Can someone help me with this?

Question

If (3,n) is a solution of Ax+3y=6 and (n,5) is a solution of 5x+5=20, what is the value of A?

Answer 1

The problem gives the equations of two lines:

     Ax+3y=6

     5x+5=20

 

Solving the second equation for x:

    5x+5=20

    5x = 15          [subtract 5 from both sides]

     x = 3             [divide both sides by 5]

The line x=3 is a vertical line and is not a function -- ANY value of y is allowed.

The value of n, however, is given by the point (n,5) which "solves" that formula.  So, n=3 since the point is (3,5).

 

Now, Ax+3y=6 contains the point (3,n).  Since n=3, we can find A.

    (A)(3) + (3)(n) = 6                    [use point (3,n)]

     3A + 9 = 6                              [remember, n=3]

      3A = -3                                   [subtract 3 from both sides]

       A = -1                                   [divide both sides by 3]

 

The equations really are:

    -x + 3y = 6

    5x + 5 = 20     which is also     x=3

 

Check (very important):

     Does (3,n) satisfies x-3y=-6    ?

         3-3n = -6   ?

         3-3(3)=-6   ?

          3-9=-6  ?  yes, when n=3

      Does (n,5) satisfy  5x+5=20   ?

          5(3)+5=20   ?

            15 + 5 = 20   ?  yes, when n=3

Answer 2

I think you meant the second equation to be     5x + 5y = 20.

 

First, plug in the second coordinate into the second equation to solve for n.  This value of n will be substituted for both coordinate points.

 

5n + 5(5) = 20

 

5n + 25 = 20

 

5n = -5

 

  n = -1

 

 

The first coordinate is then (3, -1).  Plug in this coordinate into the first equation to solve for A.

 

3A + 3(-1) = 6

 

 

I let you finish up here.

Answer 3

This is a tricky one [not to be confused with a trick question]!

 

Look at the second equation. If you solve for x...

 

5x + 5 = 20

       -5    -5   subtract 5 from each side.

5x       = 15

 

5x = 15

 5      5          divide both sides by 5

 

x = 3

 

So x is ALWAYS 3. This is a vertical line that goes through x=3. Can you see it? 

 

That means y can be ANY REAL NUMBER. That's probably what's messing you up. 

 

So Ax + 3y = 6    - let's solve for A using the fact that x is always 3

 

A(3) + 3y = 6

 3        3     3      -divide by 3 because it's the coefficient of A

 

A + y = 6

     -y   -y         -subtract y from both sides

  A = 6 - y

 

That's your answer A - 6 - y