Asked by Esther
(Systems) how do I solve 5x+3y=2 and x+2y=6 using substition
This is what I did
X+2y=6 eq.1
5x+3y=2 eq.2
X=-3y/5
Sub(x=-3y/5) into eq.1
-3y/5 + 2y=6
7/5y=6
y=30/7
Sub(y=30/7) into eq.2
5x+3(30/7)=3
5x+90/7=3
5x=-69/7
x= -68/35
The answer I got was (-68/35, 30/7) but I know it's wrong because there were no fractions in the possible answers.
I know how to solve it with elimination but I can't figure out how to do it with substitution.
This is what I did
X+2y=6 eq.1
5x+3y=2 eq.2
X=-3y/5
Sub(x=-3y/5) into eq.1
-3y/5 + 2y=6
7/5y=6
y=30/7
Sub(y=30/7) into eq.2
5x+3(30/7)=3
5x+90/7=3
5x=-69/7
x= -68/35
The answer I got was (-68/35, 30/7) but I know it's wrong because there were no fractions in the possible answers.
I know how to solve it with elimination but I can't figure out how to do it with substitution.
Answers
Answered by
Reiny
Why not take the easier equation?
x+2y=6 ---> x = 6-2y
now plug that into the other,
5x+3y=2
5(6-2y) + 3y = 2
30 - 10y + 3y = 2
-7y = -28
y = 4
then x = 6 -2(4) = -2
see how easy it is ?
x+2y=6 ---> x = 6-2y
now plug that into the other,
5x+3y=2
5(6-2y) + 3y = 2
30 - 10y + 3y = 2
-7y = -28
y = 4
then x = 6 -2(4) = -2
see how easy it is ?
Answered by
Esther
Thank u!
Answered by
Steve
Your mistake happened early. You said
5x+3y=2 eq.2
X=-3y/5
but really
x = (2-3y)/5
5x+3y=2 eq.2
X=-3y/5
but really
x = (2-3y)/5
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