(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.
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 ?
Thank u!
Your mistake happened early. You said
5x+3y=2 eq.2
X=-3y/5
but really
x = (2-3y)/5
To solve the system of equations using substitution, you need to correctly substitute the value of one variable into the other equation. Let's go through the steps again:
1. Start with the given system of equations:
5x + 3y = 2 -- (equation 1)
x + 2y = 6 -- (equation 2)
2. Solve equation 2 for x:
x = 6 - 2y
3. Substitute the expression for x into equation 1:
5(6 - 2y) + 3y = 2
4. Simplify and solve for y:
30 - 10y + 3y = 2
-7y = 2 - 30
-7y = -28
y = -28 / -7
y = 4
5. Substitute the value of y back into equation 2 to find x:
x + 2(4) = 6
x + 8 = 6
x = 6 - 8
x = -2
Therefore, the solution to the system of equations is x = -2 and y = 4.