# 10th CBSE maths solution for exercise 3.3 part 1

This page 10th cbse maths solution for exercise 3.3 part 1 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 1

(1) Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x - y = 4

Solution:

Step 1:

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

y = 14 - 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 x - y = 4

x - (14 - x) = 4

x - 14 + x  = 4

2 x - 14 = 4

2 x = 4 + 14

2 x = 18

x = 18/2

x = 9

Step 3:

Now,we have to apply the value of x in the equation y = 14 - x

y = 14 - 9

y = 5

Therefore solution is

x = 9 and  y = 5

(ii) s - t = 3

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

Solution:

Step 1:

Find the value of one variable in terms of other variable,say s in terms of t

s = 3 + t

Step 2:

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

the other equation is (s/3) + (t/2) = 6

let us simplify this equation by taking L.C.M 2 s + 3 t = 36

2 (3 + t) + 3 t = 36

6 + 2 t + 3 t = 36

6 + 5 t = 36

5 t = 36 - 6

5 t = 30

t = 30/5

t = 6

Step 3:

Now,we have to apply the value of t in the equation s = 3 + t

s = 3 + 6

s = 9

Therefore solution is

s = 9 and  t = 6

(iii) 3 x - y = 3

9 x - 3 y = 9

Solution:

Step 1:

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

y = 3 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 9 x - 3 y = 9

9 x - 3 (3 x - 3)  = 9

9 x - 9 x + 9  = 9

The statement is true,from this we can decide that the pair of linear equations has infinitely many solution.

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