Let Aftab's age be x years
Let, daughter's age be y years
Seven years ago,
Aftab's age = x - 7 years
Daughter's age = y - 7 years
According to question,
bex-7=7(y-7)
x-7=7y-49
⇒x-7y=-42.........(i)
After 3 years,
Aftab's age = x + 3 years
Daughter's age = y + 3 years
According to question,
x+3=3(y+3)
x+3=3y+9
⇒x-3y = 6...........(ii)
The algebraic equation is represented by
x−7y = −42
x−3y = 6
For, x−7y = −42 or x = −42+7y
The Soluton table is
| X | -7 | 0 | 7 |
| Y | 5 | 6 | 7 |
For, x−3y = 6 or x = 6+3y
The solution table is
| X | 6 | 3 | 0 |
| Y | 0 | -1 | -2 |
The graphical representation is:
Let us assume that the cost of a bat be ‘Rs x’
And,the cost of a ball be ‘Rs y’
to the question, the algebraic representation is
3x+6y = 3900
And x+3y = 1300
For, 3x+6y = 3900
Or x = $\dfrac{3900-6y}{3}$
The solution table is
| X | -300 | 100 | -100 |
| Y | 500 | 600 | 700 |
For, x+3y = 1300
Or x = 1300-3y
The solution table is
| X | 1000 | 700 | 400 |
| Y | 100 | 200 | 300 |
The graphical representation is as follows.
Let the cost of 1 kg of apples be ‘Rs. x’
And, cost of 1 kg of grapes be ‘Rs. y’
According to the question, the algebraic representation is
2x+y = 160
And 4x+2y = 300
For, 2x+y = 160 or y = 160−2x, the solution table is;
| X | 50 | 60 | 70 |
| Y | 60 | 40 | 20 |
For 4x+2y = 300 or y = (300-4x)/2, the solution table is;
| X | 70 | 80 | 75 |
| Y | 10 | -10 | 0 |
The graphical representation is as follows;
