📝NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
EXERCISE 3.1
1. Aftab tells his daughter, “Seven years ago, I was seven times as old as
you were then. Also, three years from now, I shall be three times as old as
you will be.” (Isn’t this interesting?) Represent this situation
algebraically and graphically.
Solution:
Let the present age of the Aftab be `x` and the daughter's age be `y`.
Seven years ago
Aftab's age `= (x -7)`
Daughter's age `= (y - 7)`
`(x - 7) = 7(y - 7)`
`⇒ x - 7 = 7y -49`
`⇒ x - 7y = - 49 + 7`
`⇒ x- 7y = - 42`
Three years from now
Aftab's age `= (x +3)`
Daughter's age `= (y + 3)`
`(x + 3) = 3(y + 3)`
`⇒ x + 3 = 3y +9`
`⇒ x - 3y = 9 -3`
`⇒ x - 3y = 6`
Algebraic representation:
` x- 7y = - 42; x - 3y = 6`
Table
` x- 7y = - 42`
Table
`x - 3y = 6`
Graphical representation:
2. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later,
she buys another bat and 3 more balls of the same kind for ₹1300.
Represent this situation algebraically and geometrically.
Solution:
Let the price of the 1 bat be `₹x` and the price of the 1 ball be `₹y.`
`3x + 6y = 3900---(i)`
`x + 3y = 1300---(ii)`
Table
`3x + 6y = 3900`
3. The cost of 2 kg of apples and 1kg of grapes on a day was found to
be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes
is ₹300. Represent the situation algebraically and geometrically.
Solution
Let the cost of 1 kg apples be `₹x` and the cost of 1kg grapes be `₹y`.
`2x + y = 160 ---(i)`
`4x + 2y = 300 ---(ii)`
Table
`2x + y = 160`
Table
`4x + 2y = 300 `
Video Solution (Click here) 🎥
EXERCISE 3.2
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) Let the number of boys be `x` and the number of girls be `y`.
`x + y = 10---(i)`
If the number of girls is 4 more than the number of boys
`y = x + 4`
`⇒ -x+ y = 4`
or, `x - y = -4 --(ii)`
`x + y = 10`
Table
Table
Both the lines intersect at (3,7)
`therefore` the solution is (3,7)
Girls = 7, Boys = 3.
Video solution (Click here)🎥
(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
(ii) Let the cost of 1 pencil be `₹x` and the cost of 1 pen be `₹y`.
`5x + 7y = ₹50---(i)`
`7x + 5y = ₹46---(ii)`
`5x + 7y = ₹50`
Table
`7x + 5y = ₹46`
Table
Both the lines intersect at (3,5)
`therefore` the solution is (3,5)
Cost of one pencil = ₹3, Cost of one pen = ₹5
Video Solution (Click here) 🎥
2. On comparing the ratios `a_1/a_2, b_1/b_2, 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) 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`
Solutions
`(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0`
`a_1 = 5, b_1 = -4, c_1 = 8`
`a_2 = 7, b_2 = 6, c_2 = -9`
`a_1/a_2 = 5/7, b_1/b_2 = -4/6 = -2/3`
`therefore a_1/a_2 ne b_1/b_2`
Hence, the lines intersect at a point
`(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0`
`a_1 = 9, b_1 = 3, c_1 = 12`
`a_2 = 18, b_2 = 6, c_2 = 24`
`a_1/a_2 = 9/18 = 1/2 , b_1/b_2 = 3/6 = 1/2, c_1/c_2 = 12/24 = 1/2`
`therefore a_1/a_2 = b_1/b_2 = c_1/c_2`
Hence lines are coincident
`(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0`
`a_1 = 6, b_1 = -3, c_1 = 10`
`a_2 = 2, b_2 = -1, c_2 = 9`
`a_1/a_2 = 6/2 = 3 , b_1/b_2 = frac{-3}{-1} = 3, c_1/c_2 =
10/9`
`therefore a_1/a_2 = b_1/b_2 ne c_1/c_2`
Hence lines are parallel
3. On comparing the ratios `a_1/a_2, b_1/b_2` and `c_1/c_2` find out
whether the following pair of linear equations are consistent, or
inconsistent.
