and

y = 7

in the equations: (1)

3 + 7 = 10

(True) (2)

7 = 3 + 4

(True)

Final Answer: The number of boys is 3 and the number of girls is 7.

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Let: The cost of one pencil be ₹

x

. The cost of one pen be ₹

y

.

Equations: According to the problem:

5x + 7y = 50 \quad ...(1)

7x + 5y = 46 \quad ...(2)

Graphical Solution: To solve graphically, we find two points for each equation.

For equation (1):

5x + 7y = 50

or

y = \frac{50 - 5x}{7}

$x$	3	10
$y$	5	0
Points are A(3, 5) and B(10, 0).

For equation (2):

7x + 5y = 46

or

y = \frac{46 - 7x}{5}

$x$	8	3
$y$	-2	5
Points are P(8, -2) and Q(3, 5).

Now, we plot these points on a graph paper. The line for equation (1) passes through A and B. The line for equation (2) passes through P and Q. We observe that the two lines intersect at the point (3, 5).

Verification: Substituting

x = 3

and

y = 5

in the equations: (1)

5(3) + 7(5) = 15 + 35 = 50

(True) (2)

7(3) + 5(5) = 21 + 25 = 46

(True)

Final Answer: The cost of one pencil is ₹ 3 and the cost of one pen is ₹ 5.

Q2Exercise 3.1

On comparing the ratios $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i)

$5 x-4 y+8=0$ $7 x+6 y-9=0$

(ii)

$9 x+3 y+12=0$ $18 x+6 y+24=0$

(iii)

$6 x-3 y+10=0$ $2 x-y+9=0$

Solution

For a pair of linear equations

a_1x + b_1y + c_1 = 0

and

a_2x + b_2y + c_2 = 0

:

If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ , the lines intersect at a point.
If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ , the lines are coincident.
If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ , the lines are parallel.

(i) $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$ Here,

a_1=5, b_1=-4, c_1=8

and

a_2=7, b_2=6, c_2=-9

.

\frac{a_1}{a_2} = \frac{5}{7}

\frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}

Since

\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

.

Final Answer: The lines intersect at a point.

(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$ Here,

a_1=9, b_1=3, c_1=12

and

a_2=18, b_2=6, c_2=24

.

\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}

\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

.

Final Answer: The lines are coincident.

(iii) $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$ Here,

a_1=6, b_1=-3, c_1=10

and

a_2=2, b_2=-1, c_2=9

.

\frac{a_1}{a_2} = \frac{6}{2} = 3

\frac{b_1}{b_2} = \frac{-3}{-1} = 3

\frac{c_1}{c_2} = \frac{10}{9}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

.

Final Answer: The lines are parallel.

Q3Exercise 3.1

On comparing the ratios $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ , find out whether the following pair of linear equations are consistent, or inconsistent.

(i)

$3 x+2 y=5 ; \quad 2 x-3 y=7$

(ii)

$2 x-3 y=8 ; \quad 4 x-6 y=9$

(iii)

$\frac{3}{2} x+\frac{5}{3} y=7 ; 9 x-10 y=14$

(iv)

$5 x-3 y=11 ;-10 x+6 y=-22$

(v)

$\frac{4}{3} x+2 y=8 ; 2 x+3 y=12$

Solution

A pair of linear equations is consistent if it has at least one solution (intersecting or coincident lines). It is inconsistent if it has no solution (parallel lines).

(i) $3x + 2y = 5$ and $2x - 3y = 7$ Rewriting in standard form:

3x + 2y - 5 = 0

and

2x - 3y - 7 = 0

. Here,

a_1=3, b_1=2, c_1=-5

and

a_2=2, b_2=-3, c_2=-7

.

\frac{a_1}{a_2} = \frac{3}{2}

\frac{b_1}{b_2} = \frac{2}{-3}

Since

\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

, the lines are intersecting and have a unique solution.

Final Answer: Consistent.

(ii) $2x - 3y = 8$ and $4x - 6y = 9$ Rewriting in standard form:

2x - 3y - 8 = 0

and

4x - 6y - 9 = 0

. Here,

a_1=2, b_1=-3, c_1=-8

and

a_2=4, b_2=-6, c_2=-9

.

\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}

\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

, the lines are parallel and have no solution.

Final Answer: Inconsistent.

(iii) $\frac{3}{2} x+\frac{5}{3} y=7$ and $9x - 10y = 14$ Rewriting in standard form:

\frac{3}{2}x + \frac{5}{3}y - 7 = 0

and

9x - 10y - 14 = 0

. Here,

a_1=\frac{3}{2}, b_1=\frac{5}{3}, c_1=-7

and

a_2=9, b_2=-10, c_2=-14

.

\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{3}{2 \times 9} = \frac{1}{6}

\frac{b_1}{b_2} = \frac{5/3}{-10} = \frac{5}{3 \times (-10)} = -\frac{1}{6}

Since

\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

, the lines are intersecting and have a unique solution.

