**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:

(iiI) y = 3x + 5 ⇒ 3x – y + 5 = 0 has infinitely many solutions. Because it is a linear equation in two variables. Values of y changes as we vary the values of x and vice versa.

**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

**if x = 1** then 2(1) + y = 7

⇒2 + y = 7

⇒ y = 7 – 2 = 5

**If x = 0** then 2(0) + y = 7

⇒ y = 7

**If x = 2** then 2(2) + y = 7

⇒ 4+ y = 7

⇒ y = 7 – 4 = 3

⇒ y = 3

**If x = -1** then 2(-1) + y = 7

⇒-2 + y = 7

⇒y =7 + 2 = 9

⇒y = 9

(ii) πx + y = 9

**If x = 0** then π(0) + y = 9

⇒y = 9

**If x = ^{1}/_{π}** then π(1) + y = 9

⇒(^{1}/_{π}) + y = 9

⇒y = 9 – 1 = 8

**If x = ^{2}/_{π}** then π(

^{2}/

_{π}) + y = 9

⇒2 + y = 9

⇒y = 9 – 2 = 7

If x = ^{9}/_{π} then π(^{9}/_{π}) + y = 9

⇒9 + y = 9

⇒y = 9 – 9 = 0

⇒y = 0

(iii) x = 4y

If x = 0 then 4y = 0

⇒y = 0

If x = 1 then 1 = 4y

⇒y = ^{1}/_{4}

If x = 2 then 2 = 4y

⇒y = ^{2}/_{4} = ^{1}/_{2}

If x = 4 then 4 = 4y

⇒y = 1

**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) ( 2 , 4 2)**

**(v) (1, 1)**

**Solution:**

x – 2y = 4 …………(1)

(i) (0, 2)

If x = 0 then eq(1) becomes

0 – 2y = 4

y = –^{4}/_{2 } = -2

Therefore , (0, 2) is not a solution of equation (1)

(ii) (2, 0)

If x = 2 then eq(1) becomes

2 – 2y = 4

-2y = 4 – 2 = 2

y = –^{2}/_{2} = -1

Therefore , (2, 0) is not a solution of equation (1)

(iii) (4, 0)

If x = 4 then eq(1) becomes

4 – 2y = 4

-2y = 4 – 4 = 0

y = –^{0}/_{2} = 0

Therefore , (4, 0) is a solution of equation (1)

(iv) (√2 , 4√2)

If x = √2 then eq(1) becomes

√2 – 2y = 4

-2y = 4 – √2

y = ^{4 – }^{√2}/_{-2} = ^{4}/_{-2 }– ^{√2}/_{-2} = -2 + ^{√2}/_{2}

Therefore , (√2 , 4√2) is not a solution of equation (1)

(v) (1, 1)

If x = 1 then eq(1) becomes

1 – 2y = 4

-2y = 4 – 1 = 3

y = ^{3}/_{-2}

Therefore , (1, 1) is not a solution of equation (1)

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

Solution:

If x = 2 and y = 1is a solution of 2x + 3y = k

2(2) + 3(1) = k

4 + 3 = k

k = 7

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