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How do I solve a linear system

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3 Answers

X = y +3
2X - y = 5
  X - y = 3   (1)
 2X - y = 5  ( 2)                       / first make equations in the standard form of
                                         / aX + b =c
   You see here that if you subtract equation equ  (1) from (2)
  X = 5-3 = 2
   Substitute  X =2 back to equation (1)
       2 - y = 3       Y = 2 - 3 = -1
You'll learn better techniques for higher order systems, but the 2nd order system (2 eqtns with 2 unknowns) is straightforward to solve by hand.  Here are the general steps
1) Place both equations in the form aX + bY = c
2) Inspect the constants of both equations to find a GCM for either X or Y
3) Multiply Either equation (or both) to achieve the GCM in both eqtns for that X or Y
4) Add or subtract the equations to get RID of that X or Y
5) Solve for that remaining Y or X
6) Substitute that first answer (YorX) to compute the other (XorY)
Let's do your example
2)Looking at the above, you'll note that if we subtract the 1st equation from the 2nd, we can make the Y-variable disappear.  To be clear, I'll rewrite them
-( x-y)=3
x = (5-3)=2
6) Now just pick an equation to substitute back to find y (doesn't matter which one you use)
-y=3-x    (moved x to the right on 1st eqtn)
 y=x-3    (multiplied by -1 on both sides)
 y=(2)-3= -1  (substituted 2 into x from above)-----> (x,y)=(2,-1)
6) Lets check the other eqtn just to verify
-y =-2x + 5
y= 2x - 5  (multiplied by -1 on both sides)
y=2(2)-5= -1 (substituted 2 into x from above)-----> (x,y)=(2,-1)  VERIFIED!


did you mean 2x-y=5 for the 2nd eqtn?  If so, this is correct.


Ok as you have said this is linear equation. The first step is to take "x" and "y" to one side for both the equation so you get: x-y = 3 (subtracting y on both side) and 2x-y = -5 ( subtracting 5 both side). Now you subtract the two equation you get : -x = 8, so x = -8 and putting this "x" in first equation gives -8 - y =3. So y= -11. Thank You