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## How do I solve a linear system

In Algebra 1 we're working on solving linear systems, to problems like "x=y+3 2x-y+5" And I don't even know where to start.

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)

1)
x-y=3
2x-y=5

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

2x-y=5
-( 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!