`(i) 3x + 2y = 5; 2x – 3y = 7`
`(ii) 2x – 3y = 8 ; 4x – 6y = 9`
`(iii) 3/2x + 5/3y = 7; 9x – 10y = 14`
`(iv) 5x – 3y = 11 ; – 10x + 6y = –22`
`(v) 4/3x + 2y = 8 ; 2x + 3y = 12`
Solutions
`(i) 3x + 2y = 5; 2x – 3y = 7`
`a_1 = 3, b_1 = 2, c_1 = -5`
`a_2 = 2, b_2 = -3, c_2 = -7`
`a_1/a_2 = 3/2, b_1/b_2 = frac{2}{-3}`
`therefore a_1/a_2 ne b_1/b_2`
Hence, pair of linear equation is consistent.
`(ii) 2x – 3y = 8 ; 4x – 6y = 9`
`a_1 = 2, b_1 = -3, c_1 = -8`
`a_2 = 4, b_2 = -6, c_2 = -9`
`a_1/a_2 = 2/4 = 1/2, b_1/b_2 = frac{-3}{-6} = 1/2, c_1/c_2 =
frac{-8}{-9}`
`therefore a_1/a_2 = b_1/b_2 ne c_1/c_2`
Hence, pair of linear equation is inconsistent.
`(iii) 3/2x + 5/3y = 7; 9x – 10y = 14`
`a_1 = 3/2, b_1 = 5/3, c_1 = -7`
`a_2 = 9, b_2 = -10, c_2 = -14`
`a_1/a_2 = frac{3}{2 × 9} = 1/6, b_1/b_2 = frac{5}{3 × -10} =
frac{-1}{6}`
`therefore a_1/a_2 ne b_1/b_2 `
Hence, pair of linear equation is consistent.
`(iv) 5x – 3y = 11 ; – 10x + 6y = –22`
`a_1 = 5, b_1 = -3, c_1 = -11`
`a_2 = -10, b_2 = 6, c_2 = 22`
`a_1/a_2 = frac{5}{-10} = -1/2, b_1/b_2 = frac{-3}{6} = -1/2, c_1/c_2
= frac{-11}{22} = -1/2`
`therefore a_1/a_2 = b_1/b_2 = c_1/c_2`
Hence, pair of linear equation is consistent.
`(v) 4/3x + 2y = 8 ; 2x + 3y = 12`
`a_1 = 4/3, b_1 = 2, c_1 = -8`
`a_2 = 2, b_2 = 3, c_2 = -12`
`a_1/a_2 = frac{4}{3 × 2} = 2/3, b_1/b_2 = 2/3, c_1/c_2 =
frac{-8}{-12} = 2/3`
`therefore a_1/a_2 = b_1/b_2 = c_1/c_2`
Hence, pair of linear equation 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 = 0, 4x – 2y – 4 = 0`
`(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0`
Solutions
`(i) x + y = 5, 2x + 2y = 10`
`a_1 = 1, b_1 = 1, c_1 = -5`
`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 = frac{-5}{-10} = 1/2`
`therefore a_1/a_2 = b_1/b_2 = c_1/c_2`
Hence, pair of linear equation is consistent.
Table
`(i) x + y = 5, 2x + 2y = 10`
`(ii) x – y = 8, 3x – 3y = 16`
`a_1 = 1, b_1 = -1, c_1 = -8`
`a_2 = 3, b_2 = -3, c_2 = -16`
`a_1/a_2 = 1/3, b_1/b_2 = frac{-1}{-3} = 1/3, c_1/c_2 = frac{-8}{-16} =
1/2`
`therefore a_1/a_2 = b_1/b_2 ne c_1/c_2`
Hence, pair of linear equation is inconsistent.
