In Exercises 29–42, solve each system by the method of your ch...

Question

In Exercises 29–42, solve each system by the method of your choice. $\begin{cases}y = (x + 3)^2 \x + 2y = -2\end{cases}$

Accepted Answer

Welcome back. I am so glad you're here. We are given a system of equations and asked to solve it. The first equation that we're given is X equals why minus five squared. The second equation we're given is x minus four, Y equals negative eight. And I would like to solve this equation Using substitution were given that X equals y minus five squared in the first equation. So I can substitute for X in the second equation by putting in a y minus five squared where there's an X. So no longer x minus four Y. But now y minus five squared minus four, Y equals negative eight. And now I need to foil out this y minus five times y minus five. Bringing down the minus four, Y equals negative eight. And our firsts y times y is y squared, outsides Y times negative five is minus five, Y insides negative five times wise again negative five Y and lasts negative five times negative five is plus 25. Bring down that minus for why? And then I am going to bring this negative eight over to the left hand side of the equation. So that will be plus eight and then it will be equal to zero because we have a quadratic that Y is squared. And that way hopefully we can factor after we combine like terms. So now combining like terms there's only the one Y squared. Now combining all the wise we've got a minus five, y minus another five Y and then minus four Y. So that will be minus 14 Y and then 25 plus eight will be plus equals zero. And now we can factor and what times what gives us Y squared. That's why times why And what two numbers When multiplied together give us a positive 33. But add up to a negative 14, that would be a minus three minus 11. And so now we can set both of these factors equal equal to zero, Y minus three equals zero and y minus equals zero, Adding 3 to both sides for the first one and adding 11 to both sides for the 2nd 1. So that will give us a Y equals three for one solution and a Y equals 11 for another solution. So we can start up a solution set and plug in our y values. One y solution is three and one Y solution is 11. And now we want to go back in we can use our second equation. The second equation is x minus four, Y equals negative eight. And we can plug in our y values to solve for X. So for the first one will be x minus four times three equals negative eight. So x minus four times three is 12 equals negative eight and will add 12 to both sides. So this will be X equals negative eight plus 12 is a positive four. So when y equals three X equals four. Now plugging in 11 for our y value. So x minus four times 11 instead of Y equals negative eight. X minus four times 11 is 44 equals negative eight, adding 44 to both sides. We get x equals negative eight plus 44 is positive 36. Alright, so when X equals four, Y equals three, and when X equals 36 Y equals 11, 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 29–42, solve each system by the method of your choice. $\begin{cases}y = (x + 3)^2 \\x + 2y = -2\end{cases}$

Key Concepts

Systems of Equations

Substitution Method

Quadratic Functions and Their Graphs

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