Class 10 Exam > Class 10 Videos > Elimination Method Part 1
Elimination Method Part 1 Video Lecture - Class 10
FAQs on Elimination Method Part 1 Video Lecture - Class 10
| 1. What is the elimination method in algebra? |  |
Ans. The elimination method, also known as the addition-subtraction method, is a technique used in algebra to solve a system of linear equations. In this method, we eliminate one variable by adding or subtracting the equations together, resulting in a new equation with only one variable. This new equation can then be solved to find the value of the variable, which can be substituted back into the original equations to find the values of the other variables.
| 2. When is the elimination method useful in solving equations? | |
Ans. The elimination method is useful when we have a system of linear equations with the same variables and we want to find their common solution. It is particularly effective when the coefficients of one of the variables in both equations are additive inverses (i.e., one is positive and the other is negative). By adding or subtracting the equations, we can eliminate one variable and solve for the other variable.
| 3. Can the elimination method be used to solve non-linear equations? | |
Ans. No, the elimination method is specifically designed to solve systems of linear equations. Non-linear equations involve variables raised to a power, such as squared or cubed variables. The elimination method relies on the linearity of the equations, which means that the variables are raised only to the power of 1. For non-linear equations, other methods like substitution or graphing are typically used.
| 4. What are the limitations of the elimination method? | |
Ans. The elimination method has a few limitations. First, it can only be applied to systems of linear equations. If the equations are non-linear, the elimination method cannot be used. Second, the elimination method may not work if the coefficients of one of the variables in both equations are not additive inverses. In such cases, other methods like substitution or graphing may be more suitable. Lastly, the elimination method can become more time-consuming and complex when dealing with systems of equations with three or more variables.
| 5. Can the elimination method be used to solve equations with fractions? | |
Ans. Yes, the elimination method can be used to solve equations with fractions. To do this, we need to eliminate the denominators by multiplying each equation by the least common multiple (LCM) of the denominators. This will result in equations with whole numbers, which can then be solved using the elimination method. After finding the values of the variables, we can simplify the fractions if necessary.
Text Transcript from Video
Let's explore a few
more methods for
solving systems of equations.
Let's say I have the equation,
3x plus 4y is equal to 2.5.
And I have another equation, 5x
minus 4y is equal to 25.5.
And we want to find an x and y
value that satisfies both of
these equations.
If you think of it graphically,
this would be the
intersection of the lines that
represent the solution sets to
both of these equations.
So how can we proceed?
We saw in substitution,
we like to
eliminate one of the variables.
We did it through substitution
last time.
But is there anything we can add
or subtract-- let's focus
on this yellow, on this top
equation right here-- is there
anything that we can add
or subtract to both
sides of this equation?
Remember, any time you deal with
an equation you have to
add or subtract the same
thing to both sides.
But is there anything that we
could add or subtract to both
sides of this equation
that might
eliminate one of the variables?
And then we would have one
equation in one variable, and
we can solve for it.
And it's probably not obvious,
even though it's sitting right
in front of your face.
Well, what if we
just added this
equation to that equation?
What I mean by that is, what if
we were to add 5x minus 4y
to the left-hand side, and add
25.5 to the right-hand side?
So if I were to literally add
this to the left-hand side,
and add that to the
right-hand side.
And you're probably saying, Sal,
hold on, how can you just
add two equations like that?
And remember, when you're doing
any equation, if I have
any equation of the form-- well,
really, any equation--
Ax plus By is equal to C, if I
want to do something to this
equation, I just have to add the
same thing to both sides
of the equation.
So I could, for example, I could
add D to both sides of
the equation.
Because D is equal to
D, so I won't be
changing the equation.
You would get Ax plus By, plus
D is equal to C plus D.
And we've seen that multiple,
multiple times.
Anything you do to one side of
the equation, you have to do
to the other side.
But you're saying, hey, Sal,
wait, on the left-hand side,
you're adding 5x minus
4y to the equation.
On the right-hand side, you're
adding 25.5 to the equation.
Aren't you adding two different
things to both sides
of the equation?
And my answer would be no.
We know that 5x minus
4y is 25.5.
