How do you solve systems of equations by elimination using multiplication?
1 Answer
There are certain steps to follow before finding the solution.
Let's say we have two equations:
The first thing you have to do is determine which of the variables that, when added, can be canceled to just get one term. In this case, we don't have any variables to be canceled out, so we need to first multiply one system. Let's begin with the first one, where the whole system is multiplied by -2
Notice that with
. . . . . . . . . . . . . . . . . . .
By doing so, we get a sum of equations in terms of
Then you would substitute the
Let's do both equations and see if we get the same answer:
Huzzah! We got the same x-value! Therefore,
In most of these problems, there are other ways that can be multiplied easily to get the answer. For the system of equations above, we could also solve for
Sometimes there will be cases where one of the variables does not equal the other (if you follow the steps carefully). This means that the system of equations cannot equal each other, thus there is no solution.
Be careful when doing some fractional equations; I would first multiply each one by the most common denominator to get whole integers, thus allowing the equations to be easier to use elimination. For more information on how to do this, see this link:
http://www.regentsprep.org/regents/math/algebra/AV5/Fequations.htm
And finally, here's another example:
Hopefully all of these help and good luck with algebra!
