| 
 
 
 
Question 145383:  Could someone tell me if I got these problems right? The first two are the Substitution Method, and the second two are using the Elimination Method.
 
1)  x + y = 15   
   4x + 3y = 38
 
2)-3x + y = -4 
    x - y = 0
 
 
The Elimination Method:
 
1)  5x - y = 12 
    3x + y = 4
 
 
2)  2x - y = -6 
    2x - 2y = -4
 
Thank you in advance. Elle J 
 Answer by jim_thompson5910(35256)      (Show Source): 
You can  put this solution on YOUR website! Well it's hard to tell what you got if you don't post your solutions =)
 
 
 
I'll do one of each to help you in the right direction.
 
 
 
 
# 1
 
Jump to problem #3
 
 
Start with the given system of equations:
 
 
 
 
 
 
 
Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.
 
 
 
 
 
So let's isolate y in the first equation
 
 
  Start with the first equation
 
 
 
   Subtract   from both sides
 
 
 
  Rearrange the equation
 
 
 
  Divide both sides by  
 
 
 
  Break up the fraction
 
 
 
  Reduce
 
 
 
 
---------------------
 
 
Since  , we can now replace each   in the second equation with   to solve for  
 
 
 
 
  Plug in   into the first equation. In other words, replace each   with  . Notice we've eliminated the   variables. So we now have a simple equation with one unknown.
 
 
 
 
  Distribute   to  
 
 
 
  Multiply
 
 
 
  Combine like terms on the left side
 
 
 
 Subtract 45 from both sides
 
 
 
  Combine like terms on the right side
 
 
 
 
 
 
-----------------First Answer------------------------------
 
 
 
So the first part of our answer is:  
 
 
 
 
 
 
 
 
 
 
Since we know that   we can plug it into the equation   (remember we previously solved for   in the first equation).
 
 
 
 
  Start with the equation where   was previously isolated.
 
 
 
  Plug in  
 
 
 
  Multiply
 
 
 
  Combine like terms 
 
 
 
 
-----------------Second Answer------------------------------
 
 
 
So the second part of our answer is:  
 
 
 
 
 
 
 
 
 
 
-----------------Summary------------------------------
 
 
So our answers are:
 
 
  and  
 
 
which form the point   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
# 3
 
Jump to problem #1
 
 
 
 
 
 
Start with the given system of equations:
 
 
 
 
 
 
 
Now in order to solve this system by using elimination/addition, we need to solve (or isolate) one variable. I'm going to solve for y.
 
 
 
 
 
 
In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for  , we would have to eliminate   (or vice versa).
 
 
 
So lets eliminate  . In order to do that, we need to have both   coefficients that are equal in magnitude but have opposite signs (for instance 2 and -2 are equal in magnitude but have opposite signs). This way they will add to zero. By adding to zero, they can be eliminated.
 
 
 
 
So to make the   coefficients equal in magnitude but opposite in sign, we need to multiply both   coefficients by some number to get them to an common number. So if we wanted to get   and   to some equal number, we could try to get them to the LCM.
 
 
 
 
Since the LCM of   and   is  , we need to multiply both sides of the top equation by   and multiply both sides of the bottom equation by   like this:
 
 
 
 
 
   Multiply the top equation (both sides) by   
   Multiply the bottom equation (both sides) by  
 
 
 
 
 
Distribute and multiply
 
 
  
 
 
 
 
Now add the equations together. In order to add 2 equations, group like terms and combine them
 
 
 
 
 
Combine like terms and simplify
 
 
 
 
  Notice how the x terms cancel out
 
 
 
 
 
  Simplify
 
 
 
 
 
  Divide both sides by   to isolate y
 
 
 
 
 
  Reduce
 
 
 
 
Now plug this answer into the top equation   to solve for x
 
 
  Start with the first equation
 
 
 
 
  Plug in  
 
 
 
 
 
  Multiply
 
 
 
 
 Subtract 2 from both sides
 
 
 
  Combine like terms on the right side
 
 
 
  Divide both sides by 5 to isolate x
 
 
 
 
  Divide
 
 
 
 
 
So our answer is
 
  and  
 
 
 
 
which also looks like  
 
 
 
 
 
Now let's graph the two equations (if you need help with graphing, check out this solver)
 
 
 
From the graph, we can see that the two equations intersect at  . This visually verifies our answer.
 
 
 
 
 
  graph of   (red) and   (green)  and the intersection of the lines (blue circle).
 
 
 
  | 
 
  
 
 |   
 
 |   
 |  |