This quantity and this quantity
are the same.
They're both 25.5.
This second equation is telling
me that explicitly.
So I can add this to
the left-hand side.
I'm essentially adding
25.5 to it.
And I could add 25.5 to
the right-hand side.
So let's do that.
If we were to add the
left-hand side,
3x plus 5x is 8x.
And then what is 4y minus 4y?
And this was the whole point.
When I looked at these two
equations, I said, oh, I have
a 4y, I have a negative 4y.
If you just add these two
together, they are going to
cancel out.
They're going to be plus 0y.
Or that whole term is just
going to go away.
And that's going to be equal
to 2.5 plus 25.5 is 28.
So you divide both sides.
So you get 8x is equal to 28.
And you divide both sides by 8,
and we get x is equal to 28
over 8, or you divide the
numerator and the
denominator by 4.
That's equal to 7 over 2.
That's our x value.
Now we want to solve
for our y value.
And we could substitute
this back into either
of these two equations.
Let's use the top one.
You could do it with the
bottom one as well.
So we know that 3 times x, 3
times 7 over 2-- I'm just
substituting the x value we
figured out into this top
equation-- 3 times 7 over 2,
plus 4y is equal to 2.5.
Let me just write that as 5/2.
We're going to stay in
the fraction world.
So this is going to be 21 over
2 plus 4y is equal to 5/2.
Subtract 21 over 2
from both sides.
So minus 21 over 2,
minus 21 over 2.
The left-hand side-- you're just
left with a 4y, because
these two guys cancel out--
is equal to-- this is 5
minus 21 over 2.
That's negative 16 over 2.
So that's negative 16 over 2,
which is the same thing--
well, I'll write it out
as negative 16 over 2.
Or we could write that-- let's
continue up here-- 4y-- I'm
just continuing this train of
thought up here-- 4y is equal
to negative 8.
Divide both sides by
4, and you get y is
equal to negative 2.
So the solution to this equation
is x is equal to 7/2,
y is equal to negative 2.
This would be the coordinate
of their intersection.
And you could try it out
on both of these
equations right here.
So let's verify that it also
satisfies this bottom equation.
5 times 7/2 is 35 over 2 minus
4 times negative 2, so minus
negative 8.
That's equivalent to-- let's
see, this is 17.5 plus 8.
And that indeed does
equal 25.5.
So this satisfies
both equations.
Now let's see if we can use
our newly found skills to
tackle a word problem, our
newly found skills in
elimination.
So here it says, Nadia and Peter
visit the candy store.
Nadia buys 3 candy bars and 4
Fruit Roll-Ups for $2.84.
Peter also buys 3 candy bars,
but can only afford 1
additional Fruit Roll-Up.
His purchase costs $1.79.
What is the cost of each candy
bar and each Fruit Roll-Up?
So let's define some
variables.
Let's just use x and y.
Let's let x equal cost of candy
bar-- I was going to do
a c and a f for Fruit Roll-Up,
but I'll just stick with x and
y-- cost of candy bar.
And let y equal the cost
of a Fruit Roll-Up.
All right.
So what does this first
statement tell us?
Nadia buys 3 candy bars, so the
cost of 3 candy bars is
going to be 3x.
And 4 Fruit Roll-Ups.
Plus 4 times y, the cost
of a Fruit Roll-Up.
This is how much Nadia spends.
3 candy bars, 4 Fruit
Roll-Ups.
And it's going to cost $2.84.
That's what this first
statement tells us.
It translates into
that equation.
The second statement.
Peter also buys 3 candy bars,
but could only afford 1
additional Fruit Roll-Up.
So plus 1 additional
Fruit Roll-Up.
His purchase cost is
equal to $1.79.
What is the cost of each candy
bar and each Fruit Roll-Up?
And we're going to solve
this using elimination.
You could solve this using any
of the techniques we've seen
so far-- substitution,
elimination, even graphing,
although it's kind of
hard to eyeball
things with the graphing.
So how can we do this?
Remember, with elimination,
you're going to add-- let's
focus on this top equation
right here.
Is there something we could
add to both sides of this
equation that'll help us
eliminate one of the variables?