Final Answer: Consistent.

(iv) $5x - 3y = 11$ and $-10x + 6y = -22$ Rewriting in standard form:

5x - 3y - 11 = 0

and

-10x + 6y + 22 = 0

. Here,

a_1=5, b_1=-3, c_1=-11

and

a_2=-10, b_2=6, c_2=22

.

\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}

\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}

\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

, the lines are coincident and have infinitely many solutions.

Final Answer: Consistent.

(v) $\frac{4}{3} x+2 y=8$ and $2x + 3y = 12$ Rewriting in standard form:

\frac{4}{3}x + 2y - 8 = 0

and

2x + 3y - 12 = 0

. Here,

a_1=\frac{4}{3}, b_1=2, c_1=-8

and

a_2=2, b_2=3, c_2=-12

.

\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{4}{3 \times 2} = \frac{2}{3}

\frac{b_1}{b_2} = \frac{2}{3}

\frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

, the lines are coincident and have infinitely many solutions.

Final Answer: Consistent.

Q4Exercise 3.1

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)

$x+y=5, \quad 2 x+2 y=10$

(ii)

$x-y=8, \quad 3 x-3 y=16$

(iii)

$2 x+y-6=0, \quad 4 x-2 y-4=0$

(iv)

$2 x-2 y-2=0, \quad 4 x-4 y-5=0$

Solution

(i) $x+y=5$ and $2x+2y=10$ Comparing ratios:

a_1=1, b_1=1, c_1=-5

and

a_2=2, b_2=2, c_2=-10

.

\frac{a_1}{a_2} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{1}{2}

\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

, the pair of equations is consistent with infinitely many solutions (coincident lines).

Graphical Solution: Both equations are equivalent. We can find points for

x+y=5

.

$x$	0	5	2
$y$	5	0	3
Points are (0, 5), (5, 0), (2, 3), etc. Any point on the line $x+y=5$ is a solution.

(ii) $x-y=8$ and $3x-3y=16$ Comparing ratios:

a_1=1, b_1=-1, c_1=-8

and

a_2=3, b_2=-3, c_2=-16

.

\frac{a_1}{a_2} = \frac{1}{3}

\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}

\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

, the pair of equations is inconsistent (parallel lines) and has no solution.

(iii) $2x+y-6=0$ and $4x-2y-4=0$ Comparing ratios:

a_1=2, b_1=1, c_1=-6

and

a_2=4, b_2=-2, c_2=-4

.

\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}

Since

\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

, the pair of equations is consistent with a unique solution (intersecting lines).

Graphical Solution: For

2x+y-6=0

or

y=6-2x

:

$x$	0	3	2
$y$	6	0	2
Points are (0, 6), (3, 0), (2, 2).

For

4x-2y-4=0

or

2x-y-2=0

or

y=2x-2

:

$x$	0	1	2
$y$	-2	0	2
Points are (0, -2), (1, 0), (2, 2).

The common point is (2, 2). Plotting the lines shows they intersect at (2, 2).

Final Answer: The solution is

x=2, y=2

.

(iv) $2x-2y-2=0$ and $4x-4y-5=0$ Comparing ratios:

a_1=2, b_1=-2, c_1=-2

and

a_2=4, b_2=-4, c_2=-5

.

\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}

\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}

Since

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

, the pair of equations is inconsistent (parallel lines) and has no solution.

Q5Exercise 3.1

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.

Solution

Let: The length of the rectangular garden be

l

meters. The width of the rectangular garden be

w

meters.

Given:

The length is 4 m more than its width. $l = w + 4 \quad ...(1)$
Half the perimeter is 36 m. The perimeter of a rectangle is $2(l+w)$ . Half the perimeter is $l+w$ . $l + w = 36 \quad ...(2)$

Solution: We have a pair of linear equations:

l - w = 4

l + w = 36

We can solve this system. Let's use the substitution method. Substitute the value of

l

from equation (1) into equation (2):

(w + 4) + w = 36

2w + 4 = 36

2w = 36 - 4

2w = 32

w = 16

Now, substitute the value of

w

back into equation (1) to find

l

:

l = 16 + 4

l = 20

Verification: Length (20 m) is 4 m more than width (16 m). Correct. Half the perimeter =

l+w = 20 + 16 = 36

m. Correct.

Final Answer: The dimensions of the garden are length = 20 meters and width = 16 meters.

Q6Exercise 3.1

Given the linear equation $2 x+3 y-8=0$ , 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

Given Equation:

2x + 3y - 8 = 0

Here,

a_1=2, b_1=3, c_1=-8

.

