CLASS:- 9 EXERCISE:- 4.2 LINEAR EQUATIONS IN TWO VARIABLES ALL SOLUTION

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EXERCISE 4.1 EXERCISE 4.3 EXERCISE 4.4

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1.    Which one of the following options is true, and why?

       `\y=3x+5\ ` has

(i)   a unique solution,

(ii)   only two solutions,

(iii)   infinitely many solutions

solution :

(i)   a unique solution,

`\y=3x+5\ ` , Every value of x is a corresponding value of y

So, this equation has no unique solution.

(ii)   only two solutions,

`\y=3x+5\ ` , Every value of x is a corresponding value of y

So, this equation has no only two solution.

(iii)   infinitely many solutions

(ii)   only two solutions,

`\y=3x+5\ ` , Every value of x is a corresponding value of y

So, this equation has infinitely many solution.

2.    Write four solutions for each of the following equations:

(i)  `\2x+y=7\ `

(ii)  `\πx+y=9\ `

(iii)  `\x=4y\ `

solution :

(i)  `\2x+y=7\ `

Taking y = 0,

We get,

`\2x+0=7\ `

`\2x=7\ `

`\x=\frac{7} {2} \ `

`\(\frac{7} {2}, 0)\ `

Taking y = 1,

We get,

`\2x+1=7\ `

`\2x=7-1\ `

`\2x=6 \ `

`\x=\frac{6} {2}\ `

`\x=3\ `

`\(3, 1)\ `

Taking y = 2,

We get,

`\2x+2=7\ `

`\2x=7-2\ `

` \2x=5

`\x=\frac{5} {2} \ `

`\(\frac{5} {2}, 2)\ `

Taking y = 3,

We get,

`\2x+3=7\ `

`\2x=7-3\ `

`\2x=4\ `

`\x=\frac{4} {2} \ `

`\x=2\ `

`\(2, 3)\ `

The solution are `\(\frac{7} {2}, 0)\ `, `\(3, 1)\ `, `\(\frac{5} {2}, 2)\ `, `\(2, 3)\ `.

(ii)  `\πx+y=9\ `

Taking x = 0,

We get,

`\π×0+y=9\ `

`\y=9\ `

`\(0, 9)\ `

Taking x = 1,

We get,

`\π×1+y=9\ `

`\π+y=9\ `

`\y=\frac{9} {π}\ `

`\(1, \frac{9} {π})\ `

Taking x = 2,

We get,

`\π×2+y=9\ `

`\2π+y=9\ `

`\y=\frac{9} {2π}\ `

`\(2, \frac{9} {2π})\ `

Taking x = 3,

We get,

`\π×3+y=9\ `

`\3π+y=9\ `

`\y=\frac{9} {3π}\ `

`\(3, \frac{9} {3π})\ `

The solution are `\(0, 9)\ `, `\(1, \frac{9} {π})\ ` ,`\(2, \frac{9} {2π})\ `, `\(3, \frac{9} {3π})\ `.

(iii)  `\x=4y\ `

Taking y = 0,

Wer get,

`\x=4×0\ `

`\x=0\ `

`\(0, 0)\ `

Taking y = 1,

Wer get,

`\x=4×1\ `

`\x=4\ `

`\(4, 1)\ `

Taking y = 2,

Wer get,

`\x=4×2\ `

`\x=8\ `

`\(8, 2)\ `

Taking y = 3,

Wer get,

`\x=4×3\ `

`\x=12\ `

`\(12, 3)\ `

The solution are `\(0, 0)\ `, `\(4, 1)\ `, `\(8, 2)\ `, `(12, 3)\ `.

3.    Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(i)  `\(0,2)\ `

(ii)  `\(2,0)\ `

(iii)  `\(4,0)\ `

(iv)  `\(\sqrt2 , 4\sqrt2)\ `

(v)  `\(1,1)\ `

solution :

(i)  `\(0,2)\ `

We know, x = 0, y = 2

Substituting the values of x and y in the equation `\x–2y = 4\ `,

We get,

`\0-2×2=4\ `

−4 = 4

`\(0, 2)\ ` is not a solution of the equation `\x–2y = 4\ `.

(ii)  `\(2,0)\ `

We know, x = 2, y = 0

Substituting the values of x and y in the equation `\x–2y = 4\ `,

We get,

`\2-2×0=4\ `

2 = 4

`\(2, 0)\ ` is not a solution of the equation `\x–2y = 4\ `.

(iii)  `\(4,0)\ `

We know, x = 4, y = 0

Substituting the values of x and y in the equation `\x–2y = 4\ `,

We get,

`\4-2×0=4\ `

4 = 4

`\(4, 0)\ ` is a solution of the equation `\x–2y = 4\ `.

(iv)  `\(\sqrt2 , 4\sqrt2)\ `

We know, x = `\sqrt2\ `, y = `\4\sqrt2\ `

Substituting the values of x and y in the equation `\x–2y = 4\ `,

We get,

`\sqrt2-2×4\sqrt2=4\ `

`\sqrt2-8\sqrt2=4\ `

`\−7\sqrt2 = 4 \ `

`\(\sqrt2, 4\sqrt2)\ ` is not a solution of the equation `\x–2y = 4\ `.

(v)  `\(1,1)\ `

We know, x = 1, y = 1

Substituting the values of x and y in the equation `\x–2y = 4\ `,

We get,

`\1-2×1=4\ `

−1 = 4

`\(1, 1)\ ` is not a solution of the equation `\x–2y = 4\ `.

4.    Find the value of k, if x = 2, y = 1 is a solution of the equation `\2x + 3y = k\ `.

solution :

Given equation,

`\2x+3y=k\ `

According to the question, x = 2 and y = 1.

We get,

`\2×2+3×1=k\ `

`\ 4+3=k\ `

`\7=k\ `

`\k=7\ `

So, the value of k = 7.

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EXERCISE 4.1 EXERCISE 4.3 EXERCISE 4.4