In the method of solving systems of linear equations...

Question

In the method of solving systems of linear equations by substitution, which of the following is done.

Select one:

a.

None of these

b.

The value 0 is substituted for both variables.

c.

A variable is substituted in terms of the other variable.

d.

An arbitrary value is substituted for both variables.

Answer 1

Let's look what would happen if you do each of these choices. Let's say y=2x and y= x-5.

What happens if we do choice d.? We substitute an arbitrary variable for x and y. "Arbitrary" means we can choose any value. Let's choose 10 for both. That gives us

10=2*10 ==> 10=20 (false)

and 10=10-5 ==> 10=5 (false)

So that one doesn't get us a solution and gets us two false equations. Not an impressive approach.

Let's try option b. With these equations, that gets us

0=2*0 ==> 0=0 (true)

0=0-5 ==> 0=-5 (false)

This option gets us a true equation and a false equation and still tells us nothing about x and y (except that they're not both 0)

Let's try option c. We can substitute what we know about y in terms of x for y. We know y = 2x. Let's plug that in to the other equation.

2x=x-5

==> x=x/2+5/2

==> x-x/2=5/2

==> x/2=5/2

==> x=5

Wow, we know what x is. Let's plug that in to what we know about y.

y=2*5 ==> y=10

Aha. A solution.

Answer 2

Let's consider an easy example. Here, we have a system of two equations:

y = x + 1

y + 2x = 7

The nice thing about this one is that we already have y in terms of x. All that's left to do is sub in (x+1) for y:

x + 1 + 2x = 7

3x + 1 = 7.

3x = 6

x = 2.

And then substitute back in to obtain y = 3 for the other part of our answer.

So what have we done, here? We've taken one variable, solved for it in terms of the others, then substituted this expression in.

Therefore, we pick (c).