How would I solve this problem using the substitution method. Any help woud be greatly appreciated. :~}

x=y+3

4x+y=2

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- Feb 26th 2008, 09:06 AMimbadatmathsystem of equations using substitution method
How would I solve this problem using the substitution method. Any help woud be greatly appreciated. :~}

x=y+3

4x+y=2 - Feb 26th 2008, 09:31 AMimbadatmathIm not sure if i did it right
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 - Feb 26th 2008, 09:48 AMmathcelebAssistance
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. - Feb 26th 2008, 09:56 AMimbadatmath
How would i enter this into my calculator? I see an x button but no y button

- Feb 26th 2008, 10:00 AMbobak
I dont think it is appropriate to suggest this sort of method for what is a very easy problem.

Quote:

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

- Feb 26th 2008, 10:09 AMimbadatmath
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

- Feb 26th 2008, 10:14 AMmathceleb
- Feb 26th 2008, 10:56 AMtopsquark