`(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0`
`a_1 = 2, b_1 = 1, c_1 = -6`
`a_2 = 4, b_2 = -2, c_2 = -4`
`a_1/a_2 = 2/4 = 1/2, b_1/b_2 = frac{1}{-2} = -1/2`
`therefore a_1/a_2 ne b_1/b_2`
Hence, pair of linear equation is consistent.
`(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0`
`a_1 = 2, b_1 = -2, c_1 = -2`
`a_2 = 4, b_2 = -4, c_2 = -5`
`a_1/a_2 = 2/4 = 1/2, b_1/b_2 = frac{-2}{-4} = 1/2, c_1/c_2 = frac{-2}{-5}
= 2/5`
`therefore a_1/a_2 = b_1/b_2 ne c_1/c_2`
Hence, pair of linear equation is inconsistent.
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.
Solution
Let the breadth be `x` and the length be `y`.
Half the perimeter of a rectangular garden is 36
`1/2 × 2(y + x) = 36`
or, `x+ y = 36---(i)`
Length is 4 m more than its width.
`y = x+ 4`
or, `-x + y = 4---(ii)`
Table
`x+ y = 36`
Table
`-x + y = 4`
`therefore l = 20 m, b = 16 m`
Video Solutions (Click here)🎥
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
Solutions
(i) intersecting lines
`⇒ a_1/a_2 ne b_1/b_2`
Given the linear equation `2x + 3y – 8 = 0`
So, 3x +2y -7 = 0
`a_1/a_2 = 2/3, b_1/b_2 = frac{3}{2}`
`therefore a_1/a_2 ne b_1/b_2`
(ii) parallel lines
`⇒ a_1/a_2 = b_1/b_2 ne c_1/c_2`
Given the linear equation `2x + 3y – 8 = 0`
So, `2x + 3y -12 = 0`
`a_1/a_2 = 2/2, b_1/b_2 = frac{3}{3}, c_1/c_2 = frac{-8}{-12}`
`therefore a_1/a_2 = b_1/b_2 ne c_1/c_2`
(iii) coincident lines
`⇒ a_1/a_2 = b_1/b_2 = c_1/c_2`
Given the linear equation `2x + 3y – 8 = 0`
So, `4x + 6y -16 = 0`
`a_1/a_2 = 2/4, b_1/b_2 = 3/6, c_1/c_2 = frac{-8}{-16}`
`therefore a_1/a_2 = b_1/b_2 = c_1/c_2`
Video solution Click here 🎥
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.
Solution
`x-y + 1 = 0`
`⇒ -y = -1 -x`
or, `y = x +1`
Table
`3x + 2y – 12 = 0`
`⇒ 2y = 12 -3x`
`⇒ y = frac{12 -3x}{2}`
Table
Video solution (Click here) 🎥
EXERCISE 3.3
1. Solve the following pair of linear equations by the substitution
method.
`(i) x + y = 14; x-y = 4`
Solution
`x + y = 14 --(i)`
`x - y = 4--(ii)`
`⇒x = y +4`
Putting the value of `x` in equation `(i)`
`(y + 4) + y = 14`
`⇒ 2y = 10`
`therefore y = 10/2 = 5`
Putting the value of `y` in equation `(ii)`
`x - 5 = 4`
`therefore x = 5+ 4 = 9`
Video Solution (Click here) 🎥
`(ii) s – t = 3`
`s/3 + t/2 = 6`
Solution
`s – t = 3---(i)`
`s/3 + t/2 = 6---(ii)`
Solving equation (i)
`s = t+3`
Putting the value of `s` in equation (ii)
`s/3 + t/2 = 6`
`⇒ frac{(t+3)}{3} + t/2 = 6`
Multiply both the side by 6
`2(t + 3) + 3t = 36`
`2t + 6 + 3t = 36`
`5t = 30`
`therefore t = 30/5 = 6`
Putting the value of `t` in equation (i)
`s -6 = 3`
`⇒s = 9`
Video Solution (Click here) 🎥
`(iii) 3x – y = 3 ; 9x - 3y = 9`
Solution
`3x – y = 3---(i)`
`9x - 3y = 9---(ii)`
Solving equation (i)
`3x - y = 3`
`⇒ -y = 3 - 3x`
or, `y = 3x -3`
Putting the value of y in equation (ii)
`9x - 3(3x -3) = 9`
`9x - 9x + 9 = 9`
`9 = 9`
`therefore y` have infinite values.
