SOLUTION: When solving a linear system using linear combinations, is there a right way to multiply both equations to get opposite variables?

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Question 146: When solving a linear system using linear combinations, is there a right way to multiply both equations to get opposite variables?
Answer by KMST(5484) About Me  (Show Source):
You can put this solution on YOUR website!
Linear combinations of linear equations can "eliminate" one of the variables,
to give you an equation having only the other variable.
That is why solving a linear system using linear combinations use to be called solving by "the elimination method".

I will show you an example.
Let's say your linear system is system%289x%2B5y=-6%2C4x%2B3y=2%29

If you wanted to be left with just x as a variable,
you would want to multiply each equation times a different number,
so that the new coefficients of y add up to zero.
In this case, I would multiply the first equation times 3 ,
and the second equation times -5 :
From there, I would substitute -4 for x in one of the original equations:
4%28-4%29%2B3y=2-->-16%2B3y=2-->3y=2%2B16-->3y=18-->y=18%2F3-->highlight%28y=6%29

If you wanted to be left with just y as a variable,
you would want to multiply each equation times a different number,
so that the new coefficients of x add up to zero.
In this case, I would multiply the first equation times -4 ,
and the second equation times 9 :
system%289x%2B5y=-6%2C4x%2B3y=2%29-->-36x-20y=24 %22%2B%22 36x%2B27y=18-->%28-36%2B36%29x%2B%28-20%2B27%29y=24%2B18-->7y=42-->highlight%28y=6%29
From there, I would substitute 6 for y in one of the original equations:
4x%2B3%286%29=2-->4x%2B18=2-->4x=2-18-->4x=-16-->highlight%28x=-4%29 .

You could also do
system%289x%2B5y=-6%2C4x%2B3y=2%29-->27x%2B15y=-18 %22%2B%22 -20x-15y=-10-->%2827-20%29x%2B%2815-15%29x=-18%2B%28-10%29-->7x=-28-->highlight%28x=-4%29
and
system%289x%2B5y=-6%2C4x%2B3y=2%29-->-36x-20y=24 %22%2B%22 36x%2B27y=18-->%28-36%2B36%29x%2B%28-20%2B27%29y=24%2B18-->7y=42-->highlight%28y=6%29 .