[sheron neal]

Hello, with 3x-y=95
x+5y=33 how do I solve by elimination method?

[Perito]

3x-y=95 (equation 1)
x+5y=33 (equation 2)

Multiply equation 2 by "3"

3x-y=95 (equation 1)
3x+15y=99 (equation 2)

Subtract equation 2 from equation 1

-y-15y=95-99

-16y=-4

y=4/16=1/4

Plug y back into one of the other equations to solve for x:

3x-1/4=95 (equation 1)
12x-1=380
12x=381
x=31 3/4

Check with the other equation

31 3/4 + 5(1/4)=33 (equation 2)

33 = 33 (qed)

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You could also multiply equation 1 by 5 and add the two equations. It doesn't matter what you multiply the equations by, just as long as you multiply one or both equations with numbers such that when the two equations are added or subtracted, one of the variables adds or subtracts to zero.

3x-y=95 (equation 1)
x+5y=33 (equation 2)

15x-5y=475 (equation 1)
x+5y=33 (equation 2)

16x = 508
x=31.75