Question

1) Solve the system by substitution.

{7x−7y=−42

{y=2x+7

a) One solution:

b) No solution

c) Infinite number of solutions

2) Solve the system by substitution.

{5x+4y=27

{y=2x-16

a) One solution:

b) No solution

c) Infinite number of solutions

Answer 1

1). The given equations are 7x−7y=−42, or, on dividing both the sides by 7, x -y = -6 or, y = x+6…(1) and y = 2x+7…(2).

On substituting y = 2x+7 in the 1^st equation, we get (2x+7) = x+6 or, 2x-x =6-7 or, x = -1.

Now, on substituting x = -1 in the 1^st equation, we get y = -1+6 = 5

Thus, there is one solution which is x = -1 and y = 5.

Option a) is the correct answer.

2). The given equations are 5x+4y = 27…(1) and y = 2x-16…(2)

On substituting y = 2x-16 in the 1^st equation, we get 5x+4(2x-16) = 27 or, 5x+8x -64 = 27 or, 13x = 27+64 = 91 so that x = 91/13 = 7.

Now, on substituting x = 7 in the 2^nd equation, we get y = 2*7-16 = 14-16 = -2.

Thus, there is one solution which is x = 7 and y = -2.

Option a) is the correct answer.