Use substitution to solve the following system of linear equations.4x+y=14x+y=14x+y=1x−y=4x-y=4x−y=4

( 1 , − 3 ) \left(1,-3\right) ( 1 , − 3 )

Use substitution to solve the following system of linear equations. 4 x + y = 1 4x+y=1 4 x + y = 1 x − y = 4 x-y=4 x − y = 4

everyone. So let's take a look at this problem here. We're gonna use the substitution method to solve the following system of equations. Remember, the whole idea with solving a system of equations is that you're trying to find values of x and y, that when you plug them back into these equations, they satisfy them, or you just get true statements for them. Alright? So how do we use the substitution method? We're just gonna walk through the steps here. The first thing we have to do is just choose the easiest equations to isolate x or y, or to label that as a. Normally, what happens is if you look through your equations, one of them is gonna have smaller coefficients, and that's usually gonna be the one that you can isolate for x or y. So in this case, what happens here is I'm gonna pick this equation to be a, and that means that by default, this equation is gonna be b. Alright? So that's the first thing. I basically just have to make a choice. It actually doesn't matter which one you choose. 1 is usually gonna be a little bit easier. Alright? So this is gonna be my equation a. The second thing I have to do is actually solve that equation for x or y. Again, it doesn't really matter here. Usually, one is gonna be easier than the other. So if I take a look here, I can try to solve either for x or y in this equation. If I try to solve for y, what happens is I'm gonna have to move the x over to the other side, and then this is gonna have a minus sign, and then the x is also gonna pick up a minus sign. There's going to be lots of minus signs there. Instead, what I can do is I can actually just try to solve for x instead because now what I can do is I can just add y to both sides and cancel this out, and then I'm gonna get fewer negative signs. If you notice what happens here is this equation really just becomes x equals y plus 4. Alright? So there's less negative signs, less chances for you to mess up and maybe make a mistake there. Alright? So this is equation a, x equals y+4, and that is my step number 2. I've solved the equation for x or y. Now we're going to take a look at step number 3. We're going to substitute the equation that we just solved into b, and then we're gonna solve for that new equation. So in other words, if I rewrite b, remember, b is just equal to 4x+yequals1. But now what happens is, remember, we've now come up with an expression for x. So remember, we're not gonna put x anymore. We're gonna have to replace this x with this expression. So this is just gonna be the expression y+4. And then I can just do plus y equals 1. Alright? So there's no more x's in this equation anymore. Now we're just gonna carry out the distribution because we have 4 times y+4. So this is gonna be 4 y+ixteen+y equals 1. Now we actually have to solve this equation here because you notice now there's only 1 variable. There's only just y. So I can group together the variables. This is gonna be 5 y+16equals1, and I can just subtract 16 from both sides like this. And what you'll get here is that 5 y is equal to negative 15. And if you just divide by 5 from both sides, what you'll see is that y is equal to negative 3. Alright? So now we've fully solved that equation, and we've gotten one of our, like, magic numbers here, y equals negative 3. Alright? So that was all of that was actually step number 3. Alright? So we substituted one equation into the other, and then we solved for that new equation. Remember, the whole idea here is that now we have one of our variables, y equals negative 3. That's only part of the puzzle. Sorry. That's already part of the numbers that will make this these equations true. Now we have to figure out what x is. And the way the way that we do that is we're gonna take this y equals negative 3, and we can plug it back into either one of our original equations, either a or b, and now we're gonna solve for x instead of y. Alright? Now, again, it doesn't really matter which one you plug it back into, but because it just was working with equation b, I'm gonna go ahead and sort of look at equation a. It's also just a little bit simpler. Right? So let's do that. So my equation a, remember, is just equal to is x plus y or sorry, is x equals y+4. But now I don't do y because now I actually have a number for y. Y is equal to negative 3. So this just becomes negative 3, negative 3 plus 4. Now what does this turn into? Well, this really just means x is equal to negative 3 plus 4, which is just 1. Alright? So that's my other answer. This is x equals 1. So I've gotten my two numbers now. I've gotten x equals 1 and y equals negative 3. Now if you may actually see them written this way with their x's and y's, but you may actually also see them written as a coordinate pair, this is really just the same thing as the coordinate 1 comma negative 3. So I just wanna let you know this is another way that you may see this. Alright? So this is really step number 4. That's what we did here. We plugged the value back into either equation, and then we solved for the other variable that we didn't know already. Right? So now I'm just done. The last thing we have to do here, and you don't really have to do this, but you can just go ahead and pause, and I heavily encourage you to do this, is to check your answer by plugging the values into both equations. Alright? So I'm just going to rewrite both of my equations over here. Remember, I've got b. And it could be the original equations, so I'm just gonna use the 4x+yequals1. And then for the a equation, it actually doesn't matter which one you use. You could use this one or you can use this one because they're really just the same thing. So we've got x equals y, or we can actually just use the original equation, x minus y equals 4. Alright? What we're trying to do here is we wanna check the points, 1 comma negative 3. So I'm just gonna check that real quick. And we can see here is that if you bring that to the down this equation and replace the x's and y's with their respective numbers, what you're gonna get is you're gonna get 4 times 1 plus negative 3 is equal to 1. And is that a true statement? Well, this just turns into 4 plus negative 3. So in other words, this is 4 minus 3, and that does in fact equal 1. So you just get 1 is equals 1, and that is a true statement. Alright? Now just do the same thing for the other equation. So I'm just gonna bring this blue equation down, and we're gonna do the same exact thing. Just replace the letters with numbers. And what this is saying is that 1 plus or sorry. 1 minus and then y is equal to negative 3 is equal to 4. Remember, you've got a double negative here, you already had a minus sign, so just be careful with that, but this is basically just gonna turn into 1 +3 equals 4, and 1+3 does, in fact, equal 4. So you just get true statements for both of them. This is really just a way to double check that you got the correct answer, especially if there's lots of, you know, different substitutions flying around. Alright? So we can see here that this is is indeed our correct solution. We've gone through all of the steps. This is how to solve any kind of substitution equation. Hopefully, this helps. Thanks for watching.

Use substitution to solve the following system of linear equation...

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Use substitution to solve the following system of linear equations.
$4x+y=1$
$x-y=4$