SOLUTION: USE THE SUBSTITUTION METHOD TO SOLVE a-2b=a, 2a+3b=4

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Question 1210543: USE THE SUBSTITUTION METHOD TO SOLVE a-2b=a, 2a+3b=4
Found 3 solutions by ikleyn, CPhill, n2:
Answer by ikleyn(53558)   (Show Source): You can put this solution on YOUR website!
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USE THE SUBSTITUTION METHOD TO SOLVE a-2b=a, 2a+3b=4
~~~~~~~~~~~~~~~~~~~~~~~~ 

Your starting equations are

     a - 2b = a,    (1)

    2a + 3b = 4.    (2)


In equation (1), cancel the identical terms 'a' in both sides.


You will get  -2b = 0,  which tells you that  b = 0.


Substitute b= 0 into equation (2)

    2a + 3*0 = 4,

    2a       = 4

     a       = 4/2 = 2.


ANSWER.  a = 2, b = 0.


You may check it on your own that (a,b) = (2,0) satisfies the given equations.

At this point, the problem is solved completely.
We found the unique solution to the given system of equations.



Answer by CPhill(2189)   (Show Source): You can put this solution on YOUR website!
To solve the system of equations using the **substitution method**, follow these steps:
### Given Equations:
1.
2.
---
### Step 1: Isolate one variable from the first equation
We start with Equation (1) and solve for one of the variables. In this case, it is easiest to simplify the equation first.
Subtract from both sides:

Divide by :

### Step 2: Substitute the value into the second equation
Now that we know , we substitute this value into Equation (2):
### Step 3: Solve for the remaining variable
Divide both sides by :
---
### Final Solution:
*
*
### Verification:
* **Check Equation 1:** . (Correct)
* **Check Equation 2:** . (Correct)
Answer by n2(35)   (Show Source): You can put this solution on YOUR website!
.
USE THE SUBSTITUTION METHOD TO SOLVE a-2b=a, 2a+3b=4
~~~~~~~~~~~~~~~~~~~~~~~~~~


The post by @CPhill does not contain a real solution,
so the reason WHY it was created and posted to the forum is dark and unclear,
while a complete and correct solution by @ikleyn was just placed at this spot.

I recommend a reader do not spend his valuable time for nothing
and to ignore the post by @CPhill.



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