If you answer ALL questions listed, you get a good amount of points (3 questions, 5 points each, 15 points total) :D



1) Which pair are solutions to the equation?


4xy + 8 = 36

A

(7, 1) and (3, 2)

B

(1, 7) and (7, 1)

C

(1, 7) and (4, 9)

D

(4, 9) and (3, 2)


2)

Which of the ordered pairs in the form (x, y) is a solution of this equation?


5x - y over 3 = 13,

(2, -9) (3, -6)

A

The first is not a solution, but the second is.

B

Both are solutions.

C

The first is a solution, but the second is not.

D

Neither is a solution.


3)

Which ordered pairs in the form (x, y) are solutions to the equation


7x – 5y = 28?




Choose all answers that are correct.

A

(−6, −14)

B

(−1, −7)

C

(4, 10)

D

(7, 9

Verified Answer :

1)

\(4xy+8=36\\4xy=28\\xy=7\)

We know that each coordinate (x,y) must therefore multiply to equal 7. The only numbers that multiply to 7 are 7 and 1, so B) is the correct answer.



2)

For this one, just plug in the coordinates they give you for x and y and see if either one works:

\(5x-\frac{y}{3}=13\ \ \ \ \ \ \ (2,-9)\\5(2)-\frac{(-9)}{3}=13\\10-(-3)=13\\13=13\)

So the first one is a solution. Now plug in the second one:

\(5x-\frac{y}{3}=13\ \ \ \ \ \ (3,-6)\\5(3)-\frac{(-6)}{3}=13\\15-(-2)=13\\17=13\)

Therefore, the first is a solution and the second isn't making C) the correct answer.



3)

For this one, just plug in the ordered pairs they give you and see which ones end up working:

\(7x-5y=28\\\\A. (-6,-14)\\7(-6)-5(-14)=28\\-42+70=28\\28=28\\\\B.(-1,-7)\\7(-1)-5(-7)=28\\-7+35=28\\28=28\\\\C.(4,10)\\7(4)-5(10)=28\\28-50=28\\-22=28\\\\D.(7,9)\\7(7)-5(9)=28\\49-45=28\\4=28\)

A) and B) are the only ones that work out, so those are the correct answers.

1.  4xy+8=36

   4xy=28

   xy=7


2. 5(2)-(-9)/3=13

    10-(-3)=13

     13=13

     5(3)-(-6)/3=13

     15-(-2)=13

     17≠13


3. 7(-6)-5(-14)=28

   -42+70=28

     28=28


    7(-1)-5(-7)=28

    -7+35=28

      28=28

     

   7(4)-5(19)=28

   28-50=28

    -22≠28


  7(7)-5(9)=28

  49-45=28

    4≠28

Hope this helps... :)