`x` can take any value.
Video Solution (Click here ) 🎥
`(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3`
Solution
`0.2x + 0.3y = 1.3---(i)`
`0.4x + 0.5y = 2.3---(ii)`
Solving equation (i)
0.2x = 1.3 - 0.3y
`x = frac{1.3 - 0.3y}{0.2}`
Putting the value of x in equation (ii)
`0.4( frac{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`
`⇒ -0.1y = 2.3 - 2.6`
`⇒ -0.1 y = -0.3`
`⇒y = 0.3/0.1`
`therefore y = 3`
`x = frac{1.3 - 0.3× 3}{0.2}`
`x = frac{1.3 - 0.9}{0.2}`
` therefore x = 0.4/0.2 = 2`
Video solution (Click here) 🎥
`(v) sqrt2x + sqrt3y = 0; sqrt3x - sqrt8y = 0`
Solution
`sqrt2x + sqrt3y = 0---(i)`
`sqrt3x - sqrt8y = 0---(ii)`
Solving equation (i)
`sqrt2x + sqrt3y = 0`
`⇒sqrt2x = -sqrt3y`
`x = frac{-sqrt3y}{sqrt2}`
Putting the value of x in equation (ii)
`sqrt3 (frac{-sqrt3y}{sqrt2}) - sqrt8y = 0`
` ⇒-3/sqrt2y - sqrt8y = 0`
` ⇒ -3y - 4 y = 0`
`⇒ - 7y = 0`
` therefore y = 0`
Putting the value of y in equation (i)
`sqrt2x + 0 = 0`
`therefore x = 0`
Video Solution (Click here) 🎥
`(vi) frac{3x}{2} - frac{5y}{3} = -2; x/3 + y/2 =
13/6`
Solution
`frac{3x}{2} - frac{5y}{3} = -2---(i)`
`x/3 + y/2 = 13/6---(ii)`
Solving equation (ii)
`x/3 + y/2 = 13/6`
Multiply both the side by 6
`2x + 3y = 13`
`⇒ 2x = 13 - 3y`
`⇒ x = frac{13 - 3y}{2}`
Putting the value of x in equation (i)
`frac{3(13- 3y)}{4} - frac{5y}{3} = -2`
Multiply both the side by 12
`9(13 - 3y) -20y = -24`
`⇒ 117 - 27y -20y = -24`
`⇒ -47y = -24 -117`
`⇒ y = 141/47`
`⇒ y = 3`
Putting the value of y in equation (ii)
`x/3 + 3/2 =13/6`
`⇒ x/3 = 13/6 - 3/2`
`⇒ x/3 = frac{13 - 9}{6}`
`⇒ x = 4/6 × 3`
`⇒x = 2`
Video Solution (Click here) 🎥
2. Solve `2x + 3y = 11` and `2x – 4y = – 24` and hence find the value of
‘m’ for which `y = mx + 3.`
Solution
`2x + 3y = 11---(i)`
`2x – 4y = – 24---(ii)`
Solving equation (i)
`2x + 3y = 11`
`⇒2x = 11 -3y`
`⇒ x = frac{11-3y}{2}`
Putting the value of x in equation (ii)
`2frac{(11 -3y)}{2} - 4y = -24`
`⇒11 - 3y - 4y = -24`
`⇒-7y = -24-11`
`⇒ -7y = -35`
`⇒ y = 35/7 = 5`
Putting the value of y in equation (i)
`2x + 3y = 11`
`⇒ 2x + 3 × 5 = 11`
`⇒ 2x = 11 -15`
`⇒ 2x = -4`
`⇒ x = -2`
`y = mx + 3`
⇒ `5 = -2m + 3`
`⇒ -2m = 5 -3`
`⇒ -2m = 2`
`⇒ m = -1`
Video Solution (Click here) 🎥
3. 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.
