4(y+3)+y=2
4y+12+y=2
x=y+3
5y+12=2 (-12 from both sides)
5y=-10 (/ both sides by 5)
y=-2
Then I found x
x=-2+3
x=1
2 Unknowns by Substitution Calculator
Enter 4x+y=2 for Equation 1 and x-y=3 for Equation 2 and Press Calculate.
I rearranged your Equation 1.
Solution: (x, y) = (1, -2)
if you want to see the opposite way of substitution method, switch equation 1 and equation 2 in the entry boxes. :-)
Sometime later this week, I'll clean up the math portion to not show denominators of 1 and coefficients of 1. It looks like you missed a sign. Look at the section where it says:
"Now, group like terms by getting the y variables on one side of the equation and the constants on the other side:"
I hope this helps, let me know if you have questions.
I dont think it is appropriate to suggest this sort of method for what is a very easy problem.
Keep up the good work. ( and possibly change your username )4(y+3)+y=2
4y+12+y=2
x=y+3
5y+12=2 (-12 from both sides)
5y=-10 (/ both sides by 5)
y=-2
Then I found x
x=-2+3
x=1
I wish i could change my user name but i am not that good at math i do all right on the home work and stuff but when the tests come around i forget how to solve each individual problem i use one type of method on the wronge type of problem and get totaly confused lol