Question 1: Which one of the following statements is true and why? `y = 3x + 5` has

(i) a unique solution

(ii) only two solutions

(iii) infinitely many solutions.

**Solution:** Since; `y = 3x + 5` is a linear equation in two variables. We know that a linear equation in two variables has infinite solutions.

Hence option (iii) is true.

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

**Solution:** (i) `2x + y = 7`

Or, `y = 7 – 2x` …………equation (1)

When x = 0, Putting the value in equation (1)

`y = 7 – 2 xx 0`

Or, `y = 7 – 0`

Or, `y = 7`

Putting the value `x = 1` in equation (1)

`y = 7 – 2 xx 1`

Or, `y = 7 – 2`

Or, `y = 5`

Putting the value `x = 2` in equation (1)

`y = 7 – 2 xx 2`

Or, `y = 7 - 4`

Or, `y = 3`

Putting the value `x = 3` in equation (1)

`y = 7 - 2 xx 3`

Or, `y = 7 - 6`

Or, `y = 1`

Hence, four solutions for equation

`2x + y = 7` are (0,7), (1,5), (2,3), (3,1).

(ii) `π x + y = 9`

Or, `y = 9 - πx` ……………… (1)

When x = 0, putting the value in equation (1)

`y = 9 - π0`

Or, `y = 9 – 0 = 9`

Putting the value `x = 1` in equation (1)

`Y = 9 - π`

Putting the value `x = - 1` in equation (1)

`y = 9 + π`

Putting the value `x = 2` in equation (1)

`y = 9 - 2π`

Hence, four solutions for the given equation are; (0, 9), (1, 9 - π), (- 1, 9 + π), (2, 9 - 2π)

(iii) `x = 4y` …………equation (1)

When `y = 0`, Putting the value in equation (1)

`x = 4 xx 0`

Or, `x = 0`

Putting the value `y = 1` in equation (1)

`x = 4 xx 1`

Or, `x = 4`

Putting the value `y = -1` in equation (1)

`x = 4 xx -1`

Or, `x = -4`

Putting the value `y = 2` in equation (1)

`x = 4 xx 2`

Or, `x = 8`

Hence, four solutions for equation `x = 4y` are (0, 0), (1, 4), (-4,-1), (8, 2).

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

**Solution:** (i) (0, 2)

The given equation is `x – 2y = 4`

L.H.S `= x – 2y`

Putting the value `x = 0` and `y = 2` to verify the solution (0, 2)

`= 0 – 2 xx 2 = 0 – 4`

`= -4 ≠` R.H.S.

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

(ii) (2, 0)

**Solution:** The given equation is; `x – 2y = 4`

L.H.S `= x – 2y`

Putting the value `x = 2` and `y = 0` to verify the solution (2, 0)

`= 2 – 2 xx 0 = 2 – 0`

`= 2 ≠` R.H.S.

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

(iii) (4, 0)

**Solution:** The given equation is; `x – 2y = 4`

L.H.S `= x – 2y`

Putting the value `x = 4` and `y = 0` to verify the solution (4, 0)

`= 4 – 2 xx 0 = 4 – 0`

= 4 = R.H.S.

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

(iv) `(sqrt2, 4sqrt2)`

Putting the value `x = sqrt2` and `y = 4sqrt2` to verify the solution:

`= sqrt2 - 2 xx 4sqrt2`

`= sqrt2 - 8sqrt2`

`= 7sqrt2 ≠` RHS

Or, `(sqrt2, 4sqrt2)` is not a solution for the equation `x – 2y = 4`

(v) (1, 1)

**Solution:** The given equation is; `x – 2y = 4`

L.H.S `= x – 2y`

Putting the value `x = 1` and `y = 1` to verify the solution (1, 1)

`= 1 – 2 xx 1 = 1 – 2 = -1 ≠` R.H.S.

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

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

**Solution:** The given equation is; `2x + 3y = k`

Putting the given value `x = 2` and `y = 1` in the equation

`2 xx 2 + 3 xx 1 = k`

`4 + 3 = k`

`7 = k`

Hence, `k = 7`

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