Solution
Let the 1st number be x and the 2nd number be y.
Let `x>y`
1st condition
`x - y = 26---(i)`
2nd condition
`x = 3y---(ii)`
Putting the value of x in the equation (i)
`3y - y = 26`
`⇒ 2y = 26`
`⇒ y = 26/2 = 13`
then, `x = 3 × 13 = 39`
Video Solution (Click here) 🎥
(ii) The larger of two supplementary angles exceeds the smaller by
18 degrees. Find them.
Solution
Let the one angle be x and its supplementary angle be y.
Let `x> y`.
1st condition
`x + y = 180°---(i)`
2nd condition
`x - y = 18°---(ii)`
or `x = y + 18°`
Putting the value of x in equation (i)
`y + 18° + y = 180°`
`⇒ 2y = 180° - 18°`
`⇒ y = frac{162°}{2} = 81°`
`x = y + 18°`
⇒ x = 81° + 18° = 99°
Video Solution (Click here) 🎥
(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.
Solution
Let the price of the one bat be ₹x and the price of the one ball be ₹y
1st condition
`7x + 6y = 3800---(i)`
2nd condition
`3x + 5y = 1750 ---(ii)`
Solving equation (ii)
`3x = 1750 - 5y`
`x = frac{1750 -5y}{3}`
Putting the value of x in equation (i)
`7(frac{1750 -5y}{3}) + 6y = 3800`
Multiply both the side by 3
`7(1750 - 5y) + 18y = 11400`
`⇒ 12250 - 35y + 18y = 11400`
`⇒ -17y = 11400 -12250`
`⇒ -17y = -850`
`⇒ y = 850/17 = 50`
Putting the value of y in equation (ii)
`3x + 5 × 50 = 1750`
`⇒ 3x = 1750 -250`
`⇒ 3x = 1500`
`⇒ x = 1500/3 = 500`
The price of the 1 bat = ₹500
1 ball = ₹50
Video Solution (Click here) 🎥
(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?
Solution
Let the fixed charge be ₹x and charge per km be ₹y.
1st condition
`x + 10y = 105---(i)`
2nd condition
`x + 15y = 155---(ii)`
Solving equation (i)
`x + 10y = 105`
`⇒ x = 105 - 10y`
Putting the value of x in equation (ii)
`105 - 10y + 15y = 155`
`⇒ 5y = 155-105`
`⇒ y = 50/5 = 10`
Putting the value of y in equation (i)
`x + 10 × 10 = 105`
`⇒ x = 105 -100`
`⇒ x = 5`
Fixed charge = ₹5
Charge per km = ₹10
3rd condition
`x+ 25y = 5 + 25 × 10 = 255`
The person has to pay ₹255
Video Solution (Click here) 🎥
(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 numerator be x and the denominator be y
1st condition
`frac{x +2}{y + 2} = 9/11`
`⇒ 11(x + 2) = 9 (y + 2)`
`⇒ 11x+ 22 = 9y + 18`
`⇒ 11x - 9y = 18 -22`
`⇒ 11x - 9y = -4---(i)`
2nd condition
`frac{x +3}{y + 3} = 5/6`
`⇒ 6(x + 3) = 5(y + 3)`
`⇒ 6x + 18 = 5y + 15`
`⇒ 6x - 5y = 15 -18`
`⇒ 6x - 5y = -3---(ii)`
Solving equation (i)
`11x - 9y = -4`
`⇒ 11x = 9y -4`
`⇒ x = frac{9y -4}{11}`
Putting the value of x in equation (ii)
`6x - 5y = -3`
`⇒ 6(frac{9y -4}{11}) -5y = -3`