(i) Intersecting lines For intersecting lines, the condition is

\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

. We can choose any values for

a_2

and

b_2

that do not satisfy

\frac{2}{a_2} = \frac{3}{b_2}

. Let's choose

a_2 = 3

and

b_2 = 2

. Then

\frac{2}{3} \neq \frac{3}{2}

. We can choose any value for

c_2

, say

c_2 = -7

. An example of such an equation is:

3x + 2y - 7 = 0

(ii) Parallel lines For parallel lines, the condition is

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

. We need

\frac{2}{a_2} = \frac{3}{b_2}

. Let's multiply

a_1

and

b_1

by the same non-zero constant, say 2. So,

a_2 = 2 \times 2 = 4

and

b_2 = 3 \times 2 = 6

. Now we have

\frac{2}{4} = \frac{3}{6} = \frac{1}{2}

. We need

\frac{c_1}{c_2} \neq \frac{1}{2}

, so

\frac{-8}{c_2} \neq \frac{1}{2}

. This means

c_2 \neq -16

. We can choose any other value, say

c_2 = -5

. An example of such an equation is:

4x + 6y - 5 = 0

(iii) Coincident lines For coincident lines, the condition is

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

. We can multiply the entire given equation by a non-zero constant, say 2.

2(2x + 3y - 8) = 2(0)

4x + 6y - 16 = 0

Here,

a_2=4, b_2=6, c_2=-16

, and

\frac{2}{4} = \frac{3}{6} = \frac{-8}{-16} = \frac{1}{2}

.

Final Answer:

(i)

For intersecting lines:

3x + 2y - 7 = 0

(many other answers are possible)

(ii)

For parallel lines:

4x + 6y - 5 = 0

(many other answers are possible)

(iii)

For coincident lines:

4x + 6y - 16 = 0

(any non-zero multiple of the original equation is correct)

Q7Exercise 3.1

Draw the graphs of the equations $x-y+1=0$ and $3 x+2 y-12=0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and shade the triangular region.

Solution

Given Equations:

$x - y + 1 = 0 \implies y = x + 1$
$3x + 2y - 12 = 0 \implies 2y = 12 - 3x \implies y = \frac{12 - 3x}{2}$

Step 1: Find points to plot the lines. For line (1):

y = x + 1

$x$	0	-1	2
$y$	1	0	3
Points are A(0, 1), B(-1, 0), C(2, 3).

For line (2):

y = \frac{12 - 3x}{2}

$x$	0	4	2
$y$	6	0	3
Points are P(0, 6), Q(4, 0), R(2, 3).

Step 2: Describe the graph and find the vertices. When we plot these points and draw the lines, we observe the following:

The line $x - y + 1 = 0$ intersects the x-axis at point B(-1, 0). (A vertex of the triangle)
The line $3x + 2y - 12 = 0$ intersects the x-axis at point Q(4, 0). (A second vertex of the triangle)
The two lines intersect each other at point C(2, 3), which is also R(2, 3). (The third vertex of the triangle)

Step 3: Identify the vertices and describe the shaded region. The triangle is formed by the two lines and the x-axis. The vertices of this triangle are the intersection points found above.

Intersection of line 1 and line 2: (2, 3)
Intersection of line 1 and x-axis (where y=0): (-1, 0)
Intersection of line 2 and x-axis (where y=0): (4, 0)

The triangular region to be shaded is the area enclosed by the points (2, 3), (-1, 0), and (4, 0).

Final Answer: The coordinates of the vertices of the triangle are (2, 3), (-1, 0), and (4, 0). The student should shade the region bounded by these three points on the graph.

Q1Exercise 3.2

Solve the following pair of linear equations by the substitution method.

(i)

$x+y=14$ $x-y=4$

(ii)

$s-t=3$ $\frac{s}{3}+\frac{t}{2}=6$

(iii)

$3 x-y=3$ $9 x-3 y=9$

(iv)

$0.2 x+0.3 y=1.3$ $0.4 x+0.5 y=2.3$

(v)

$\sqrt{2} x+\sqrt{3} y=0$ $\sqrt{3} x-\sqrt{8} y=0$

(vi)

$\frac{3 x}{2}-\frac{5 y}{3}=-2$ $\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Solution

(i) $x+y=14$ and $x-y=4$

Given:

x+y=14 \quad ...(1)

x-y=4 \quad ...(2)

Solution: From equation (2), we can write

x = 4+y

. ...(3) Substitute this value of

x

into equation (1):

(4+y) + y = 14

4 + 2y = 14

2y = 10

y = 5

Substitute

y=5

into equation (3):

x = 4 + 5

x = 9

Final Answer:

x=9, y=5

.

(ii) $s-t=3$ and $\frac{s}{3}+\frac{t}{2}=6$

Given:

s-t=3 \quad ...(1)

\frac{s}{3}+\frac{t}{2}=6 \quad ...(2)

Solution: From equation (1),

s = 3+t

. ...(3) Substitute this into equation (2):

\frac{3+t}{3} + \frac{t}{2} = 6

Multiply the entire equation by 6 (LCM of 3 and 2) to clear the denominators:

2(3+t) + 3(t) = 6(6)

6 + 2t + 3t = 36

5t = 30

t = 6

Substitute

t=6

into equation (3):

s = 3 + 6

s = 9

Final Answer:

s=9, t=6

.

