Today we'll discuss the third method for solving systems of equations, and that method will be the elimination or addition method. Let's take a look at how this is going to work. Essentially what we're going to do is add these equations together. The reason we're allowed to do that is because if we take a look at the second equation, since this expression is equal to 7, we could add this expression to the left side of this equation and add 7 to the right side of this equation since they are equal. Now let's see what happens when we do this.
If we add the x terms together we'd have 5x. Now when we add the y terms together since they're opposites we get 0. So these terms are eliminated and on the right 8 plus 7 would give us 15. We have an equation with one variable we can solve that by dividing both sides by 5. We have x equal to 3. Remember one solution consists of an x and a y value. We know the x value is 3, but we still have to find the y value. So what we do now is we replace x with 3 in either equation and solve for y.
I'll go ahead and use the first equation for this purpose. So we know that x is 3, so we'd have 4 times 3 plus 3y is equal to 8. So we'll simplify and solve. Minus 12 on both sides.
3y is equal to negative 4. Divide both sides by 3. y is equal to negative 4 thirds. So we have one solution for this system. Therefore, the system is consistent and independent.
You can see this method works very nicely. It's a lot faster than trying to graph these two linear equations and estimate a point of intersection. Let's take a look at, in general, how we're going to solve a system using the elimination method. Step 1. We'll multiply one or both of the equations by the appropriate numbers so that one of the variables will have the same coefficient with opposite signs. Then we'll add the two equations together.
Because one of the variables had the same coefficient with the opposite sign, it will be eliminated when added to the other equation. Then we'll solve the resulting equation. And then substitute this answer back into one of the original equations to find the remaining variable. So the key here, if we want to solve this using elimination, either the x terms or the y terms must be opposites. Since the x terms are already opposite signs, I'm going to go ahead and try to eliminate the x terms.
Now the least common multiple of 2 and 3 would be 6. So I'm going to convert this to a positive 6x and this to a negative 6x. The way I'm going to do that is I'm going to take this first equation and multiply by a positive 2. Now multiply the second equation by a positive 3. When I do this, I do have to multiply the entire equation by 2 to maintain the equality. So we'd have 6x plus 10y equal to 8. Here we'd have negative 6x plus 9y equal to 30. Now we'll add the two equations together. That's sometimes why it's called the addition method. Of course when we add a positive 6 and a negative 6, that would be 0. So the result would be 19y is equal to 38. Divided by 19, we have a y value equal to positive 2. We're going to take this value, sub it into one of these equations, and then solve for x.
I'll go ahead and just use the first equation. We have 3x plus 5y, but we now know y is 2, so that would be a 10. So we have 3x plus 5y is equal to 4. Subtracting 10 on both sides, dividing by 3, we have x is equal to negative 2. Therefore, our solution to this system would be x equals negative 2, y equals positive 2. So we can classify the system as consistent and independent. Let's take a look at another. This time, I'll try to eliminate the y terms. Since we have a minus 3y and a positive 6y, If we can make this minus 6y, they'd be opposites.
So in this case, we only have to multiply the first equation by positive 2. We can leave the second equation exactly the same. So the result would be 4x minus 6y is equal to negative 2. Second equation remains the same. Now when we try to add these two equations together, notice how the x terms and the y terms are opposites. So on the left side of this equation, we'd have 0 is equal to positive 3. Now if you recall from the previous video, algebraically if this statement is false, which it is, that's an indication that this system has no solution, which also means this system is inconsistent and independent.
So you can see the elimination is a nice method. It's much faster than trying to graph these linear equations. Let's go ahead and take a look at an application.
Sunset Phone offers two long distance plans. Plan A is $5 per month and 8 cents a minute. Plan B has no monthly fee but charges 12 cents a minute. For what number of minutes will the two plans cost the same?
So let's let x equal the number of minutes. So if we let C equal the total cost for the month for plan A, we have a fixed cost of $5 and a variable cost of 8 cents per minute. So to find the total cost for the month, we'd have to take the cost per minute, multiply it by the number of minutes, which is x, and then add the fixed cost of $5.
For plan B, the total cost for the month is a little bit simpler. it's just 12 cents per minute, which we'd represent as 0.12 times x. Again, there's no fixed cost for plan B.
Now the question is, for what number of minutes will these two plans be the same? So we want to know when these two c's will be equal, and that will occur when .08x plus five is equal to 0.12x. So what we're really doing is we're performing substitution. We'll take this expression and sub it in for the c or vice versa. The result would be 0.08x plus 5.00 is equal to 0.12x.
Now if we don't like to work with decimals, we can multiply this entire equation by 100. Remember multiplying by 100 moves the decimal to the right two places. So we could solve the equation 8x plus 500 is equal to 12x. Then we would subtract 8x on both sides to give us 4x. Divide by 4, x is equal to 125. Therefore these plans would cost the same when 125 minutes are used for the same month.
I hope that helps explain how to use the elimination method. Thank you for watching.