Please see your third line: It should be
4y-3y=20-5 or y=15, by subtracting first equation from second.
Now x+3*15=5
Or x=5-45=-40. Check:-40+4*15=20 by putting values in second equation.
This -40=15-60=-40 OK.
X+3y=5
4x+4y=4 I tried to work this problem and
this is what I came up with.
x+3y=5
x+4y=20
7y=15
y=7/15
and x= (5+y)/4
(5+7/15)/2
2/15
x=2/15
x=2/15, y=7/15 I would like for someone to check to see if it is correct.
4y-3y=20-5 or y=15, by subtracting first equation from second.
Now x+3*15=5
Or x=5-45=-40. Check:-40+4*15=20 by putting values in second equation.
This -40=15-60=-40 OK.
Given system of equations:
1) x + 3y = 5
2) 4x + 4y = 4
To eliminate the variable x, we need to manipulate the equations so that the coefficients of x are the same or the additive inverse of each other.
To achieve this, we can multiply equation 1 by 4 and equation 2 by 1, so that the coefficients of x become equal:
4 * (x + 3y) = 4 * 5
1 * (4x + 4y) = 1 * 4
Simplifying these equations:
4x + 12y = 20
4x + 4y = 4
Now, subtract equation 2 from equation 1 (4x - 4x = 0x cancels out):
(4x + 12y) - (4x + 4y) = 20 - 4
Simplifying and solving for y:
8y = 16
y = 16/8
y = 2
Now that we have a value for y, we can substitute it back into either of the original equations to find the value of x. Let's use equation 1:
x + 3(2) = 5
x + 6 = 5
x = 5 - 6
x = -1
The solution to the system of equations is:
x = -1, y = 2.
Therefore, it seems like there was an error in your calculations. The correct solution to the system is x = -1 and y = 2.