(iii) $3x-y=3$ and $9x-3y=9$

Given:

3x-y=3 \quad ...(1)

9x-3y=9 \quad ...(2)

Solution: From equation (1),

y = 3x-3

. ...(3) Substitute this into equation (2):

9x - 3(3x-3) = 9

9x - 9x + 9 = 9

9 = 9

This is a true statement for all values of

x

. Therefore, the pair of linear equations has infinitely many solutions. The equations are dependent.

Final Answer: Infinitely many solutions.

(iv) $0.2x+0.3y=1.3$ and $0.4x+0.5y=2.3$

Given:

0.2x+0.3y=1.3 \quad ...(1)

0.4x+0.5y=2.3 \quad ...(2)

Multiply both equations by 10 to remove decimals:

2x+3y=13 \quad ...(3)

4x+5y=23 \quad ...(4)

Solution: From equation (3),

2x = 13-3y \implies x = \frac{13-3y}{2}

. ...(5) Substitute this into equation (4):

4(\frac{13-3y}{2}) + 5y = 23

2(13-3y) + 5y = 23

26 - 6y + 5y = 23

26 - y = 23

y = 3

Substitute

y=3

into equation (5):

x = \frac{13-3(3)}{2} = \frac{13-9}{2} = \frac{4}{2} = 2

Final Answer:

x=2, y=3

.

(v) $\sqrt{2}x+\sqrt{3}y=0$ and $\sqrt{3}x-\sqrt{8}y=0$

Given:

\sqrt{2}x+\sqrt{3}y=0 \quad ...(1)

\sqrt{3}x-\sqrt{8}y=0 \quad ...(2)

Solution: From equation (1),

\sqrt{2}x = -\sqrt{3}y \implies x = -\frac{\sqrt{3}}{\sqrt{2}}y

. ...(3) Substitute this into equation (2):

\sqrt{3}(-\frac{\sqrt{3}}{\sqrt{2}}y) - \sqrt{8}y = 0

-\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0

y(-\frac{3}{\sqrt{2}} - \sqrt{8}) = 0

Since

(-\frac{3}{\sqrt{2}} - \sqrt{8}) \neq 0

, we must have

y=0

. Substitute

y=0

into equation (3):

x = -\frac{\sqrt{3}}{\sqrt{2}}(0) = 0

Final Answer:

x=0, y=0

.

(vi) $\frac{3x}{2}-\frac{5y}{3}=-2$ and $\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Given:

\frac{3x}{2}-\frac{5y}{3}=-2 \quad ...(1)

\frac{x}{3}+\frac{y}{2}=\frac{13}{6} \quad ...(2)

Multiply equation (1) by 6 (LCM of 2, 3):

9x - 10y = -12 \quad ...(3)

Multiply equation (2) by 6 (LCM of 3, 2, 6):

2x + 3y = 13 \quad ...(4)

Solution: From equation (4),

2x = 13-3y \implies x = \frac{13-3y}{2}

. ...(5) Substitute this into equation (3):

9(\frac{13-3y}{2}) - 10y = -12

Multiply by 2:

9(13-3y) - 20y = -24

117 - 27y - 20y = -24

117 - 47y = -24

47y = 117 + 24

47y = 141

y = 3

Substitute

y=3

into equation (5):

x = \frac{13-3(3)}{2} = \frac{13-9}{2} = \frac{4}{2} = 2

Final Answer:

x=2, y=3

.

Q2Exercise 3.2

Solve $2 x+3 y=11$ and $2 x-4 y=-24$ and hence find the value of ' $m$ ' for which $y=m x+3$ .

Solution

Given Equations:

2x + 3y = 11 \quad ...(1)

2x - 4y = -24 \quad ...(2)

And

y = mx + 3 \quad ...(3)

Step 1: Solve the pair of linear equations. We can use the elimination method as the coefficient of

x

is the same. Subtract equation (2) from equation (1):

(2x + 3y) - (2x - 4y) = 11 - (-24)

2x + 3y - 2x + 4y = 11 + 24

7y = 35

y = 5

Substitute

y=5

into equation (1):

2x + 3(5) = 11

2x + 15 = 11

2x = 11 - 15

2x = -4

x = -2

So, the solution to the pair of equations is

x=-2, y=5

.

Step 2: Find the value of 'm'. Substitute the values of

x

and

y

into the equation

y = mx + 3

.

5 = m(-2) + 3

5 = -2m + 3

5 - 3 = -2m

2 = -2m

m = \frac{2}{-2}

m = -1

Final Answer: The solution is

x = -2, y = 5

, and the value of

m

is -1.

