10th CBSE maths solution for exercise 3.3 part 2

This page 10th cbse maths solution for exercise 3.3 part 2 is going to provide you solution for every problems that you find in the exercise no 3.3

10th CBSE maths solution for exercise 3.3 part 2

(iv)  0.2 x + 0.3 y = 1.3

       0.4 x + 0.5 y = 2.3

Solution:

we are going to multiply both equations by 10.

2 x + 3 y = 13 ----(1)

4 x + 5 y = 23 -----(2)

Step 1:

Find the value of one variable in terms of other variable,say y in terms of x

 3 y = 13 - 2 x

       y = (13 - 2 x)/3

Step 2:

Now we have to substitute the value of y in the other equation,and reduce it to an equation of one variable.

 the other equation is 4 x + 5 y = 23

                   4 x + 5 [(13 -2x)/3] = 23

                   12 x + [5 (13 -2 x)]/3 = 23

                   12 x + 65 -10 x = 69

                   2 x = 69 - 65

                    2 x = 4

                      x = 4/2

                      x = 2

Step 3:

Now,we have to apply the value of x in the equation y = (13 -2x)/3

                         y = (13 -2(2))/3

                         y = (13 -4)/3

                         y = 9/3

                         y = 3

Therefore solution is

  x = 2 and  y = 3


(v)  √2 x + √3y =0

      √3 x - √8 y = 0

Solution:

Step 1:

Find the value of one variable in terms of other variable,say y in terms of x

   √3 y = - √2 x

       y = - (√2/√3) x

Step 2:

Now we have to substitute the value of y in the other equation,and reduce it to an equation of one variable.

 the other equation is √3 x - √8 y = 0

                    √3 x - √8 (- (√2/√3) x) = 0

                    √3 x + (√16/√3) x) = 0

                    (3 x + 4x)/√3 = 0

                    7 x/√3 = 0

                     7 x = 0

                       x = 0

Step 3:

Now,we have to apply the value of x in the equation y = - (√2/√3) x

                         y = - (√2/√3) (0)

                         y = 0

Therefore solution is

  x = 0 and  y = 0


(vi) (3x/2) - (5y/3) = -2

      (x/3) + (y/2) = 13/6

Solution:

we are going to take L.C.M for both equations.


9 x - 10 y = -12 ---- (1)

2 x + 3 y = 13 -----(2)

Step 1:

Find the value of one variable in terms of other variable,say y in terms of x

 10 y = 9 x + 12 

       y = (9 x + 12)/10

Step 2:

Now we have to substitute the value of y in the other equation,and reduce it to an equation of one variable.

 the other equation is 2 x + 3 y = 13

                   2 x + 3 [(9 x + 12)/10] = 13

                   (20 x + 27 x + 36)/10 = 13

                   47 x + 36 = 130

                   47 x = 130 - 36

                    47 x = 94

                        x = 94/47

                        x = 2

Step 3:

Now,we have to apply the value of x in the equation

y = (9 x + 12)/10

                         y = (9(2) + 12)/10

                         y = (18 + 12)/10

                         y = 30/10

                         y = 3

Therefore solution is

  x = 2 and  y = 3       10th CBSE maths solution for exercise 3.3 part 2 10th CBSE maths solution for exercise 3.3 part 2




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