In Exercises 1–18, solve each system by the substitution metho...

Question

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}xy = 6 \2x - y = 1\end{cases}$

Accepted Answer

everyone in this problem. We are asked to solve the system given below. Using the substitution method. The system we're given is X. Y equals eight and x minus two Y equals six. Okay, so this is a system of two equations and they are nonlinear. All right, so we have equation one. We're just gonna go ahead and label our equation so that it's easier to refer to them later. So we have X. Y. Is equal to eight. An equation two is x minus two. Y. Is equal to six. Okay. We're told to use the substitution method and the substitution method. But we want to do is we want to choose one of the equations. We want to isolate one of the variables from that equation and substitute it into the other equation. Okay, so we can choose either variable to to isolate and we can choose to do it in either equation. In this case it's gonna be easier to do it in equation too because the X. Is already has a coefficient of one. We don't need to do any division of variables. Okay, so let's go ahead and choose equation too. We're gonna get X -2. I is equal to six and rearranging. We're gonna call this equation to star and it's going to be X is equal to two. Y plus six. Okay, so it's the exact same as equation too. It's just been rearranged. Alright now our substitution method, we want to substitute that into the other equation. So we're going to substitute that into equation one which is gonna allow us to find solutions that satisfy both of those equations. Alright so substituting equation to Star into equation one we have X which is now two Y plus six times Y is equal to eight. And you'll notice that now we have an equation with just why? Okay so we can solve for y expanding on the left, we get to I squared plus six Y. Okay. And on the right we still have eight. Yes we get two Y squared plus six. Y minus eight is equal to zero. We can divide by two. Across the equation we get y squared plus three, Y minus four is equal to zero, divided by two is still zero. I'm just gonna move up so we have some more room. Alright factoring this we get Y plus four times Y minus one is equal to zero. Okay? Which gives us y value of -4 or positive one. So these are y values now. In order to solve the system. We need more than just y values, we need x values as well. Okay so we're gonna substitute these y values back into either equation one or two. Okay you can choose to do either equation in this case we're going to choose equation to star because we've already rearranged it and it's just easier to use. Okay so let's go ahead and substitute Y equals and we'll choose the first fly value of negative four into equation two sir. Alright so we get X is equal to two times Y. Which is negative four plus six. So your ex is equal to negative eight plus six which is equal to negative two. So we get the 0.-2 -4. Alright so we found one point. Now let's go ahead and find another. We have a second Y value Y equals one. We're gonna do the same thing. Substitute it into equation two star. So we get X is equal to two times one which is or sorry two times Y which is one plus six. So in this case we get two plus six which is equal to eight. And so we get the point eight comma one. Alright so writing this as a solution set we found two points. And so the solution will be the set containing these two points negative two negative and 8:01 which is going to correspond with answer. D Hope this video helped see you in the next one.

In Exercises 1–18, solve each system by the substitution method. $\begin{cases}\)xy = 6 \\2x - y = 1\(\end{cases}$

Key Concepts

System of Equations

Substitution Method

Solving Nonlinear Equations

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