In Exercises 5–18, solve each system by the substitution method.2...

Question

In Exercises 5–18, solve each system by the substitution method.2x + 5y = - 43x - y = 11

Two linear equations for solving a system by substitution method in college algebra.

Accepted Answer

Hey everyone welcome back in this problem. We are asked to use the substitution method to solve the system. And the system we're given is five X plus three. Y equals one and two, X minus Y equals minus four. Okay so let's go ahead and write these equations and we're going to label them so that we can refer to them later. So the first equation will be five, X plus three, Y equals one. And equation two will be two, X minus Y equals minus four. Now with the substitution method, what we wanna do is we want to isolate one of the variables from one of the equations. Okay, we can choose either variable X or Y. And we can choose either equation one or two. In this case we're going to choose equation too because the Y doesn't have just has a coefficient of one. Okay so that's going to be the simplest. So we'll choose equation to but you can choose anything that you'd like. Alright, so working with equation two, we have two X minus Y equals minus four. And again we want to isolate one of the variables. In this case we're going to choose Y. So if we move the Y to the right hand side and the four to the left hand side we're gonna have Y. Is equal to two X plus four. Alright now let's call this two star. It's the exact same equation as to it tells us the same information it's just been rearranged to be more convenient. Okay so we'll call that two star. What we're gonna do is we're gonna substitute to star. Okay this is our substitution method. So we're going to use that information from equation to we're gonna substitute it into equation one. Okay so that we're finding a solution that satisfies both equations. Alright so now we have five x plus three Y. But in this case y is going to be two, X plus four. Okay so three times two X plus four is equal to one. And you'll notice that we just have an equation of X. And constant terms here. So we can go ahead and solve for X expanding on the left. We have five X plus six X plus 12 equals one. Okay so we have 11 X. And if we subtract the 12 minus 11 on the right hand side, I'm just gonna scroll down so we have a bit more room here And this is gonna give us X. is equal to -1. Alright so we get X is equal to minus one. That's great. We have an X value but we need a solution which means we need a corresponding y value. Okay so let's go ahead and work on this side now And we're gonna substitute x equals -1 into one of these equations. Ok to find the corresponding y. Now we can choose either equation. Okay we can choose equation one or we can choose equation to we're gonna seven an X. Value and we want to solve for y. Okay well we already have equation to Star here that we've rearranged for why? So that's gonna be the easiest choice. So let's go ahead and choose equation two Star. Okay so we have Y. Is equal to two times X, which is minus one plus four. This is going to give us minus two plus four and we get Y. Is equal to two. So now we have our X. Value and our Y value. So we can write our solution set As the set of the single .12. Okay this is going to correspond with answers. See I hope this video helped. Thanks everyone for watching.

In Exercises 5–18, solve each system by the substitution method.2x + 5y = - 43x - y = 11

Key Concepts

Systems of Linear Equations

Substitution Method

Linear Equations

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