1) If Sx +2y = 18 and 2x + Sy = 24 then x+y= (2, 4, 6, 42)
2) The solution of the pair of linear equations 13x 7y
27 and 7x-13y=-19 is x=1
and y (2,-2, 7, 14)
=
3) If 3x+ 2y - 16 and 2x + 3y =19 then x +y_3,-3,7,35)
4)If there is no solution for the linear equation 5x-ky = 12 and x-y=2 then the
value of k = (-7,-5, 5, 7)
5) Sarika is 'x' years old and her daughter is years aid. If twice Sarika's age is added to thrice her daughter's age then the expression will be I X+y+5, 2x + 3y, 3x + 2y)
6) For what value of 'k will the following system of linear equations have infinitely many solutions. (k+2) x + (2k + 1) y = 21-1) and 2x + 3y = 2
7) 5 years ago Amar was x years old and three years hence Akbar will be y years old. Therefore, the sum of their present ages is
(x+ y-2, x+ y+2, x+y+ 8, x+ y-8)
7, then x+y (2, 3, 18 ,32)
8) If 11x + Sy = 25 and 5x + 11y s 9) The present ages of Ram and Ramesh are x and y years respectively. Therefore the sum of their ages 6 years ago was _ years, (x+y+12, x+y+6, x+y-6, x+y-1)
10)OF 5x +y = 31 and 7x + 29 then xy= (1,4,5,-5)
11) If there is no solution for the linear equations 4x ky =10 and x-y = 5, then the value of k= (2.4,-2,-4)
12)) if the given pair of linear equations lay -6= 0 and 10x + 14y- 12 =0 has infinitely many solutions, then the value at k' is(2,3,4, 5) 13}f 4x + 3y = 9 and 3x + 4y =-8, then x-y=_(1, -1,17, - 17)
14)The solution of the equation 2y=4 and 2x - 3y = 1 is_ 1282112
(-2,11 15) 3+ y = 15 and x + 3y - 11, then x _ 24, 13, 26) 16)The solution of the equation 2x +3y = 2 and x + y=2 is s_ 2414222 X4,2)
17) 3x y= 13 and x- 3y = 7 then the value of (x + vi is 3,5,6,20 ) 18)The solution of the equation 5x2y=18 and 2x + 5y =24 is 24422424
19)if 13x + 12y = 7 and 12x + 13y = 5 then xy 1221212
20)The solution of the equations 2x + y = 5 and 3x + 2y = 8s_ 12111221 /-1,2)
answer all questions
Answers
Answer:
Step-by-step explanation:
ANSWER
(i) x + y =5 and 2x –3y = 4
By elimination method
x + y =5 ... (i)
2x –3y = 4 ... (ii)
Multiplying equation (i) by (ii), we get
2x + 2y = 10 ... (iii)
2x –3y = 4 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 - (6/5) = 19/5
Hence, x = 19/5 and y = 6/5
By substitution methodx + y = 5 ... (i)
Subtracting y both side, we get
x = 5 - y ... (iv)
Putting the value of x in equation (ii) we get
2(5 – y) – 3y = 4
-5y = - 6
y = -6/-5 = 6/5
Putting the value of y in equation (iv) we get
x = 5 – 6/5
x = 19/5
Hence, x = 19/5 and y = 6/5 again
(ii) 3x + 4y = 10 and 2x – 2y = 2
By elimination method
3x + 4y = 10 .... (i)
2x – 2y = 2 ... (ii)
Multiplying equation (ii) by 2, we get
4x – 4y = 4 ... (iii)
3x + 4y = 10 ... (i)
Adding equation (i) and (iii), we get
7x + 0 = 14
Dividing both side by 7, we get
x = 14/7 = 2
Putting in equation (i), we get
3x + 4y = 10
3(2) + 4y = 10
6 + 4y = 10
4y = 10 – 6
4y = 4
y = 4/4 = 1
Hence, answer is x = 2, y = 1
By substitution method
3x + 4y = 10 ... (i)
Subtract 3x both side, we get
4y = 10 – 3x
Divide by 4 we get
y = (10 - 3x )/4
Putting this value in equation (ii), we get
2x – 2y = 2 ... (i)
2x – 2(10 - 3x )/4) = 2
Multiply by 4 we get
8x - 2(10 – 3x) = 8
8x - 20 + 6x = 8
14x = 28
x = 28/14 = 2
y = (10 - 3x)/4
y = 4/4 = 1
Hence, answer is x = 2, y = 1 again.
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By elimination method
3x – 5y – 4 = 0
3x – 5y = 4 ...(i)
9x = 2y + 7
9x – 2y = 7 ... (ii)
Multiplying equation (i) by 3, we get
9 x – 15 y = 11 ... (iii)
9x – 2y = 7 ... (ii)
Subtracting equation (ii) from equation (iii), we get
-13y = 5
y = -5/13
Putting value in equation (i), we get
3x – 5y = 4 ... (i)
3x - 5(-5/13) = 4
Multiplying by 13 we get
39x + 25 = 52
39x = 27
x =27/39 = 9/13
Hence our answer is x = 9/13 and y = - 5/13
By substitution method
3x – 5y = 4 ... (i)
Adding 5y both side we get
3x = 4 + 5y
Dividing by 3 we get
x = (4 + 5y )/3 ... (iv)
Putting this value in equation (ii) we get
9x – 2y = 7 ... (ii)
9 ((4 + 5y )/3) – 2y = 7
Solve it we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = - 5
y = -5/13
x = 4 + 5 ( -5/13)/ 3
= 4 - 25/13 / 3
= 4 × 13 - 25/13 / 3
= 27/13×3
= 27/39
= 9/13
Hence we get x = 9/13 and y = - 5/13 again.
(iv) x/2 + 2y/3 = - 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = -1 ... (i)
x – y/3 = 3 ... (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = - 2 ... (iii)
x – y/3 = 3 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = -5
Dividing by 5 and multiplying by 3, we get
y = -15/5
y = - 3
Putting this value in equation (ii), we get
x – y/3 = 3 ... (ii)
x – (-3)/3 = 3
x + 1 = 3
x = 2
Hence our answer is x = 2 and y = −3.
By substitution method
x – y/3 = 3 ... (ii)
Add y/3 both side, we get
x = 3 + y/3 ... (iv)
Putting this value in equation (i) we get
x/2 + 2y/3 = - 1 ... (i)
(3+ y/3)/2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = - 1
Multiplying by 6, we get
9 + y + 4y = - 6
5y = -15
y = - 3
Hence our answer is x = 2 and y = −3.