# 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

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(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