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.

## How do I solve a linear system

Tutors, please sign in to answer this question.

# 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

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!

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

## Comments

Comment