Q3Exercise 3.2

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i)

The difference between two numbers is 26 and one number is three times the other. Find them.

(ii)

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii)

The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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?

(v)

A fraction becomes $\frac{9}{11}$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$ . Find the fraction.

(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?

Solution

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Let: The two numbers be

x

and

y

, with

x > y

.

Equations:

x - y = 26 \quad ...(1)

x = 3y \quad ...(2)

Solution (Substitution): Substitute equation (2) into equation (1):

3y - y = 26

2y = 26

y = 13

Substitute

y=13

into equation (2):

x = 3(13) = 39

Final Answer: The numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let: The two supplementary angles be

x

and

y

, with

x > y

.

Equations: Supplementary angles add up to 180 degrees.

x + y = 180 \quad ...(1)

The larger exceeds the smaller by 18.

x = y + 18 \quad ...(2)

Solution (Substitution): Substitute equation (2) into equation (1):

(y + 18) + y = 180

2y + 18 = 180

2y = 162

y = 81

Substitute

y=81

into equation (2):

x = 81 + 18 = 99

Final Answer: The angles are 99 degrees and 81 degrees.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

Let: Cost of one bat be ₹

x

and cost of one ball be ₹

y

.

Equations:

7x + 6y = 3800 \quad ...(1)

3x + 5y = 1750 \quad ...(2)

Solution (Substitution): From equation (2),

3x = 1750 - 5y \implies x = \frac{1750 - 5y}{3}

. ...(3) Substitute this into equation (1):

7(\frac{1750 - 5y}{3}) + 6y = 3800

Multiply by 3:

7(1750 - 5y) + 18y = 11400

12250 - 35y + 18y = 11400

12250 - 17y = 11400

17y = 12250 - 11400 = 850

y = 50

Substitute

y=50

into equation (3):

x = \frac{1750 - 5(50)}{3} = \frac{1750 - 250}{3} = \frac{1500}{3} = 500

Final Answer: The cost of a bat is ₹ 500 and the cost of a ball is ₹ 50.

(iv) Taxi charges...

Let: The fixed charge be ₹

x

and the charge per km be ₹

y

.

Equations: For 10 km:

x + 10y = 105 \quad ...(1)

For 15 km:

x + 15y = 155 \quad ...(2)

Solution (Substitution): From equation (1),

x = 105 - 10y

. ...(3) Substitute this into equation (2):

(105 - 10y) + 15y = 155

105 + 5y = 155

5y = 50

y = 10

Substitute

y=10

into equation (3):

x = 105 - 10(10) = 105 - 100 = 5

Charges for 25 km: Total charge = Fixed charge + (Charge per km)

\times

distance Total charge =

x + 25y = 5 + 25(10) = 5 + 250 = 255

Final Answer: The fixed charge is ₹ 5, the charge per km is ₹ 10, and the cost for travelling 25 km is ₹ 255.

(v) A fraction becomes $\frac{9}{11}$ ...

Let: The numerator be

x

and the denominator be

y

. The fraction is

\frac{x}{y}

.

Equations:

\frac{x+2}{y+2} = \frac{9}{11} \implies 11(x+2) = 9(y+2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4 \quad ...(1)

\frac{x+3}{y+3} = \frac{5}{6} \implies 6(x+3) = 5(y+3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3 \quad ...(2)

Solution (Substitution): From equation (2),

6x = 5y - 3 \implies x = \frac{5y - 3}{6}

. ...(3) Substitute this into equation (1):

11(\frac{5y - 3}{6}) - 9y = -4

Multiply by 6:

11(5y - 3) - 54y = -24

55y - 33 - 54y = -24

y - 33 = -24

y = 9

Substitute

y=9

into equation (3):

x = \frac{5(9) - 3}{6} = \frac{45 - 3}{6} = \frac{42}{6} = 7

Final Answer: The fraction is

\frac{7}{9}

.

(vi) Five years hence, the age of Jacob...

Let: Jacob's present age be

x

years and his son's present age be

y

years.

Equations: Five years hence: Jacob's age =

x+5

, Son's age =

y+5

x+5 = 3(y+5) \implies x+5 = 3y+15 \implies x - 3y = 10 \quad ...(1)

Five years ago: Jacob's age =

x-5

, Son's age =

y-5

x-5 = 7(y-5) \implies x-5 = 7y-35 \implies x - 7y = -30 \quad ...(2)

Solution (Substitution): From equation (1),

x = 3y + 10

. ...(3) Substitute this into equation (2):

(3y + 10) - 7y = -30

-4y + 10 = -30

-4y = -40

y = 10

Substitute

y=10

into equation (3):

x = 3(10) + 10 = 30 + 10 = 40

Final Answer: Jacob's present age is 40 years and his son's present age is 10 years.

