How to Solve Systems of Equations
Solving Systems of Equations
Hey guys! Welcome to this video over systems of equations.
A system of equations is a group of two or more equations, and each of the equations within the group have an unknown variable.
When given a system of equation, the goal is to find the value for each of the unknown variables.
In this video we will discuss two tools to help you solve for the unknown variables: substitution, and Elimination.
Substitution is a way of solving the system by getting rid of all of the variables, except for one, and then solving for that equation.
The best time to use substitution is when you have a variable that has a coefficient of 1 or negative one. The reason why is, because if it has a coefficient of one or negative one, then you don’t need to undo multiplication or division, you just need to undo addition or subtraction in order to isolate a variable.
There are three steps that you need to follow in order to be able to solve the system using substitution.
First, solve for x or y in one equation.
Second, plug the x or y that you solved for into the other equation and then solve.
Finally, use the number that you get when you solve to solve for the other variable or variables, depending on how many equations you have.
So, let’s get started.
In this example, we can see that our second equation has a variable with one as the coefficient. So, that let’s us know to solve for x. Now, you can actually solve for any of the variables… It will just always be easier to solve for one that has a coefficient of one or negative one.
Now, to do this we just add 9y to both sides, to get x= 9y -19. So, we’ve done what our first step tells us to do by solving for one of our variables. Step two now tells us to plug in the variable that we have solved for into our other equations. So, we are going to plug our x into 4x+3y = 2 and solve. Because we know that x is equal to 9y -19 we can substitute this in for the x in our first equation. We now have 4(9y- 19) +3y =2. Now that we have it down to one variable, we are able to solve for the value of that variable, so in this case y.
Let’s rewrite this by multiply our 4 by everything inside the parentheses. 36y – 76 + 3y = 2.
Now we can add 76 to both sides… you can do this multiple ways. If you wanted to add your y’s together first then do that. I’m just doing what is easier in my mind.
Once I add 76 to both sides and add my y’s together I get 39y= 78. Now, I divide both sides by 39 to get y is equal to 2.
So, that was step 2. Now, what does step 3 tell us to do? “Use the number that you get when you solve to solve for the other variable or variables.”
All we have to do here is take our y value, so 2, and plug it into either of our original equations to solve for x.
You can plug it into either the first our second, but I am going to plug it into the second equation.
So, x -9(2) = -19. Let me rewrite this as x – 18 = -19. Now we can add 18 to both sides to get x= -1… and we’re done! We have found the value of both of our variables using substitution.
Now, let’s take a look at how to solve for a system using elimination.
The reason this tool is called elimination, is because you add together the two equations in order to eliminate one of your variables.
Look at this example:
3x – 4y = -27
7x + 4y = 57
We can tell just by looking, that our y’s will cancel out. So, once we add the two equations together. We have 10x = 30, and we divide both sides by 10 to give us x is equal to 3.
Now, we can take our x value and plug it into either of our original equations. I’ll plug it into the first one here. So, we have 3(3) – 4y = -27, and I’ll rewrite this as 9 – 4y = -27.
To solve we need to move the -4y and the -27; and to do this will add a positive 4y to both sides, and a positive 27 to both sides giving us 9+27 which is 36 which is equal to 4y (36=4y).
So we divide both sides by 4, and that gives us y is equal to 9.
That was a good example to learn how elimination works, but it may not always be that our terms cancel so easily.
Like in this example:
9x – 3y = -57
2x + 6y = 34
None of our terms cancel right off the bat, so we will need to do a little manipulation in order to get them to do what we want them to do.
So, what could we do to get our terms to cancel? Well, there a couple of things, but the easiest, as it appears to me, is to multiply our first equations by 2. That will allow us to cancel our y’s, because we will be adding a negative six and a positive six. So, let’s try that.
We have (9x – 3y = -57) x 2 , so let’s rewrite this as 18x – 6y = -114. Now, we need to take this and add it to our other equations to cancel the y term.
18x – 6y = -114
2x + 6y = 34
20x = -80
Now, we need to divide both sides by 20 to give us x is equal to -4.
So, now that we know that x is equal to -4, we can plug -4 into where are x is in either one of the original equations.
I’ll plug it into the second one.
2(-4) + 6y = 34
I’ll simplify this as -8 + 6y =34, then add 8 to both sides to get 6y = 42, and finally divide by 6 to get y is equal to 7.
So, our final answer to this problem is x= -4, and y = 7.
I hope that this video on how to solve a system of equations using substitution and elimination has been helpful. For further help, be sure to subscribe to our channel below.
See you next time!