Hello, MathMack!
A simple error . . .
Solve by substiution: .
I found y for the 2nd one and got: .
It should have been: .
Here is the problem:
solve these equations using substiution:
2x+4y=8
3x+2y+0
I figured Y for the 2nd one first so I got
2y=0+3x which equals y=3/2x
substitute that into the first one
2x +4(3/2x)=8
2x+6x=8
x=1
put that into y=3/2x which I got 3/2 for Y
then I went to check and only one worked, where am I going wrong? Thanks in advance!
2x + 4y = 8
2x = 8 - 4y
x = 4 - 2y
Substitute into the other equation:
3(4 - 2y) + 2y = 0
12 - 6y + 2y = 0
12 - 4y = 0
12 = 4y
y = 3
And go back to the other equation:
2x + 4(3) = 8
2x + 12 = 8
2x = -4
x = -2
You want to start by making the coefficients in front of one of the variables the same by multiplying each equation by a constant. I am going to make the coefficients of x equivalent.this one is by elimination
3x+4y=2
2x-3y= -10
Where do I start?
2(3x + 4y = 2) => 6x + 8y = 4
3(2x - 3y = -10) => 6x - 9y = -30
Now, subtract these two equations.
(6x + 8y = 4)
- (6x - 9y = -30) [Remember that subtracting a negative is equivalent to addition]
--------------------
0x + 17y = 34
17y = 34
y = 2
Now, just like in substitution, use one of the original equations to solve for x.
3x + 4(2) = 2
3x + 8 = 2
3x = -6
x = -2
Good luck. :>