Q1Exercise 3.3

Solve the following pair of linear equations by the elimination method and the substitution method:

(i)

$x+y=5$ and $2 x-3 y=4$

(ii)

$3 x+4 y=10$ and $2 x-2 y=2$

(iii)

$3 x-5 y-4=0$ and $9 x=2 y+7$

(iv)

$\frac{x}{2}+\frac{2 y}{3}=-1$ and $x-\frac{y}{3}=3$

Solution

(i) $x+y=5$ and $2x-3y=4$

Elimination Method:

x+y=5 \quad ...(1)

2x-3y=4 \quad ...(2)

Multiply equation (1) by 3:

3x+3y=15 \quad ...(3)

Add equation (2) and (3):

(2x-3y) + (3x+3y) = 4 + 15

5x = 19 \implies x = \frac{19}{5}

Substitute

x=\frac{19}{5}

into (1):

\frac{19}{5} + y = 5 \implies y = 5 - \frac{19}{5} = \frac{25-19}{5} = \frac{6}{5}

Substitution Method: From (1),

y = 5-x

. Substitute into (2):

2x - 3(5-x) = 4

2x - 15 + 3x = 4

5x = 19 \implies x = \frac{19}{5}

y = 5 - \frac{19}{5} = \frac{6}{5}

Final Answer:

x=\frac{19}{5}, y=\frac{6}{5}

.

(ii) $3x+4y=10$ and $2x-2y=2$

Elimination Method:

3x+4y=10 \quad ...(1)

2x-2y=2 \quad ...(2)

Multiply equation (2) by 2:

4x-4y=4 \quad ...(3)

Add equation (1) and (3):

(3x+4y) + (4x-4y) = 10 + 4

7x = 14 \implies x = 2

Substitute

x=2

into (2):

2(2) - 2y = 2 \implies 4 - 2y = 2 \implies 2y = 2 \implies y=1

Substitution Method: From (2),

2(x-y)=2 \implies x-y=1 \implies y=x-1

. Substitute into (1):

3x + 4(x-1) = 10

3x + 4x - 4 = 10

7x = 14 \implies x = 2

y = 2 - 1 = 1

Final Answer:

x=2, y=1

.

(iii) $3x-5y-4=0$ and $9x=2y+7$

Rewrite as:

3x-5y=4 \quad ...(1)

9x-2y=7 \quad ...(2)

Elimination Method: Multiply equation (1) by 3:

9x-15y=12 \quad ...(3)

Subtract equation (3) from (2):

(9x-2y) - (9x-15y) = 7 - 12

13y = -5 \implies y = -\frac{5}{13}

Substitute

y = -\frac{5}{13}

into (1):

3x - 5(-\frac{5}{13}) = 4 \implies 3x + \frac{25}{13} = 4 \implies 3x = 4 - \frac{25}{13} = \frac{52-25}{13} = \frac{27}{13}

x = \frac{27}{13 \times 3} = \frac{9}{13}

Substitution Method: From (2),

x = \frac{2y+7}{9}

. Substitute into (1):

3(\frac{2y+7}{9}) - 5y = 4

\frac{2y+7}{3} - 5y = 4

2y+7 - 15y = 12

-13y = 5 \implies y = -\frac{5}{13}

x = \frac{2(-5/13)+7}{9} = \frac{-10/13 + 91/13}{9} = \frac{81/13}{9} = \frac{9}{13}

Final Answer:

x=\frac{9}{13}, y=-\frac{5}{13}

.

(iv) $\frac{x}{2}+\frac{2y}{3}=-1$ and $x-\frac{y}{3}=3$

Clear denominators: Multiply first equation by 6:

3x+4y=-6 \quad ...(1)

Multiply second equation by 3:

3x-y=9 \quad ...(2)

Elimination Method: Subtract equation (2) from (1):

(3x+4y) - (3x-y) = -6 - 9

5y = -15 \implies y = -3

Substitute

y=-3

into (2):

3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x=2

Substitution Method: From (2),

y = 3x-9

. Substitute into (1):

3x + 4(3x-9) = -6

3x + 12x - 36 = -6

15x = 30 \implies x = 2

y = 3(2) - 9 = 6 - 9 = -3

Final Answer:

x=2, y=-3

.

Q2Exercise 3.3

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

(i)

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

(ii)

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii)

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv)

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v)

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution

(i) Fraction problem

Let: The fraction be

\frac{x}{y}

.

Equations:

\frac{x+1}{y-1} = 1 \implies x+1 = y-1 \implies x-y = -2 \quad ...(1)

\frac{x}{y+1} = \frac{1}{2} \implies 2x = y+1 \implies 2x-y = 1 \quad ...(2)

Solution (Elimination): Subtract equation (1) from (2):

(2x-y) - (x-y) = 1 - (-2)

x = 3

Substitute

x=3

into (1):

3-y = -2 \implies y = 5

Final Answer: The fraction is

\frac{3}{5}

.