Or let me put it this way, is
there something we could add
or subtract to both sides of
this equation that will help
us eliminate one of
the variables?
Well, like in the problem we
did a little bit earlier in
the video, what if we were to
subtract this equation, or
what if we were to subtract 3x
plus y from 3x plus 4y on the
left-hand side, and subtract
$1.79 from
the right-hand side?
And remember, by doing that, I
would be subtracting the same
thing from both sides
of the equation.
This is $1.79.
How do I know?
Because it says this
is equal to $1.79.
So if we did that we would be
subtracting the same thing
from both sides of
the equation.
So let's subtract 3x
plus y from the
left-hand side of the equation.
And let me just do this
over on the right.
If I subtract 3x plus y, that is
the same thing as negative
3x minus y, if you just
distribute the negative sign.
So let's subtract it.
So you get negative 3x minus
y-- maybe I should make it
very clear this is not a plus
sign; you could imagine I'm
multiplying the second equation
by negative 1-- is
equal to negative $1.79.
I'm just taking the
second equation.
You could imagine I'm
multiplying it by negative 1,
and now I'm going to add the
left-hand side to the
left-hand side of this equation,
and the right-hand
side to the right-hand side
of that equation.
And what do we get?
When you add 3x plus 4y,
minus 3x, minus y,
the 3x's cancel out.
3x minus 3x is 0x.
I won't even write it down.
You get 4x minus-- sorry,
4y minus y.
That is 3y.
And that is going to be equal
to $2.84 minus $1.79.
What is that?
That's $1.05.
So 3y is equal to $1.05.
Divide both sides by 3.
y is equal to-- what's
$1.05 divided by 3?
So 3 goes into $1.05.
It goes into 1 zero times.
0 times 3 is 0.
1 minus 0 is 1.
Bring down a 0.
3 goes into 10 three times.
3 times 3 is 9.
Subtract.
10 minus 9 is 1.
Bring down the 5.
3 goes into 15 five times.
5 times 3 is 15.
Subtract.
We have no remainder.
So y is equal to $0.35.
So the cost of a Fruit
Roll-Up is $0.35.
Now we can substitute back into
either of these equations
to figure out the cost
of a candy bar.
So let's use this bottom
equation right here.
Which was originally, if you
remember before I multiplied
it by negative 1, it was 3x
plus y is equal to $1.79.
So that means that 3x plus the
cost of a Fruit Roll-Up, 0.35
is equal to $1.79.
If we subtract 0.35 from both
sides, what do we get?
The left-hand side-- you're just
left with the 3x; these
cancel out-- is equal
to-- let's see, this
is $1.79 minus $0.35.
That's $1.44.
And 3 goes into $1.44, I think
it goes-- well, 3 goes into
$1.44, it goes into
1 zero times.
1 times 3 is 0.
Bring down the 1.
Subtract.
Bring down the 4.
3 goes into 14 four times.
4 times 3 is 12.
I'm making this messy.
14 minus 12 is 2.
Bring down the 4.
3 goes into 24 eight times.
8 times 3 is 24.
No remainder.
So x is equal to 0.48.
So there you have it.
We figured out, using
elimination, that the cost of
a candy bar is equal to $0.48,
and that the cost of a Fruit
Roll-Up is equal to $0.35.
more methods for
solving systems of equations.
Let's say I have the equation,
3x plus 4y is equal to 2.5.
And I have another equation, 5x
minus 4y is equal to 25.5.
And we want to find an x and y
value that satisfies both of
these equations.
If you think of it graphically,
this would be the
intersection of the lines that
represent the solution sets to
both of these equations.
So how can we proceed?
We saw in substitution,
we like to
eliminate one of the variables.
We did it through substitution
last time.
But is there anything we can add
or subtract-- let's focus
on this yellow, on this top
equation right here-- is there
anything that we can add
or subtract to both
sides of this equation?
Remember, any time you deal with
an equation you have to
add or subtract the same
thing to both sides.
But is there anything that we
could add or subtract to both
sides of this equation
that might
eliminate one of the variables?
And then we would have one
equation in one variable, and
we can solve for it.
And it's probably not obvious,
even though it's sitting right
in front of your face.
