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## Introduction to systems of equations

Current time:0:00Total duration:2:38

# Testing a solution to a system of equations

CCSS Math: HSA.REI.C.6, HSA.REI.D.11

## Video transcript

Is negative 1 comma 7 a
solution for the system of linear equations below? And they give us
the first equation is x plus 2y is equal to 13. Second equation is 3x minus
y is equal to negative 11. In order for negative
1 comma 7 to be a solution for the
system, it needs to satisfy both equations. Or another way of thinking about
it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and
y is equal to 7, need to satisfy both of
these equations in order for it to be a Solution. So let's try it out. Let's try it out with
the first equation. So we have x plus
2y is equal to 13. So if we're thinking
about that, we're testing to see if when x
is equal to negative 1, and y is equal to 7,
will x plus 2y equals 13? So we have negative
1 plus 2 times 7-- y should be 7-- this
needs to be equal to 13. And I'll put a
question mark there because we don't
know whether it does. So this is the same
thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least,
satisfy this first equation. This point does sit on the
graph of this first equation, or on the line of
this first equation. Now let's look at
the second equation. I'll do that one in blue. We have 3 times negative
1 minus y, so minus 7, needs to be equal
to negative 11. I'll put a question
mark here because we don't know whether
it's true or not. So let's see, we have 3 times
negative 1 is negative 3. And then we have minus 7 needs
to be equal to negative 11-- I put the question mark there. Negative 3 minus 7,
that's negative 10. So we get negative 10
equaling negative 11. No, negative 10 does
not equal a negative 11. So x equaling negative
1, and y equaling 7 does not satisfy
the second equation. So it does not sit on its graph. So this over here is not
a solution for the system. So the answer is no. It satisfies the
first equation, but it doesn't satisfy the second. In order to be a
solution for the system, it has to satisfy
both equations.