# Substitution Method Review | Expressões & Equações | Grau 8

Agora, when we’re using the substitution method, we notice that one of the equations that we are going to use is already solved for a variable, whether it’s y or it can be solved for x. So what we’ll do let’s go ahead and box this expression on the right hand, side of the equation and we’re going to use that in a little bit, because we are going to substitute negative x plus 5 into the second equation. So if i write the second equation of 2x plus y equals 11, i know wherever i see a y, i am going to plug in a negative x plus 5., so let’s go ahead and try it so 2x plus and wherever i see y i’m, going to Open up with parentheses, equals 11. and in those parentheses i am going to add or substitute a negative x plus 5. let’s go ahead and simplify. Now i have 2x and positive times. A negative is a negative x, mais 5 equals 11.. Now it looks like we’re going to have to combine like terms here a little bit and so 2x minus 1x is going to give me x, mais 5 equals 11.. So our next step is to isolate this variable x and when we isolate the variable x, we are going to add its inverse of a positive 5 with a negative 5 and what i do on one side of the equation. I’M going to have to complete.

On the second side, assim 5 minus 5 é 0, so i’m going to bring my x down and 11 minus 5 é 6., so our x value is 6. now something that i say is like goes from box to box because to find the y value. I am going to plug in this 6 into the negative x and then we will and add five and we will find what the y value is. So y equals a negative x plus five, wherever i see i’m going to put the negative. But wherever i see the x i’m going to open up with parentheses, now i’m not going to open up with parentheses and lose this negative, because this negative is important, então isso é. Why i brought it down then opened up with parentheses and then i’m going to write. The remainder of the problem, so here i’m, going to plug in 6 for the value of x, so y equals negative times. A positive six would be a negative six plus five, and so my y value is going to be a negative one. So the solution of this system is going to be the ordered pair of six negative one. Agora, once again, i just want to review – and this is important right here – is that when i have a negative x, i do not lose this negative sign. I need to bring that negative sign down and then plug in the value of x, which was six, and this is kind of like this is a positive six, so a negative times a positive still is a negative and then negative.

Six then negative. Six plus five is that negative 1., so let’s go ahead and try it again, try one more time. Let’S have another example: let’s go ahead and see here we have. Our first example is 4x. Minus 3y equals a negative 1, and then i also have 3x minus y is equal to a negative 2. now what’s interesting about this is neither one of them is in the slope, intercept, form or it’s solved for a a variable x or a variable y. So i’m going to have to isolate one of these variables and i’m going to choose this second problem or the second equation right here to isolate, and so what i’m going to do is i’m going to move this y i’m going to add a y to both Sides of the equation this gives 0y on this side, the left side and i have 3x equals y minus 2.. Now the y is not necessary isolated, but you see what i did is now. I have a positive y on the right hand side. So when i add a 2 to both sides now i can’t combine like terms because this is 3x and this is a positive 2. Agora, if this 2 had an x to it, i could combine like terms, but since it isn’t, i have to create the expression. Three x plus two is equal to y, because this negative, two and positive two is equal to zero and so it’s no longer there.

So when i change it around, i shouldn’t say change it around. So when i see the y is equal to three x plus two, so in this equation into the other equation, wherever i see a y let’s go ahead and write 4x minus 3., i see a y so i’m going to open up with parentheses and write. The remainder of the problem and so i’m going to plug it into the equation, which is 3x plus 2. vamos, go ahead and get rid of these parentheses. Simplify this equation and solve for the variable x, so i’m, going to bring down 4x don’t need to do anything with it. Yet i have a negative 3 that is going to be distributed to 3x, so negative, 3 times 3x is a negative 9x, a negative 3 vezes. A positive 2 is going to be a negative 6 and then i’ll write the right side of the equation down. Trying to keep my equal signs in line with each other to make it look as simple and organized as possible, so it looks like i need to combine these terms of 4x and a negative 9x, which gives me a negative 5x. Minus 6 equals a negative 1.. I need to add six to both sides, which gives me zero here. A negative five x is equal to a positive five. Agora, if i go ahead and divide both sides by a negative five, my x value is going to be a negative one.

So a positive and a negative is a negative, and five divided by five is one so my final solution, my final solution in this system let’s go ahead and put it right here, my x value being a negative one and my y value. I have no idea yet so what i’m going to do is plug it into right here bam, because i know that y equals three x plus two so three times a negative one plus two is equal to y so 3 times a negative. A negative 1 is a negative 3 mais 2 is equal to y, so y is going to equal a negative one as well. So my solution is my y value is negative. Um. My x value is a negative one, so the point of intersection is negative. One negative one all right guys go ahead and make sure you subscribe to our youtube channel hit a like. So the algorithm knows that these are the types of videos that you like to watch. Um hit your bell notifications. So you know the next time a mathis, simple video is produced and you can get extra assistance with your mouth work.