Stephen W. answered • 02/24/15

Tutor

4.9
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Academic and independent Algebra 1&2 experience 10+ years

You will need to choose one of the equations, and solve for one of the variables. Then, plug the new information into the equation for the variable. Here is an example:

3x - 6y = 12

x + 3y = 3

I choose to work with the second equation, then solve for 'x', which then looks like this:

x = -3y + 3

Now. If we had a value for 'x', say 2, then we could just plug 2 into the equation wherever there is an 'x'. In this case, we don't have a number value to equal 'x', but we do have an expression: 3y + 3. We will plug this into the other equation for 'x', like this:

3

**(-3y + 3)**- 6y = 12Now we solve the expression for the only variable in it - 'y':

-9y + 9 - 6y = 12

-15y +9 = 12

-15y = 3

y = 3/-15

**y = -1/5**

Now that we have an actual number value for 'y', we can then plug it into either of the original equations, then solve for 'x' I choose to use the second equation:

x + 3

**(-1/5)**= 3x - 3/5 = 3

x = 3 + 3/5

**x = 3 3/5**

The solution to the sample problem is x = 3 3/5, y = -1/5, or (3 3/5, -1/5)

I hope this was helpful!