# Video: Solving Systems of Linear and Quadratic Equations

Find all of the solutions to the simultaneous equations 𝑦 + 2𝑥 = 3 and 𝑥² + 𝑥𝑦 + 𝑦² = 3.

03:58

### Video Transcript

Find all of the solutions to the simultaneous equations 𝑦 plus two 𝑥 equals three and 𝑥 squared plus 𝑥𝑦 plus 𝑦 squared equals three.

To solve simultaneous equations, there’s a few different ways: substitution, elimination, graphing, and so on. Let’s go ahead and use substitution. So we’ll take one of the equations, solve for a variable, and then substitute that into the other equation. Since this equation is shorter, let’s go ahead and solve for 𝑦. It seems the simplest. And by subtracting two 𝑥 over to the right-hand side, we get 𝑦 equals negative two 𝑥 plus three. So now that we have this value for 𝑦, we can plug it in for 𝑦 into the other equation.

Here, we’ve taken negative two 𝑥 plus three and we’ve substituted that in for 𝑦 into our second equation. And now we need to evaluate. So we need to distribute 𝑥 and square our binomial. So we’ve brought down the 𝑥 squared. 𝑥 times negative two 𝑥 is negative two 𝑥 squared and 𝑥 times three is three 𝑥. When we square something, that means we’re multiplying it to itself. So we’re gonna have to foil.

So let’s go ahead and foil. Negative two 𝑥 times negative two 𝑥 is positive four 𝑥 squared. Negative two 𝑥 times three is negative six 𝑥. Three times negative two 𝑥 is negative six 𝑥. And three times three is nine. Now we need to bring everything else down. So we’d bring down our equal sign and the three and then everything that we had before. Now we combine like terms. So our highest power is two in 𝑥 squared. So let’s go ahead and put all of those together. So that would be 𝑥 squared minus two 𝑥 squared plus four 𝑥 squared is three 𝑥 squared. Three 𝑥 minus six 𝑥 minus six 𝑥 would be negative nine 𝑥. And to put the nine and the three together, let’s subtract three from both sides. So we get six and then the three is cancelled. So now we have zero.

So we need to solve this. We can figure out what 𝑥 is and then once we have 𝑥, we can plug it in and find 𝑦. So first, we have three, negative nine, and six as our numbers. We can divide everything by three. That’s our greatest common factor. And we take three out. We have 𝑥 squared minus three 𝑥 plus two left because we divided everything by three. Now we need to factor this quadratic. So what two numbers multiply to be two and add to be negative three. That would be negative two and negative one.

So we have our GCF, our greatest common factor, of three and then 𝑥 minus two and 𝑥 minus one. So we say each of these factors is equal to zero. And now, three being equal to zero doesn’t do anything; that’s not even true. So we can ignore that. If it had a variable with it, like 𝑥, then we would actually solve. Now 𝑥 minus two equals zero. We add two to both sides and get 𝑥 equals two. And then, now we need to add one to our next equation and we get 𝑥 equals one.

So we can either have two for 𝑥 or one for 𝑥. So now we can take those values and plug them in to our original equations, either one or both if you’d like. So let’s go ahead and plug it in to the smaller one of the two, which would be this one. However, these equations are exactly the same. All we did was move the 𝑥 term over to the right-hand side. So since that’s already having 𝑦 isolated, let’s go ahead and use the one that we moved.

So plugging in the value 𝑥 equals two, we have negative two times two which is negative four plus three which is negative one. So when 𝑥 is two, 𝑦 is negative one. And now, we plug in 𝑥 equals one. And negative two times one is negative two, plus three is positive one. So another solution to the simultaneous equations would be one, one.

Therefore, the solutions to the simultaneous equations would be two, negative one and one, one.