use either substitution or elimination. im in algebra 2

First, I will assume that the two equations in question are:

Eqn. 1) x=5y-11

Eqn. 2) 3x+20y=30

Solving this system does not yield whole number results, but the process is the same nonetheless. Let's begin.

Let's solve this problem using each method. We have two unknown variables, therefore it is necessary there are two equations. The goal is to use one of the two methods to reduce either equation into an equation with only one variable. Once one of the unkowns has been solved for, it is substituted into the other equation to find the other unknown.

1) Substitution. In this case, substitution is a bit simpler since eqn. 1 is already solved for x in terms of y. Therefore, we simply substitute eqn. 1 for the
**x** variable in **eqn. 2**:

3x+20y=30

3(**5y-11**)+20y=30; now you see that the equation is only in terms of
**y**. Solve for y.

15y-33+20y=30

35y=66

y=66/35, or approx. 1.89

Now that we know the value of y, substitute for y in eqn 1 and solve for x.

x=5(**1.89**)-11

x=5.89-11

x= -5.11

So the solution is approx. the ordered pair (-5.11, 1.89).

Now by elimination. The goal is the same, reduce one equation so it has only one variable. First, write each equation in standard form (Ax+By=C, where A,B,C are the constants) in a column:

x - 5y = -11

3x+20y = 30

Now, we have to manipulate one of the equations by multiplying it by a factor so that when added together, one of the variables cancels out. Let's manipulate eqn 1 by asking the question, "what do I need to multiply eqn 1 by so when added together, one of the variables adds to zero?" The x variable looks like a good candidate to focus on. Let's multiply eqn 1 by negative 3.

-3[x - 5y = -11]

-3x+15y = 33

Now we have,

-3x+15y = 33

3x+20y = 30

Addding you get,

35y=66; look familiar?

y=66/35, or approx. 1.89

Now, substitute into either of the original equations to find x, and you'll find that x= -5.11 as before.

If you graph both eqations, you will see that this is the approx. location where the graphs of the two equations intersect, the point at which they share a common solution.

## Comments

this helped so much. thank you

You're very welcome!