Question 126657This question is from textbook Beginning Algebra
: 8x-4y=16
y=2x-4
Solve by substition
4y=8x-16
divide both sides by 4y
y=8x-16/4
y=8x/4+16/4
y=4x/2+16/4
reduce
y=2x+4
Upon plugging top equation of y into bottom equation, I became confused I came up with
y=2x-4 and y=2x+4
I must have done something wrong along the line
This question is from textbook Beginning Algebra
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given the following pair of equations:
.
8x - 4y = 16
y = 2x - 4
.
To solve a pair of equations, you need to solve one of the equations for one variable in
terms of the other variable. Then you substitute that equivalent value into the other equation
and solve it.
.
Notice that the second equation has already been solved for y in terms of x. Therefore,
you already know that y is equal to 2x - 4.
.
Now go to the first equation and substitute that value (2x - 4) for y in that equation.
When you do that substitution the first equation becomes:
.
8x - 4(2x - 4) = 16
.
You now have an equation that contains only one variable, so it can be solved to get that
variable. Begin the solution by multiplying the -4 times each of the terms in the parentheses.
When you do that multiplication the left side of the equation becomes:
.
8x - 8x + 16 = 16
.
Notice that the 8x and the -8x cancel each other out and the equation reduces to:
.
16 = 16
.
Well, that is an interesting development. What does it mean. To find out, let's go back
to the original pair of equations. Let's take the first equation:
.
8x - 4y = 16
.
and let's try to work it into the form of the second equation ... with just a y on the left
side and everything else on the right side. Begin by subtracting 8x from both sides to
get rid of the 8x on the left side. When you do this subtraction the equation becomes:
.
-4y = -8x + 16
.
Now divide both sides of the equation (all terms) by -4 so that you just have y on the left
side. When you divide all terms on both sides by -4 the equation becomes:
.
y = 2x - 4
.
Wow! Notice that this is the exact same equation as the second equation. Therefore,
if you graphed both equations you would find that the two graphs coincide ... one graph is
identical to the other graph. This means that there is not a unique solution to the two original
equations. Every solution pair of one equation is also a solution pair of the other equation.
.
Let's try a point just to verify this. For example, suppose in the first equation we
assume that x = 0. This means that in the equation
.
8x - 4y = 16
.
we set x equal to zero and the equation reduces to
.
-4y = 16
.
Divide both sides by -4 and the equation reduces to
.
y = -4
.
This tells us that the point (0, -4) is on that graph
.
Now go to the second equation and set x = 0. The second equation equation is
.
y = 2x - 4
.
and when you set x = 0 it reduces to
.
y = -4
.
This tells you that the coordinate point (0, -4) also in the solution set for this equation.
.
You can do the same sort of exercise for other values of x and you will always find the
same thing to be true. For example if you set x equal to 5 in both equations you will find the
corresponding value of y will be 6. Therefore, the point (5, 6) is a solution for both of
the equations. There are an infinite number of common solutions, not a just a single solution.
.
In working this problem you came up with the same solution as I did. And you were correct.
You just needed to find out what this meant when you worked the two equations out and found
they were identical. Rest easy ... you know what you are doing.
.
Hope this helps you to understand what was going on in this problem.
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