Amanda A. answered • 03/26/14

Tutor

4.9
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Experienced Teacher and Education Professional w/ Test-Prep Experience

Melanie - I can certainly help you with this! Using the substitution method means putting one equation into the other. For example, here, we can put the first equation into the second, because we can substitute for x.

**x = 6 + 4y**

**-3(x + y) = -36**

**-3(6+4y +y) = -36**

Now I have one equation with one variable, which means I can solve it! It just takes a little moving things around. First let's simplify what's inside the parentheses.

**-3(6 + 5y) = -36**

Next, we need to take care of the -3 so we can get the 6 + 5y out of the parentheses. We have two choices: distribute or divide. If we distribute, that means we will multiple the 6 and the 5y by -3. However, I think the better option is to divide. Let's divide both sides of the equation by -3.

**-3(6 + 5y)/-3 = -36/-3**

**-3(6 + 5y)/-3 = -36/-3**

**6 + 5y = 12**

This equation looks much easier to solve! The next step is to isolate the y (get it all alone on one side of the equation).

**6 + 5y = 12**

**6 - 6 + 5y = 12 - 6**

**5y = 6**

**5y/5 = 6/5**

**5y/5 = 6/5**

**y = 6/5 OR 1 1/5 OR 1.2**

Great! Now we know what y equals! We can plug that value of y into either equation to figure out x. I'm going to choose to plug y into the first equation.

**x = 6 + 4y**

**x = 6 + 4(1.2)**

**x = 6 + 4.8**

**x = 10.8**

Great! Now we have a value for both x and y. Let's check by plugging both into the second equation.

**-3(x + y) = -36**

**-3(10.8 + 1.2) = -36**

**-3(12) = -36**

**-36 = -36**

PERFECT. Now we have values for both x and y, and we've confirmed that they work! We've solved this system of equations! I hope this was helpful, Melanie!

Melanie G.

03/26/14