Solving a system of equations

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You have copied (and high-lighted) the last line of problem #10 as if it were part of problem #11. Did you realize that? Problem #11 asks you to solve the two equations 4x- 3y= -19 and 2x+ y= 13. Multiply the second equation by 3, 6x+ 3y= 39, and add that to the first equation.
 
Jun 2018
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Home
Thank you I didn’t catch that error but that would’ve simplified everything! I appreciate the help, thank you once again
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Where did the 3 come from though like why multiply by 3?
That technique is trying to eliminate one of the variables. So, the first equation has -3y. The second equation has y. If you multiply the second equation by 3, you have a -3y in the first equation and a +3y in the modified second equation, so now when you add them, you have 0y (this gives an equation only in x).

Another method: In the second equation, solve for y in terms of x (I choose the second equation because the coefficient of y is 1):

$y = 13-2x$

Now, plug that into the first equation:

$4x-3(13-2x)=-19$

After multiplying out, you can solve for $x$ since it is an equation in one variable. Once you know x, plug it into the formula you just created for y.
 
Feb 2015
2,255
510
Ottawa Ontario
Where did the 3 come from though like why multiply by 3?
These are your equations:
4x-3y=-19 [1]
2x+y=13 [2]

Multiply [2] by 3 and you now have:
4x-3y=-19 [1]
6x+3y= 39 [2]

Add 'em up:
4x-3y=-19 [1]
6x+3y=39 [2]
-----------------
10x + 0 = 20
So x = 2

Are you saying that your teacher never taught that?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Where did the 3 come from though like why multiply by 3?
The first equation had "-3y". The second equation had "+ y". Multiplying the second equation by 3 changes that to "+ 3y" and adding "-3y+ 3y" eliminates "y" from the equation.