In Exercises 47–52, solve each system by the method of your ch...

Question

In Exercises 47–52, solve each system by the method of your choice. $\begin{cases}-4x + y = 12 \y = x^3 + 3x^2\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations. The first equation is negative. Six X plus Y equals 18. The second equation is Y equals six X cubed plus 18 X squared. And we're asked to solve this system of equations. Well I am going to solve it using substitution because we already have it set up where A. Y equals something. So anywhere there's a Y. In the first equation, I'm going to replace it with six X cubed plus 18 X squared. So equation one will now read negative six X plus not Y but six X cubed plus X squared and that equals 18. Well there are a few things to do here but first I'd like to note that everything is divisible by six. So I'm going to divide everything here by six. So this will now be negative X plus x cubed plus three, X squared equals three. And now I want to bring that three from the right hand side, over to the left hand side. So this will be set equal to zero and perhaps we can factor and then write these in descending order of their variables degrees. So starting with X cubed is the highest plus three, X squared minus x minus three. All equals zero. To try factoring by grouping grouping the first two terms together in the 3rd and 4th terms together asking is there anything the first two terms have in common? They both have an X squared. So we can factor that out. And that leaves in the parentheses X plus three. Now looking at the second group, anything they have in common, they both have a negative one, factor that out. And that leaves in parentheses X plus three equals zero. And this is great because they share this factor X plus three. So we can factor out that factor X plus three is multiplied by both X square and minus one still equals zero. Now looking at this second factor here, I recognize a difference of squares. So we can factor that, recalling previous lessons on difference of squares. We're going to take both terms and take the square root. So the squared of X squared is X. That goes in the first position for both sets of parentheses and the squared of one is one that goes in the second position for both sets of parentheses and by definition one is plus one is minus and now that is all factored and we can set all three of these factors equal to zero to solve for our x values. So for the first one we can subtract three from both sides. The second one subtract one from both sides and the third one add one to both sides. So for our three X solutions, we get x equals negative three, we get x equals negative one And we get x equals a positive one. Great. Now we have three solutions and we have to figure out what their corresponding y values are. So we're going to set up three equations. We get to go back to our original equations and we can plug in our X values to see what we get for our y corresponding y values. I'm going to use the first equation so negative six X plus Y equals 18. I'm going to write it three times because we have three X values need to find their three Y values. Okay in the first one we have negative six and we're going to solve for when X equals negative three. So negative six times negative three plus Y equals negative six times negative three is a positive 18. So 18 plus Y equals 18, interesting. So we subtract 18 from both sides. Well this one is a Y equals zero when X equals negative three. There is one solution. Now we're going to plug in a negative one for that X value to see what we get for our corresponding why value so negative six times negative one. That is a positive six. So we have six plus Y equals 18, subtracting six from both sides and we get y equals 12. When X equals negative one. Alright there is a second solution and now we're going to plug in for our X value. That last one which is X equals one. So negative six times one plus Y equals 18, negative six times one is negative six. So negative six plus Y equals 18 for this one we will add six to both sides. And we get that Y equals 24. When X equals one, there is our solution set. We look at our answer choices and this matches with answer choice C. Well done, we'll catch you on the next one.

In Exercises 47–52, solve each system by the method of your choice. $\begin{cases}\)-4x + y = 12 \\(\y\) = x^3 + 3x^2\(\end{cases}$

Key Concepts

Systems of Equations

Substitution Method

Solving Nonlinear Equations

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