Multiply both the side by 11
`⇒ 6 (9y -4) - 55y = -33`
`⇒54y - 24 - 55y = -33`
`⇒ -y = -33 + 24`
`⇒ y = 9`
Putting the value of y in equation (ii)
`6x - 5 ×9 = -3`
`⇒ 6x = - 3 + 45`
`⇒ x = 42/6 = 7`
Video Solution (Click Here) 🎥
(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
Let the present age of Jacob be `x` and his son's age be `y.`
1st condition
Five years hence
Jacob's age `= x+ 5`
Son's age `= y + 5`
`x+ 5 = 3(y + 5)`
`⇒ x + 5 = 3y + 15`
`⇒ x - 3y = 10---(i)`
2nd condition
Five years ago
Jacob's age `= x -5`
Son's age `= y -5`
`x -5 = 7(y -5)`
`⇒ x - 5 = 7y - 35`
`⇒ x - 7y = -30---(ii)`
Solving equation (i)
`⇒ x - 3y = 10`
` ⇒x = 10 + 3y`
Putting the value of x in equation (ii)
`10 + 3y - 7y = -30`
` ⇒-4y = -30 -10`
`⇒ -4y = -40`
`⇒ y = 40/4 = 10`
`x = 10 + 3y`
`⇒ x = 10 + 3 × 10 = 40`
Jacob's age = 40 years
Son's age = 10 years
Video Solution (Click Here) 🎥
EXERCISE 3.4
1. Solve the following pair of linear equations by the elimination method and the substitution
method : `(i) x + y = 5` and `2x – 3y = 4`
Solution
Elimination method
`x + y = 5---(i)`
`2x - 3y = 4---(ii)`
Multiply equation (i) by 2
2x + 2y = 10---(iii)
Substracting equation (ii) from (iii)
`2x + 2y = 10`
`2x - 3y = 4`
`- + -`
-----------------
`5y = 6`
`⇒ y = 6/5`
Putting the value of y in equation (i)
`x + y = 5`
`⇒ x + 6/5 = 5`
`⇒ x = 5 -6/5`
`⇒ x = frac{25 -6}{5}`
`⇒ x = 19/5`
Substitution method
`x + y = 5---(i)`
`2x - 3y = 4---(ii)`
Solving equation (i)
`x + y = 5`
`⇒ x = 5-y`
Putting the value of x in equation (ii)
`2x - 3y = 4`
`⇒ 2(5 - y) - 3y = 4`
`⇒ 10 - 2y - 3y = 4`
`⇒ -5y = 4 -10`
`⇒ -5y = -6`
`⇒ y = 6/5`
Putting the value of y in equation (i)
`x + y = 5`
`⇒ x + 6/5 = 5`
`⇒ x = 5-6/5`
`⇒ x = frac{25 -6}{5}`
`⇒ x = 19/5`
Video Solution (Click Here) 🎥
`(ii) 3x + 4y = 10` and `2x – 2y = 2`
Solution
Elimination method
`3x + 4y = 10---(i)`
`2x – 2y = 2---(ii)`
Multiply equation (ii) by 2
`4x - 4y = 4---(iii)`
Adding equation (i) & (iii)
`3x + 4y = 10`
`4x -4y = 4`
--------------
`7x = 14`
`⇒ x = 14/7 = 2`
Putting the value of x in equation (ii)
2x – 2y = 2
`⇒ 4 - 2y = 2`
`⇒ -2y = -2`
`⇒ y = 1`
Substitution method
`3x + 4y = 10---(i)`
`2x – 2y = 2---(ii)`
Solving equation (ii)
2x – 2y = 2
`⇒ x - y = 1`
`⇒ x =1+y`
Putting the value of x in equation (i)
3x + 4y = 10
`⇒ 3(1 + y) + 4y = 10`
`⇒ 3 + 3y + 4y = 10`
`⇒ 7y = 7`
`⇒ y = 1`
Putting the value of y in equation (ii)
2x – 2y = 2
`⇒ 2x -2 = 2`
`⇒ 2x = 4`
`⇒ x = 2`
