# Can someone please give me an easier way to understand or how to do system equation using the substitution medthod example of how it goes step by step

*print*Print*list*Cite

### 1 Answer

Suppose you are asked to solve the following system:

2x+3y=5

-3x+7y=11

You want to solve one of the equations for one of the variables. If possible, I would solve for a variable with coefficient of 1. In this case there is no such variable, so I decide to solve the first equation for x. (There is no "right" choice; some choices might be simpler algebraically or arithmetically, but the method is general.)

2x+3y=5

2x=5-3y

`x=5/2-3/2y`

We take this expression for x and substitute it into the second equation. (You must substitute into the other equation -- otherwise you have an identity i.e. you will get a statement that is always true.)

-3x+7y=11

`-3(5/2-3/2y)+7y=11` ** Substituting in place of "x" **

`-15/2+9/2y+7y=11`

`-15/2+23/2y=11`

Notice that the new equation involves only one variable, so it is amenable to standard techniques.

`23/2y=22/2+15/2`

`y=2/23*22/2+2/23*15/2`

`y=22/23+15/23`

`y=37/23`

Then `x=5/2-3/2y` ==>`x=5/2-3/2(37/23)`

`x=5/2-111/46=2/23`

The solution will be `x=2/23,y=37/23`

Check: 2x+3y=5

`2(2/23)+3(37/23)=4/23+111/23=115/3=5`

-3x+7y=11

`-3(2/23)+7(37/23)=-6/23+259/23=253/23=11`