(ii) Ages of Nuri and Sonu

Let: Nuri's present age be

n

and Sonu's present age be

s

.

Equations: Five years ago:

(n-5) = 3(s-5) \implies n-5 = 3s-15 \implies n-3s = -10 \quad ...(1)

Ten years later:

(n+10) = 2(s+10) \implies n+10 = 2s+20 \implies n-2s = 10 \quad ...(2)

Solution (Elimination): Subtract equation (1) from (2):

(n-2s) - (n-3s) = 10 - (-10)

s = 20

Substitute

s=20

into (2):

n - 2(20) = 10 \implies n - 40 = 10 \implies n = 50

Final Answer: Nuri is 50 years old and Sonu is 20 years old.

(iii) Two-digit number

Let: The ten's digit be

x

and the unit's digit be

y

. The number is

10x+y

.

Equations: Sum of digits:

x+y=9 \quad ...(1)

Condition:

9(10x+y) = 2(10y+x)

90x + 9y = 20y + 2x

88x - 11y = 0

8x - y = 0 \quad ...(2)

Solution (Elimination): Add equation (1) and (2):

(x+y) + (8x-y) = 9 + 0

9x = 9 \implies x = 1

Substitute

x=1

into (1):

1+y=9 \implies y=8

Final Answer: The number is

10(1)+8 = 18

.

(iv) Bank notes

Let: Number of ₹ 50 notes be

x

and number of ₹ 100 notes be

y

.

Equations: Total notes:

x+y = 25 \quad ...(1)

Total value:

50x + 100y = 2000

. Divide by 50:

x+2y = 40 \quad ...(2)

Solution (Elimination): Subtract equation (1) from (2):

(x+2y) - (x+y) = 40 - 25

y = 15

Substitute

y=15

into (1):

x+15=25 \implies x=10

Final Answer: Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

(v) Library charges

Let: The fixed charge for the first 3 days be ₹

x

and the additional charge per day thereafter be ₹

y

.

Equations: Saritha (7 days = 3 fixed + 4 extra days):

x + 4y = 27 \quad ...(1)

Susy (5 days = 3 fixed + 2 extra days):

x + 2y = 21 \quad ...(2)

Solution (Elimination): Subtract equation (2) from (1):

(x+4y) - (x+2y) = 27 - 21

2y = 6 \implies y = 3

Substitute

y=3

into (2):

x + 2(3) = 21 \implies x+6=21 \implies x=15

Final Answer: The fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.

Pair of Linear Equations in Two VariablesClass 10 Mathematics NCERT Solutions

Form the pair of linear equations in the following problems, and find their solutions graphically.

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.

5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Solution

On comparing the ratios a1a2,b1b2\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}a2​a1​​,b2​b1​​ and c1c2\frac{c_{1}}{c_{2}}c2​c1​​, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

5x−4y+8=05 x-4 y+8=05x−4y+8=0 7x+6y−9=07 x+6 y-9=07x+6y−9=0

9x+3y+12=09 x+3 y+12=09x+3y+12=0 18x+6y+24=018 x+6 y+24=018x+6y+24=0

6x−3y+10=06 x-3 y+10=06x−3y+10=0 2x−y+9=02 x-y+9=02x−y+9=0

Solution

On comparing the ratios a1a2,b1b2\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}a2​a1​​,b2​b1​​ and c1c2\frac{c_{1}}{c_{2}}c2​c1​​, find out whether the following pair of linear equations are consistent, or inconsistent.

3x+2y=5;2x−3y=73 x+2 y=5 ; \quad 2 x-3 y=73x+2y=5;2x−3y=7

2x−3y=8;4x−6y=92 x-3 y=8 ; \quad 4 x-6 y=92x−3y=8;4x−6y=9

32x+53y=7;9x−10y=14\frac{3}{2} x+\frac{5}{3} y=7 ; 9 x-10 y=1423​x+35​y=7;9x−10y=14

5x−3y=11;−10x+6y=−225 x-3 y=11 ;-10 x+6 y=-225x−3y=11;−10x+6y=−22

43x+2y=8;2x+3y=12\frac{4}{3} x+2 y=8 ; 2 x+3 y=1234​x+2y=8;2x+3y=12

Solution

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

x+y=5,2x+2y=10x+y=5, \quad 2 x+2 y=10x+y=5,2x+2y=10

x−y=8,3x−3y=16x-y=8, \quad 3 x-3 y=16x−y=8,3x−3y=16

2x+y−6=0,4x−2y−4=02 x+y-6=0, \quad 4 x-2 y-4=02x+y−6=0,4x−2y−4=0

2x−2y−2=0,4x−4y−5=02 x-2 y-2=0, \quad 4 x-4 y-5=02x−2y−2=0,4x−4y−5=0

Solution

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.

