+0  
 
0
860
3
avatar+1438 

What would be the easiest way to solve these two problems, and how would I solve them using those ways? (these are practice problems and they are for getting answers and understanding how it's done. It isn't a test or anything like that.)

 

4x + 7y = 6

9x - 2y = 49

 

 

and

 

 

3x - y = 4

y = 2x - 2 

 May 6, 2019
 #1
avatar+2862 
+3

When looking at systems of linear equations. You can either use substitution or elimination.

 

The question is WHICH one to use? 

Don't skim my answer for the answer

(ax with a being the constant)

-----------

 

Substitution. You should use this whenever you see a variable with a constant of 1. For example the second equation you posted.

3x - y = 4

y = 2x - 2 

 

Notice how this can EASILY be substituted

 

Solving:

 

3x - (2x-2) = 4  (remember there is a -1 multiplying at the parentheses).

 

(remove parentheses and combine like terms.)

 

x + 2 = 4

 

x = 2

 

Now substitute back into one of your equations and find y.

 

Elimination. You should use this whenever you can't divide both sides of one of the equations and get integer constants.

4x + 7y = 6

9x - 2y = 49

 

Notice how the equation will be difficult substituting. If you try to isolate a variable, you will get fraction constants.

 

So a good strategy is to find the greatest common multiple.


Solving:

4x + 7y = 6            (Mulitply 7y by 2)

9x - 2y = 49           (Multiply 2y by 7)

 

8x + 14y = 12

63x - 14y = 343

 

Now you can just add both equations: 71x = 355

 

Then divide - x=5

 

Now substitute back into one of the equations and find y

 May 6, 2019
edited by CalculatorUser  May 6, 2019
 #3
avatar+1438 
+2

Thank you! You really explained that in a way I could understand.

SmartMathMan  May 6, 2019

0 Online Users