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Solving the system of equations by the addition method?

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Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)

I think you mean the elimination method, which would ultimately require you to add the equations, but the main goal is to eliminate one of the variables first so that you can solve for the other variable then use the answer to solve for the other variable.

With this method, you want to manipulate one of the equations, or both is necessary, by multiplying it by a constant that will generate opposite coefficients on either the x or y variable so as to eliminate it when you add the equations. 

      8x  -  3y  = 9

     40x - 15y = 18

You can choose to eliminate either variable, but we'll have to manipulate the first equation since the coefficients are smaller and we need to multiply them by another constant. Let's eliminate the y-variable. To do so, we will need its coefficient to be +15 since the y coefficient in the second equation is -15 and so we multiply the first equation by -5 to generate this result:

     -5·(8x - 3y = 9)  

     (-5)·8x - (-5)·3y = (-5)·9

     -40x + 15y = -45

Now, we add the two equations:

       -40x + 15y = -45

   +   40x - 15y = 18 


         0x + 0y = -27    

                  0 = -27     ==> not true because   0 ≠ -27

Therefore, since our solution is 0=-27 and that is obviously not true because 0 does not equal -27 then we can conclude that the system has no solution and is, thus, inconsistent.