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### Course: Class 10 (Old) > Unit 3

Lesson 5: Linear equations word problems- Age word problem: Ben & William
- Age word problems
- System of equations word problem: walk & ride
- Systems of equations word problems
- Systems of equations with elimination: TV & DVD
- Forming equations with two variables
- Word problems involving pair of linear equations (advanced)

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# Systems of equations with elimination: TV & DVD

Sal solves a word problem about the weights of TVs and DVDs by creating a system of equations and solving it. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- 2:51Why do you have to multiply the equation by -1? Can't you just subtract the 2nd from the first equation? It's much easier that way than multiplying the 2nd equation by -1 and adding it to the first equation.

Subtracting is the same thing as adding something negative.(13 votes)- Yes that is completely true. You can definitely do that. But you see, the reason most people don't do this however is because it's very easy to forget which parts of the term should be negative or not; it's very easy to forget if a + or - sign should be assigned(it really depends on the person and how they do it). That's why people multiply by *-1* and then add, but you can definitely resort to just subtracting.(18 votes)

- sal says that 3t+2d is the same thing as 52, so, with that being said, can you divide 3t=52 and that giving you the answer for "t" and 52=2d giving you the answer for "d"?(4 votes)
- No. That doesn't work. You must have at least two equations to solve for two variables.(7 votes)

- anyone please do tell me the logic behind adding two equations to derive the unknown variables. is the logic similar to x + 1 =2 => x = 2 - 1 therefore, x=1 ?(4 votes)
- This is called collecting like terms and then isolating the variable x.

In your example we subtract both sides by 1 to isolate x on the other side alone.(5 votes)

- how would you solve the equation via substitution?

2x=5

x+y=7(4 votes)- What you can do is x=5/2 because you want the x by itself then sub into equation #2 like this

Since we know x=5/2 we can do 5/2+y=7. no subtract 5/2 from both sides which will be:y=7-5/2=4.5

5/2=2.5

you can also check:

2(2.5)=5

2.5+4.5=7(4 votes)

- solving the system of linear equation by the addition method

{x+2y=7

{-x+3y=18(1 vote) - When x = 3 and y = 5, by how much does the value of 3x2 – 2y exceed the value of 2x2 – 3y ? ineed answer and explanation plezzzzzz...(2 votes)
- All you have to do is plug in the giving x and y values into both equations and solve.

I assume the 2 is an exponent on 3x2.

3(3)^2-2(5) = 17

2(3)^2-3(5) = 3

So it exceeds the value by 14.(6 votes)

- 6.5 is what percent of 9(3 votes)
- 72.2(the 2s go on forever)(3 votes)

- how do I solve 3x+2y=-17

x-3y=9(3 votes)- To solve a system of equations by elimination, you have to manipulate one or both of the equations so that when you add them together, one of the variables drops out.

3x + 2y = -17

1x - 3y = 9

If you were to multiply both sides of the second equation by -3, you would get:

3x + 2y = -17

-3x + 9y = -27

When you add the two equations together, you get:

0x + 11y = -44

or simply:

11y = -44

Divide both sides by 11 and you get:

y = -44/11 = -4

Now, substitute the value for y into either equation and solve for x:

x -3(-4) = 9

x -(-12) = 9

x + 12 = 9

x = 9 - 12 = -3

Check your answer...

3(-3) + 2(-4) = -17

-9 + -8 = -17

-17 = -17

-3 - 3(-4) = 9

-3 + 12 = 9

9 = 9

Therefore: x=-3, and y=-4 is the solution to the system of equations.(4 votes)

- I am having trouble with this one : Given a word problem, how do you set it up in algebraic form and then solve it ?

Ten (10) years back, Jack’s father was four (4) times Jack’s age. But ten years from now, the father will be only 2 times Jack’s age. What are their current ages?(2 votes)- Always start with defining current ages, j for Jack and d for dad. So subtracting 10 gives f-10 = 4(j-10). Adding 10 years gives f+10 = 2(j+10). I assume you are okay with solving from here.(4 votes)

- do all systems o' equations have 2 variables(2 votes)
- No, there can be more. But for each variable, there must be an equation. For example, there's the x, y and z variable. Then there would be 3 equations to solve in order to know all three variables. Hope this helps!(2 votes)

## 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 3 televisions and 5 DVD players is 62.5 pounds, and the
weight of 3 televisions and 2 DVD players-- so they're
giving us different combinations-- is 52 pounds. Create a system of
equations that represents this situation. Then solve it to find out how
much each television and DVD player weighs. Well, the two pieces of
information they gave us in each of these statements can be
converted into an equation. The first one is is that the
weight of 3 televisions and 5 DVD players is 62.5 pounds. Then they told us that the
weight of 3 televisions and 2 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 3 times the weight
of a television, or 3 televisions, plus 5 times the
weight of a DVD player, is going to be equal
to 62.5 pounds. That's exactly what this first
statement is telling us. The second statement, the weight
of 3 televisions and 2 DVD players, so if I have 3
televisions and 2 DVD players, so the weight of 3 televisions
plus the weight of 2 DVD players, they're telling us
that that is 52 pounds. And so now we've set up the
system of equations. 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, 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 an equation, if
I add 5 to one side of an equation, I have to add 5 to
another side of the equation. So if I add this business to
this side of the equation, if I add this blue stuff 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 negative 1. And I want to multiply
it by negative 1. So negative 3t plus-- I could
write negative 2d 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
2d 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. 5d plus negative 2d is 3d. So you have a 3d is equal to
62.5 plus negative 52, or 62.5 minus 52 is 10.5. And now we can divide both sides
of this equation by 3. And you get d is equal
to 10.5 divided by 3. So let's figure out
what that is. 3 goes into 10.5-- it goes
into 10 three times. 3 times 3 is 9. Subtract. Get 1. Bring down the 5. Of course, you have your decimal
point right here. 3 goes into 15 five times. 5 times 3 is 15. You've got to subtract,
and you get a 0. So it goes exactly 3.5 times. So the weight of a DVD player--
that's what d represents-- 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 can just use that
top equation. So you get 3t 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-- needs
to be equal to 62.5. 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 62.5. Now we can subtract 17.5 from
both sides of this equation. 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 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.