Solution

Given the linear equation 2x+3y−8=02 x+3 y-8=02x+3y−8=0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

intersecting lines

parallel lines

coincident lines

Solution

Draw the graphs of the equations x−y+1=0x-y+1=0x−y+1=0 and 3x+2y−12=03 x+2 y-12=03x+2y−12=0. Determine the coordinates of the vertices of the triangle formed by these lines and the xxx-axis, and shade the triangular region.

Solution

Solve the following pair of linear equations by the substitution method.

x+y=14x+y=14x+y=14 x−y=4x-y=4x−y=4

s−t=3s-t=3s−t=3 s3+t2=6\frac{s}{3}+\frac{t}{2}=63s​+2t​=6

3x−y=33 x-y=33x−y=3 9x−3y=99 x-3 y=99x−3y=9

0.2x+0.3y=1.30.2 x+0.3 y=1.30.2x+0.3y=1.3 0.4x+0.5y=2.30.4 x+0.5 y=2.30.4x+0.5y=2.3

2x+3y=0\sqrt{2} x+\sqrt{3} y=02​x+3​y=0 3x−8y=0\sqrt{3} x-\sqrt{8} y=03​x−8​y=0

3x2−5y3=−2\frac{3 x}{2}-\frac{5 y}{3}=-223x​−35y​=−2 x3+y2=136\frac{x}{3}+\frac{y}{2}=\frac{13}{6}3x​+2y​=613​

Solution

Solve 2x+3y=112 x+3 y=112x+3y=11 and 2x−4y=−242 x-4 y=-242x−4y=−24 and hence find the value of ' mmm ' for which y=mx+3y=m x+3y=mx+3.

Solution

Form the pair of linear equations for the following problems and find their solution by substitution method.

The difference between two numbers is 26 and one number is three times the other. Find them.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

A fraction becomes 911\frac{9}{11}119​, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56\frac{5}{6}65​. Find the fraction.

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?

Solution

Solve the following pair of linear equations by the elimination method and the substitution method:

x+y=5x+y=5x+y=5 and 2x−3y=42 x-3 y=42x−3y=4

3x+4y=103 x+4 y=103x+4y=10 and 2x−2y=22 x-2 y=22x−2y=2

3x−5y−4=03 x-5 y-4=03x−5y−4=0 and 9x=2y+79 x=2 y+79x=2y+7

x2+2y3=−1\frac{x}{2}+\frac{2 y}{3}=-12x​+32y​=−1 and x−y3=3x-\frac{y}{3}=3x−3y​=3

Solution

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12\frac{1}{2}21​ if we only add 1 to the denominator. What is the fraction?

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution

On comparing the ratios $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

$5 x-4 y+8=0$ $7 x+6 y-9=0$

$9 x+3 y+12=0$ $18 x+6 y+24=0$

$6 x-3 y+10=0$ $2 x-y+9=0$

On comparing the ratios $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}$ and $\frac{c_{1}}{c_{2}}$ , find out whether the following pair of linear equations are consistent, or inconsistent.

$3 x+2 y=5 ; \quad 2 x-3 y=7$

$2 x-3 y=8 ; \quad 4 x-6 y=9$

$\frac{3}{2} x+\frac{5}{3} y=7 ; 9 x-10 y=14$

$5 x-3 y=11 ;-10 x+6 y=-22$

$\frac{4}{3} x+2 y=8 ; 2 x+3 y=12$

$x+y=5, \quad 2 x+2 y=10$

$x-y=8, \quad 3 x-3 y=16$

$2 x+y-6=0, \quad 4 x-2 y-4=0$

$2 x-2 y-2=0, \quad 4 x-4 y-5=0$

Given the linear equation $2 x+3 y-8=0$ , write another linear equation in two variables such that the geometrical representation of the pair so formed is:

Draw the graphs of the equations $x-y+1=0$ and $3 x+2 y-12=0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and shade the triangular region.

$x+y=14$ $x-y=4$

$s-t=3$ $\frac{s}{3}+\frac{t}{2}=6$

$3 x-y=3$ $9 x-3 y=9$

$0.2 x+0.3 y=1.3$ $0.4 x+0.5 y=2.3$

$\sqrt{2} x+\sqrt{3} y=0$ $\sqrt{3} x-\sqrt{8} y=0$

$\frac{3 x}{2}-\frac{5 y}{3}=-2$ $\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Solve $2 x+3 y=11$ and $2 x-4 y=-24$ and hence find the value of ' $m$ ' for which $y=m x+3$ .

A fraction becomes $\frac{9}{11}$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$ . Find the fraction.

$x+y=5$ and $2 x-3 y=4$

$3 x+4 y=10$ and $2 x-2 y=2$

$3 x-5 y-4=0$ and $9 x=2 y+7$

$\frac{x}{2}+\frac{2 y}{3}=-1$ and $x-\frac{y}{3}=3$

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?