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In this example, I am going to solve three equations of the form
a times x equals b where a and b
are numbers. The first one, three times x equal 6,
if I'm wondering what this value of x is this is what I'm trying to find
I can probably determine that X has to be two in this case because three times
two
equal six. But what are the algebra steps steps that are involved
to actually find that value. What I'm going to do
is I'm going to divide both sides of the equation
by the value that value that's multiplied times x.
So what I'm doing is creating a form of one
on the left hand side. So three divided by three
gives me one. And on the right hand side
I get my result which is six divided by three
equals two. So it's sort of like if the piece that I want
is part of a multiplication then to find that piece
I'm going to need to do a division. So if I want to do a check here,
I would check, as I kinda did at the start, multiplying 3 times 2
and does that equal 6 and it does so
there is my check. Same thing on this one, twelve times what number
equals 24, well, that number is going to have to be two
but again the official mathematical process
is that I'm going to divide both sides of the equation by the same number
and that number is a form of one
that will reduce because 12 is a common factor
for 12x and then 12 and reduce and on the right hand side give me my
result. So again we have another way to get x equals 2.
Let's do a check does 12 times two
equal 24 and yes it does so there we have
our check. Part C is a little bit different
two times what number equals twenty. We can probably figure out that that number
has to be ten so again
our process is we are going to divide both sides
by a form of one, or we're going to create a form of one I should say,
two divided by two is one and and so
the left hand side becomes one times x, 20 divided by two
is 10 so there's how we get our 10. Let's do a check. Does two times 10
equal twenty and yes it does so there we have
our check.