how do you solve a system of equations using the elimination method when both equations have no equal number associated with x or y

The method requires you to multiply the first equation by some number, the second equation by some number, in order that either the x coefficients are the same or that the y coefficients are the same. Sometimes, you get lucky and a pair of coefficients, either the x's or the y's are already equal in which case the jobs already done, or it's sufficient to multiply just one of the equations by a number in order to make a pair of coefficients equal.

 

With these equations, you have a choice. Multiply the first equation by 3 to get 6x + 9y = -3, multiply the second equation by 2 to get 6x + 8y = -8, and now subtract one equation from the other. The x's cancel out. That allows you to find the value of y, and by substituting that into one of the earlier equations allows you to find the value of x.

 

Alternatively, multiply the first equation by 4 to get 8x + 12y = -4, the second equation by 3 to get

9x + 12y = -12,  and subtract one from the other. That gets you the value of x, and by substituting that into one of the earlier equations, the value of y.

 

You should do it both ways, as an exercise, to show that you get the same results, either way.

 

There will be sets of equations where the x's or y's are opposite in sign in which case the equations have to be added rather than subtracted.

LEGITIMATE ANSWER:

1+1=2 is correct right?

2*1=2, and 2*2 = 4, right?

2+2=4, correct?

so (1+1=2)*2 = 2+2=4

SO:

2x+3y= -1 and 3x+4y= -4

(2x+3y=-1)*4 and (3x+4y=-4)*-3

8x+12y=-4 and -9x - 12y = 12

this set of equations accepts the same values for x and y as the original, and can be solved with elimination or whatever.