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i) 2x + y = 7

ii) \(\pi\)x + y = 9

iii) x = 4y

Answer :

i)By inspection, x = 2 and y = 3 is a solution because for these values of x and y,

we get,2(2) + 3 = 4 + 3 = 7

Now, if we put x = 0, then the equation reduces to an unique solution of y = 7.

So, one solution of 2x + y = 7 is (0, 7).

Similarly, if we put y = 0, then the equation reduces to an unique solution of x = \(\frac{7}{2} \).

So, another solution of 2x + y = 7 will be (\( \frac{7}{2} \) , 0).

Finally, let us put x = 1, then,

we get, 2(1) + y = 7

=> 2 + y = 7

=> y = 5

Hence, (1, 5) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are :

(2, 3) , (0, 7) , (\(\frac{7}{2} \) , 0) and (1, 5).

ii)By inspection, x = \(\frac{1}{\pi } \) and y = 8 is a solution because for these values of x and y,

we get,\(\pi (\frac{1}{\pi} \)) + 8 = 1 + 8 = 9

Now, if we put x = 0, then the equation reduces to an unique solution of y = 9.

So, one solution of \(\pi\)x + y = 9 is (0, 9)).

Similarly, if we put y = 0, then the equation reduces to an unique solution of x = \(\frac{9}{\pi } \).

So, another solution of 2x + y = 7 will be (\(\frac{9}{\pi } \), 0).

Finally, let us put x = 1, then,

we get, \(\pi\)(1) + y = 9

=> \(\pi\) + y = 9

=> y = 9 - \(\pi\)

Hence, (1, 9 - \(\pi\)) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are :

(\(\frac{1}{ \pi } \), 8) , (0, 9) , (\(\frac{9}{\pi }\), 0) and (1, 9 - \(\pi\)).

iii)By inspection, x = 0 and y = 0 is a solution because for these values of x and y,

we get,0 = 4(0) => 0 = 0

Now, if we put x = 4, then the equation reduces to solution of y = 1.

So, one solution of x = 4y is (4, 1).

Similarly, if we put y = -1, then the equation reduces to solution of x = -4.

So, another solution of x = 4y will be (-4, -1).

Finally, let us put y = \(\frac{1}{2} \) , then,

we get, x = 4(\(\frac{1}{2} \))

=> x = 2

Hence, (2, \(\frac{1}{2} \) ) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation x = 4y are :

(0, 0) , (4, 1) , (-4, -1) and (2, \(\frac{1}{2} \) ).

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

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