Algebra?? substitution problem?

Question

8x - 4y = 16
y = 2x - 4
Solve by substitution

Thanks for your help last time Punk Rocker!

Answer 1

8x - 4y = 16
y = 2x - 4

You know what y is in terms of x (the second equation).
So, substitute as follows:

8x - 4(2x - 4) = 16

Now, continue:
8x - 8x + 16 = 16

16 = 16 is what you get.

Now, this means that there is no SINGLE value of x and y that solves the system. They are in fact the SAME, and hence there are INFINITE solutions.

Answer 2

Substituting either equation to the other one will both result to 0:

8x - 4y = 16 --> 1
y = 2x - 4 --> 2

Subst. 2 to 1

8x - 4(2x - 4) = 16
8x - 8x + 16 = 16
8x - 8x = 16 - 16
0 = 0

Else subst. 1 to 2

x = 2 + y/2
y = 2(2+y/2) - 4
y = 4 + y - 4
y - y = 4 - 4
0 = 0

Only way you can obtain an answer for either equation is by substituting a value to either x or y to solve for the other variable.

Try x = 0

8 (0) - 4y = 16
-4y = 16
y = - 4

Answer 3

x=4

Answer 4

here you go: I'll get you started

=8x-4(2x-4)=16

can you do the rest?

or do you need more help from here??
distribute the 4 into 2x-4)

=8x+8x-16=16

combine like terms

=16x-16x=16
+16 +16

-16 and +16 cancell eschother out

16+16=32

now
16x=32

get the x alone

16x = 32
---------------
16 16

16 & 16 cancell eachother

32/16=2

x=2

Answer 5

8x - 4(2x-4) = 16
8x - 8x + 16 = 16
16 = 16

Answer 6

8x - 4(2x-4) = 16
8x-8x-16=16
-16=16
16-16=0

Answer 7

put the value of y in the 1st eqn.

8x - 4 * (2x-4) = 16
or, 8x - 8x +16 = 16
or, 16 = 16

I don't know.

Answer 8

8x - 4(2x-4)=16
8x-8x-16=16
0=32
no solution, at least one of the variable statements is false

Answer 9

You really have only one equation (the second equation is the same as the first equation, just rearranged).

If you wish to get a "unique" set of answers for X and Y, you need two distinct equations. Otherwise, you have infinite number of possibilities....e.g., trying X=1 and you will get Y= -2, for X= 2, Y is 0, and so on...