3. Check which of the following are the solutions of the equation 5x – 4y = 20.
(i) (4, 0) (ii) (0, 5) (iii) (−2, 52) (iv) (0, –5) (v) (2, −52)
4. Find five different solutions of each of the following equations:
(i) 2x – 3y = 6
(ii) 3y = 4x
5. If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.
6. The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y).
Answers
Answer:(i) It is given
5x−4y=20
Now by substituting (4,0) in the place of x and y We know that the LHS
=5x−4y
=5(4)−4(0)
On further calculation
=20
=RHS
Therefore (4,0) is the solution of 5x−4y=20
(ii) It is given
5x−4y=20
Now by substituting (0,5) in the place of x and y We know that the LHS
=5x−4y
=5(0)−4(5)
On further calculation
=−20
=RHS
Therefore (−2,
2
5
) in the place of x and y we know that the LHS
=5x−4y
=5(−2)−4(
2
5
)
On further calculation
=−10−10
=−20
=RHS
Therefore (−2,
2
5
)is not the solution of 5x - 4y = 20
(iv) it is given
5x−4y=20
Now by substituting (0,5) in the place of x and y We know that the LHS
=5x−4y
=5(0)−4(−5)
Our further calculation
=20
=RHS
Therefore, (0,-5) is the solution of 5x−4y=20
(v) it is given
5x−4y=20
Now by substituting (2,−
2
5
) in the place of x and y we know that the LHS
=5x−4y
=5(2)-4 (−
2
5
)
On further calculation
=10+10
=20
=RHS
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