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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.

4 Answers by Expert Tutors

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Tom D. | Very patient Math Expert who likes to teachVery patient Math Expert who likes to te...
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!
 
 
 

Comments

did you mean 2x-y=5 for the 2nd eqtn?  If so, this is correct.
Md Akib Ali S. | Tutor Who Can Tutor all Science, Math, and Engineering SubjectsTutor Who Can Tutor all Science, Math, a...
1
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
Daniel B. | I tutor algebra, calculus, and trigonometry, but not statisticsI tutor algebra, calculus, and trigonome...
0
Instead of adding equations you could also just use substitution since we know x=y+3.  Plug in x in the 2nd equation and we get
 
2(y+3)-y=5            Solve for y.
2y+6-y=5          
y+6=5
y=-1
 
Now plug in y in the 1st equation and we get
 
x=(-1)+3=2
 
Hope this helps!
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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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
 
 

Comments

would I be able to talk live about this, if so you'll need a lot of patience, it will take a bit for me to understand it
 
There's a small error: the standard form is actually ax + by = c.  You just forgot the 'y'.