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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