solving linear systems by the method of substitution

[me_is_me]

ok, how would I do this? It's substitution; that's all I know.

1) 2x - 3y = 21, y = 3 - x

2) x + 2y = 13, 3x - 5y = 6

 

[soroban]

Re: i need help

Hello, me_is_me!

You say you "know" Substitution . . . but apparently, you don't.
(1) Solve one of the equation for one of its variables.
(2) Substitute into the other equation.
(3) Solve the resulting equation.
(4) Substitute this value into either equation and solve for the second variable.

\(\displaystyle 2x\,-\,3y\:=\:21\)
\(\displaystyle y\:=\:3\,-\,x\)

The second equation is already solved for \(\displaystyle y\!:\;\;y \:=\:3\,-\,x\)

Substitute this into the first equation: \(\displaystyle \:2x\,-\,3(3\,-\,x) \:=\:21\)

Then we have: \(\displaystyle \:2x\,-\,9\,+\,3x\:=\:21\;\;\Rightarrow\;\;5x\:=\:30\;\;\Rightarrow\;\;\fbox{x\,=\,6}\)

Substitute into the second equation: \(\displaystyle \:y\:=\:3\,-\,6\;\;\Rightarrow\;\;\fbox{y\,=\,-3}\)

\(\displaystyle \begin{array}{cccc}x\,+\,2y & = & 13 & [1] \\
3x\,-\,5y & = & 6 & [2]\end{array}\)

Solve [1] for \(\displaystyle x:\;\;x \:=\:13\,-\,2y\)

Substitute into [2]: \(\displaystyle \:3(13\,-\,2y)\,-\,5y\:=\:6\)

Can you finish it now?

 

[TchrWill]

Re: i need help

ok how do u do this........... its substitution thats all i know
(1) 2x-3y=21, y=3-x

(2) x+2y=13, 3x-5y=6

1--Substitute y = 3 - x into x + 2y = 15 abd solve for x.

2--Multiply the first by 3 and subtract the second equation from the result and solve for y.

 

[stapel]

...how would I do this? It's substitution; that's all I know.

If you're not familiar with the method, then try studying some online lessons to learn! :wink:

By the way: Welcome to FreeMathHelp!

Eliz.