Well, what if we
just added this
equation to that equation?
What I mean by that is, what if
we were to add 5x minus 4y
to the left-hand side, and add
25.5 to the right-hand side?
So if I were to literally add
this to the left-hand side,
and add that to the
right-hand side.
And you're probably saying, Sal,
hold on, how can you just
add two equations like that?
And remember, when you're doing
any equation, if I have
any equation of the form-- well,
really, any equation--
Ax plus By is equal to C, if I
want to do something to this
equation, I just have to add the
same thing to both sides
of the equation.
So I could, for example, I could
add D to both sides of
the equation.
Because D is equal to
D, so I won't be
changing the equation.
You would get Ax plus By, plus
D is equal to C plus D.
And we've seen that multiple,
multiple times.
Anything you do to one side of
the equation, you have to do
to the other side.
But you're saying, hey, Sal,
wait, on the left-hand side,
you're adding 5x minus
4y to the equation.
On the right-hand side, you're
adding 25.5 to the equation.
Aren't you adding two different
things to both sides
of the equation?
And my answer would be no.
We know that 5x minus
4y is 25.5.
This quantity and this quantity
are the same.
They're both 25.5.
This second equation is telling
me that explicitly.
So I can add this to
the left-hand side.
I'm essentially adding
25.5 to it.
And I could add 25.5 to
the right-hand side.
So let's do that.
If we were to add the
left-hand side,
3x plus 5x is 8x.
And then what is 4y minus 4y?
And this was the whole point.
When I looked at these two
equations, I said, oh, I have
a 4y, I have a negative 4y.
If you just add these two
together, they are going to
cancel out.
They're going to be plus 0y.
Or that whole term is just
going to go away.
And that's going to be equal
to 2.5 plus 25.5 is 28.
So you divide both sides.
So you get 8x is equal to 28.
And you divide both sides by 8,
and we get x is equal to 28
over 8, or you divide the
numerator and the
denominator by 4.
That's equal to 7 over 2.
That's our x value.
Now we want to solve
for our y value.
And we could substitute
this back into either
of these two equations.
Let's use the top one.
You could do it with the
bottom one as well.
So we know that 3 times x, 3
times 7 over 2-- I'm just
substituting the x value we
figured out into this top
equation-- 3 times 7 over 2,
plus 4y is equal to 2.5.
Let me just write that as 5/2.
We're going to stay in
the fraction world.
So this is going to be 21 over
2 plus 4y is equal to 5/2.
Subtract 21 over 2
from both sides.
So minus 21 over 2,
minus 21 over 2.
The left-hand side-- you're just
left with a 4y, because
these two guys cancel out--
is equal to-- this is 5
minus 21 over 2.
That's negative 16 over 2.
So that's negative 16 over 2,
which is the same thing--
well, I'll write it out
as negative 16 over 2.
Or we could write that-- let's
continue up here-- 4y-- I'm
just continuing this train of
thought up here-- 4y is equal
to negative 8.
Divide both sides by
4, and you get y is
equal to negative 2.
So the solution to this equation
is x is equal to 7/2,
y is equal to negative 2.
This would be the coordinate
of their intersection.
And you could try it out
on both of these
equations right here.
So let's verify that it also
satisfies this bottom equation.
5 times 7/2 is 35 over 2 minus
4 times negative 2, so minus
negative 8.
That's equivalent to-- let's
see, this is 17.5 plus 8.
And that indeed does
equal 25.5.
So this satisfies
both equations.
Now let's see if we can use
our newly found skills to
tackle a word problem, our
newly found skills in
elimination.
So here it says, Nadia and Peter
visit the candy store.
Nadia buys 3 candy bars and 4
Fruit Roll-Ups for $2.84.
Peter also buys 3 candy bars,
but can only afford 1
additional Fruit Roll-Up.
His purchase costs $1.79.
What is the cost of each candy
bar and each Fruit Roll-Up?
So let's define some
variables.
Let's just use x and y.
Let's let x equal cost of candy
bar-- I was going to do
a c and a f for Fruit Roll-Up,
but I'll just stick with x and
y-- cost of candy bar.