Video Solution (Click here) 🎥
`(iii) 3x – 5y – 4 = 0` and `9x = 2y + 7`
Solution
Elimination method
`3x – 5y – 4 = 0---(i)`
`9x - 2y = 7---(ii)`
Multiply equation (i) by 3
`9x - 15y = 12---(iii)`
Substracting equation (ii) from (iii)
`9x - 15y = 12`
`9x - 2y = 7`
`- + -`
----------------
`-13y = 5`
`y = -5/13`
Putting the value of y in equation (i)
`3x – 5y – 4 = 0`
`3x - 5 (-5/13) -4 = 0`
`3x + 25/13 -4 = 0`
Multiply by 13 both the side
`39x + 25 - 52 = 0`
`39x - 27 = 0`
`x = 27/39 = 9/13`
Substitution method
`3x – 5y – 4 = 0---(i)`
`9x - 2y = 7---(ii)`
Solving equation (i)
`3x – 5y – 4 = 0`
`3x = 5y + 4`
`x = frac{5y+4}{3}`
Putting the value of x in (ii)
`9x - 2y = 7`
`9 (frac{5y+4}{3}) -2y = 7`
`3(5y + 4) - 2y = 7`
`15y + 12 - 2y = 7`
`13y = 7 -12`
`y = -5/13`
Putting the value of y in equation (i)
`3x – 5y – 4 = 0`
`3x - 5 (-5/13) -4 = 0`
`3x + 25/13 -4 = 0`
Multiply by 13 both the side
`39x + 25 - 52 = 0`
`39x - 27 = 0`
`x = 27/39 = 9/13`
Video Solution (Click here) 🎥
`(iv) x/2 + frac{2y}{3} = -1` and `x - y/3 = 3`
Solution
Elimination Method
` x/2 + frac{2y}{3} = -1---(i)`
`x - y/3 = 3---(ii)`
Multiply equation (i) by 6
Multiply equation (ii) by 3
`3x + 4y = -6---(iii)`
`3x - y = 9---(iv)`
Substracting (iv) from (iii)
`3x + 4y = -6`
`3x - y = 9`
`- + -`
--------------
`5y = -15`
`y = -3`
Putting the value of y in (iv)
`3x - y = 9`
`⇒ 3x - (-3) = 9`
`⇒ 3x + 3 = 9`
`⇒ 3x = 6`
`⇒ x = 6/3 = 2`
Substitution Method
` x/2 + frac{2y}{3} = -1---(i)`
`x - y/3 = 3---(ii)`
Multiply equation (i) by 6
Multiply equation (ii) by 3
`3x + 4y = -6---(iii)`
`3x - y = 9---(iv)`
Solving equation (iv)
`-y = 9 - 3x`
`y = 3x - 9`
Putting the value of y in equation (iii)
`3x + 4(3x - 9) = -6`
`3x + 12x - 36 = -6`
`15x = 30`
`x = 2`
Putting the value of x in equation (iv)
`3x - y = 9`
`6 - y = 9`
`-y = 9-6`
`y = -3`
Video Solution (Click here) 🎥
2. 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 `1/ 2` if we only add 1 to the denominator. What is the fraction?
Solution:
Let the numerator be `x` and denominator be `y `.
Then fraction is `x/y`
ATQ
`frac{x + 1}{y -1} = 1`
`⇒ x+1 = y-1`
`⇒ x - y = -2---(i)`
`frac{x}{y+1} = 1/2`
`⇒ 2x = y +1`
`⇒ 2x - y = 1---(ii)`
Subtracting equation (i) form (ii)
`2x - y = 1`
`x - y = -2`
`- + +`
------------
`x = 3`
Putting the value of x in equation (i)
`x - y = -2`
`⇒ 3 - y = -2`
`⇒ - y = -5`
`⇒ y = 5`
fraction is `3/5`
Video Solution (Click here) 🎥
Work in progress
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