If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Systems of equations with elimination: TV & DVD

CCSS.Math:

## Video transcript

an electronics warehouse ships televisions and DVD players in certain combinations to retailers throughout the country they tell us that the weight of three televisions and five DVD players is sixty two point five pounds and the weight of three televisions and two DVD players are they're giving us different combinations is 52 pounds create a system of equations that represents the situation then solve it to find out how much each television and DVD player weighs well the two pieces of information they gave us and each of these statements can be converted into an equation the first one is is that the weight of three televisions weight of three televisions and five DVD players is sixty two point five pounds then they told us that the weight of three televisions and two DVD players is 52 pounds so we can translate these directly into equations if we let T to be the weight of a television and D to be the weight of a DVD player this first statement up here says that three times the weight of a television or three televisions plus five times the weight of a DVD player is going to be equal to sixty two point five pounds that's exactly what this first statement is telling us the second statement the weight of three televisions in 2vd and two DVD players so if I have three televisions and two DVD players so the weight of three televisions plus the weight of two DVD players they're telling us that that is 52 pounds and so now we've set up the system of equation we've done the first part to create a system that represents the situation now we need to solve it now one thing that's especially tempting when you have two systems and both of them have something where it's you know you have a 3t here and you have a 3t here what we can do is we can multiply one of the systems by some factor so that if we were to add this equation to that equation we would get one of the terms to cancel out and that's what we're going to do right here and you can do this you can do this business of adding equations to each other because remember when we learned this at the beginning of algebra anything you do to one side of equation if I add five to one side of an equation I have to add five to another side of so if I add this business to this side of the equation if I had this blue stuff to this to the left side of the equation I can add this 52 to the right-hand side because this is saying that 52 is the same thing as this thing over here this thing is also 52 so if we're adding this to the left-hand side we're actually adding 52 to it we're just writing it a different way now before we do that what I want to do is multiply the second blue equation by a negative one and I want to multiply it by negative one so negative 3t plus I could write negative 2 D is equal to negative 52 so I haven't changed the information in this equation I just multiplied everything by negative 1 the reason why I did that is because now if I add these two equations these 3t terms are going to cancel out so let's do that let's add these two equations and remember all we're doing is we're adding the same thing to both sides of this top equation we're adding essentially negative 52 now now that we've multiplied everything by a negative 1 this negative 3t plus negative 2 D is the same thing as negative 52 so let's add this left-hand side over here the 3t and the negative 3t will cancel out that was the whole point 5 D plus negative a 2 D is 3 D so you have a 3 D is equal to 62 point 5 plus negative 52 or 62 point 5 minus 52 is 10 point five is ten point five and now we can divide both sides of this equation by 3 divide both sides by 3 and you get D is equal to ten point five divided by three so let's figure out what that is 3 goes into ten point five it goes into 10 three times three times three is nine subtract get 1 bring down the 5 of course you have your decimal point right here 3 goes into 15 5 times 5 times 3 is 15 you got to subtract and you get a zero so it goes exactly three point five times so the weight of a DVD player that's what dear ents is 3.5 pounds now we can substitute back into one of these equations up here to figure out the weight of a television we could just use that top equation so you get 3 T 3 T plus 5 times the weight of a DVD player which we just figured out is 3.5 remember we're just looking for values that satisfy both of these equations so 5 times 3.5 need to be equal to 60 2 point 5 needs to equal 60 2.5 so you get 3t plus what is this going to be this is going to be 15 plus 2.5 right 5 times 0.5 is 2.5 5 times 3 is 15 so it's 17.5 is equal to 62 point 5 now we can subtract 17.5 from both sides of this equation subtract 17.5 subtracting 17.5 and what do we get the left-hand side is just going to be 3t this cancels out that was the whole point of it 3t is going to be equal to let's see the 0.5 minus 0.5 that cancels out so this is the same thing as 62 minus 17 62 minus 7 is would be 55 and so we're going to subtract another 10 so it's going to be 45 so this is going to be equal to 45 now you can divide both sides of this equation by 3 and we get T is equal to 15 so we've solved our system the weight of a DVD player is 3.5 pounds and the weight of a television is 15 pounds and we're done