Math, asked by blackskull066, 1 month ago

please answer me my question


teniteraj bro in maths​

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Answers

Answered by tennetiraj86
2

Step-by-step explanation:

Solutions :-

1.

Twice the amount A = 2× x = 2x

10 more than amount B = B+10 = y+10

According to the given problem

Twice the amount of A is Rs. 10 more than the amount B

=> 2x = y+10

=> 2x-y-10 = 0

Required equation is 2x-y-10 = 0

2)

Given equation is 4x+3y = 12

(3,0) is a solution because

LHS =>

=> 4(3)+3(0)

=> 12+0

=> 12

=> RHS

LHS = RHS

3)

3x-7y = 41 passes through the point (2,-5)

Put x = 2 and y = -5 then

LHS = 3(2)-7(-5)

=>6+35

=> 41

=> RHS

=> LHS = RHS

4)

Given equation is (2/3)x -(7/5)y = 2.3

LCM of 3 and 5 = 15

=> (10x-21y)/15 = 2.3

=> 10x-21y =2.3×15

=> 10x-21y = 34.5

=> 10x-21y-34.5 = 0

On comparing with ax+by+c = 0 then

a = 10

b = -21

c = -34.5

5)

Given equation is ax+3y = 1

Point lies on it = (2,-1)

Since it lies on it then it satisfies the given equation

Put x = 2 and y = -1 then

=>a(2)+3(-1) = 1

=> 2a -3 = 1

=> 2a = 1+3

=> 2a = 4

=> a = 4/2

=> a = 2

The value of a = 2

6)

There are infinitely number of many lines passes through the point (2,3)

X+y = 5

2x+3y = 5

3x+4y = 18

and so on

7)

The present age of Ravi = x years

If 10 years added to the age then it will be

(x+10) years

According to the given problem

The age becomes = 25 years

=> x+10 = 25

=> x+10-25 = 0

=> x -15 = 0

The linear equation is x-15 = 0

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