Arwin implementations
Q.1
- The cost of a notebook ((y)) is twice the cost of a pen ((x)).
- Therefore, (y) can be represented as (2x).
Q.2
(i) 2x + 3y = 9.35 (a = 2), (b = 3), (c = 9.35)
(ii) x - 5y - 10 = 0 (a = 1), (b = -5), (c = -10)
(iii) -2x + 3y = 6 (a = -2), (b = 3), (c = 6)
(iv) x = 3y (a = 1), (b = -3), (c = 0)
(v) 2x = -5y (a = 2), (b = 5), (c = 0)
(vi) 3x + 2 = 0 (a = 3), (b = 0), (c = 2)
(vii) y - 2 = 0 (a = 0), (b = 1), (c = -2)
(viii) 5 = 2x (a = 2), (b = 0), (c = -5)
Q3
Matching Solutions to Equations
Match each equation to its correct set of solutions.
Q4
Match the below exponents
2x + y = 7
πx + y = 9
x = 4y
(x=1, y=9-π) (x=2, y=9-2π) (x=0, y=9) (x=-1, y=9+π)
(x=1, y=5) (x=2, y=3) (x=3, y=1) (x=0, y=7)
(x=0, y=0) (x=4, y=1) (x=8, y=2) (x=-4, y=-1)
Q5
(i)
Check which of the following are solutions of the equation ( x - 2y = 4 ) and which are not:
(i) (0, 2): (0 - 2 \times 2 = 0 - 4 = -4 \neq 4) => Not a solution.
(ii) (2, 0): (2 - 2 \times 0 = 2 - 0 = 2 \neq 4) => Not a solution.
(iii) (4, 0): (4 - 2 \times 0 = 4 - 0 = 4 = 4) => Solution.
(iv) ( \left( 2 , \frac{4}{2} \right) ): (2 - 2 \times 2 = 2 - 4 = -2 \neq 4) => Not a solution.
(v) (1, 1): (1 - 2 \times 1 = 1 - 2 = -1 \neq 4) => Not a solution.
(ii)
Summary of the solutions:
(i) (0, 2) => Not a solution.
(ii) (2, 0) => Not a solution.
(iii) (4, 0) => Solution.
(iv) ( \left( 2 , \frac{4}{2} \right) ) => Not a solution.
(v) (1, 1) => Not a solution.
Q.6
Check which of the following are solutions of the equation ( x – 2y = 4 ) and which are not: 1. (0, 2), 2. (2, 0), 3. (4, 0) , 4. ( \left(2, \frac{4}{2}\right) ) , 5. (1, 1)
Substitute ( (0, 2) ) into the equation: ( 0 - 2(2) = 0 - 4 = -4 \neq 4 ) (Not a solution)
Substitute ( (2, 0) ) into the equation: ( 2 - 2(0) = 2 - 0 = 2 \neq 4 ) (Not a solution)
Substitute ( (4, 0) ) into the equation: ( 4 - 2(0) = 4 - 0 = 4 = 4 ) (Solution)
Substitute ( \left(2, \frac{4}{2}\right) ) into the equation: ( 2 - 2\left(\frac{4}{2}\right) = 2 - 2(2) = 2 - 4 = -2 \neq 4 ) (Not a solution)
Substitute ( (1, 1) ) into the equation: ( 1 - 2(1) = 1 - 2 = -1 \neq 4 ) (Not a solution)
Q.6
Find the value of ( k ), if ( x = 2 ), ( y = 1 ) is a solution of the equation ( 2x + 3y = k ).
Substitute ( x = 2 ) and ( y = 1 ) into the equation ( 2x + 3y = k ).
The equation becomes ( 2(2) + 3(1) = k ).
Calculate ( 2(2) + 3(1) = 4 + 3 = 7 ).
Therefore, ( k = 7 ).
- Draw the graphs of the equations x + y = 4 and 2x - 3y = 12.
For: x + y = 4
| x | y |
|---|---|
| 0 | |
| 4 | |
| 4 | |
| 0 |
For: 2x – 3y = 12
| x | y |
|---|---|
| 0 | |
| 3 | |
| -3 | |
| -6 |
| y = 2x – 2 | x | y |
|---|---|---|
| 2 | ||
| 0 |
| y = 4x – 4 | x | y |
|---|---|---|
| 0 | ||
| 1 |