I need help with Solving systems of Linear equations by the addition method?

Question

My homework wants me to solve a system of two linear equations in two variables by the addition method, my book doesn't explain it real well, if someone can suggest a free website that I can look at some examples or that can explain it to me in plain english I would greatly appreciate it!

Answer 1

the idea is to cancel out a variable when adding the problems. You may need to multiply one of the equations by something to do this

For example add these two
x + y = 7
x - y = 3
-----------
2x + 0y = 10
2x = 10
x = 5

now plug 5 in for x in one of the equations
x + y = 7
5 + y = 7
y = 2

Sometimes it is not that easy

2x + 2y = 12
6x - y = 8

If you add these nothing will cancel out. So we should multiply both sides of the second equation by 2 so that the ys will cancel

(2*6x)-(2*y) = 2*8
12x - 2y = 16

now we can add and solve
2x + 2y = 12
12x - 2y = 16
------------------
14x + 0y = 28
14x = 28
x = 2
and now plug 2 in for x in the first equation
2x + 2y = 12
2*2 + 2y = 12
4 + 2y = 12
2y = 8
y = 4

so x-2, y=4

hope that helps

Answer 2

The first thing is to eliminate one of the variables Here: add equations 1 and 2 to get 4x + 4y = 4 ==> x + y = 1 ==> x = 1 - y then equations 1 and 3 3x + 5y =7 now substitute for x with the previous equation 3 - 3y + 5y = 7 ==> 2y = 4 ==> y = 2 and hence x = -1 Put these values for x and y into one of the original equations -1 + 6 + z = 6 ==> z = 1 test in one of the others: -3 + 2 - 1 = -2 or -2 + 4 - 1 = 1 and you know you have the right answers: x = -1 y = 2 z = 1

Answer 3

Let's take a simple system:

x + y = 4
x - y = 2
Now add the above two equations getting:
2x = 6 [Notice the y terms canceled out so we can solve fo x]
x = 3
Now put 3 for x in the 1st equation getting:
3+y = 4
y = 4-3 = 1
The solution is x = 3 and y=1

Let's try a little harder one:
2x - 3y = -5 <-- Eq 1
x - y = -1 <-- Eq 2
Now we need to eliminate x so we would like to change the x in Eq 2 to -2x so when we add the xterms will be eliminated. We can do this by multiplying EQ 2 by -2 getting:
-2x +2y = +2 <-- Eq 3
Now add Eq 3 and Eq 1 getting
-y = -3
y = 3
Substitute 3 for y in Eq 3 getting
-2x +2(3) = 2
-2x + 6 = 2
-2x = 2-6 = -4
x = 2
So solution is x=2 and y = 3

So just pick which term to wish to eliminate and the do whatever manipulations you need to do to get the terms in the two equations to cancel out. Then you can solve for the other variable.

Answer 4

Ax + By + C = 0
Rx + Sy + T = 0

The idea is to get the coeffecients of the y or the x to be opposites (SB and -SB, for instance).

If we multiply the first equation by -S, & the second one by B, we get:

-SA x-SBy - SC = 0
RBx+SBy+TB=0

Now, we just add the two equations:

x(-SA + RB) + (TB - SC) = 0
Now we have one equation in one variable. Solve for x.
Once you have x, substitute that value into either equation to solve for y.

{The principle is that you can solve both equations for SBy:
SBy=-SAx-SC
SBy=-RBx-TB
Since these are equal to the same thing, we can equate them, and then solve.}