And let y equal the cost
of a Fruit Roll-Up.
All right.
So what does this first
statement tell us?
Nadia buys 3 candy bars, so the
cost of 3 candy bars is
going to be 3x.
And 4 Fruit Roll-Ups.
Plus 4 times y, the cost
of a Fruit Roll-Up.
This is how much Nadia spends.
3 candy bars, 4 Fruit
Roll-Ups.
And it's going to cost $2.84.
That's what this first
statement tells us.
It translates into
that equation.
The second statement.
Peter also buys 3 candy bars,
but could only afford 1
additional Fruit Roll-Up.
So plus 1 additional
Fruit Roll-Up.
His purchase cost is
equal to $1.79.
What is the cost of each candy
bar and each Fruit Roll-Up?
And we're going to solve
this using elimination.
You could solve this using any
of the techniques we've seen
so far-- substitution,
elimination, even graphing,
although it's kind of
hard to eyeball
things with the graphing.
So how can we do this?
Remember, with elimination,
you're going to add-- let's
focus on this top equation
right here.
Is there something we could
add to both sides of this
equation that'll help us
eliminate one of the variables?
Or let me put it this way, is
there something we could add
or subtract to both sides of
this equation that will help
us eliminate one of
the variables?
Well, like in the problem we
did a little bit earlier in
the video, what if we were to
subtract this equation, or
what if we were to subtract 3x
plus y from 3x plus 4y on the
left-hand side, and subtract
$1.79 from
the right-hand side?
And remember, by doing that, I
would be subtracting the same
thing from both sides
of the equation.
This is $1.79.
How do I know?
Because it says this
is equal to $1.79.
So if we did that we would be
subtracting the same thing
from both sides of
the equation.
So let's subtract 3x
plus y from the
left-hand side of the equation.
And let me just do this
over on the right.
If I subtract 3x plus y, that is
the same thing as negative
3x minus y, if you just
distribute the negative sign.
So let's subtract it.
So you get negative 3x minus
y-- maybe I should make it
very clear this is not a plus
sign; you could imagine I'm
multiplying the second equation
by negative 1-- is
equal to negative $1.79.
I'm just taking the
second equation.
You could imagine I'm
multiplying it by negative 1,
and now I'm going to add the
left-hand side to the
left-hand side of this equation,
and the right-hand
side to the right-hand side
of that equation.
And what do we get?
When you add 3x plus 4y,
minus 3x, minus y,
the 3x's cancel out.
3x minus 3x is 0x.
I won't even write it down.
You get 4x minus-- sorry,
4y minus y.
That is 3y.
And that is going to be equal
to $2.84 minus $1.79.
What is that?
That's $1.05.
So 3y is equal to $1.05.
Divide both sides by 3.
y is equal to-- what's
$1.05 divided by 3?
So 3 goes into $1.05.
It goes into 1 zero times.
0 times 3 is 0.
1 minus 0 is 1.
Bring down a 0.
3 goes into 10 three times.
3 times 3 is 9.
Subtract.
10 minus 9 is 1.
Bring down the 5.
3 goes into 15 five times.
5 times 3 is 15.
Subtract.
We have no remainder.
So y is equal to $0.35.
So the cost of a Fruit
Roll-Up is $0.35.
Now we can substitute back into
either of these equations
to figure out the cost
of a candy bar.
So let's use this bottom
equation right here.
Which was originally, if you
remember before I multiplied
it by negative 1, it was 3x
plus y is equal to $1.79.
So that means that 3x plus the
cost of a Fruit Roll-Up, 0.35
is equal to $1.79.
If we subtract 0.35 from both
sides, what do we get?
The left-hand side-- you're just
left with the 3x; these
cancel out-- is equal
to-- let's see, this
is $1.79 minus $0.35.
That's $1.44.
And 3 goes into $1.44, I think
it goes-- well, 3 goes into
$1.44, it goes into
1 zero times.
1 times 3 is 0.
Bring down the 1.
Subtract.
Bring down the 4.
3 goes into 14 four times.
4 times 3 is 12.
I'm making this messy.
14 minus 12 is 2.
Bring down the 4.
3 goes into 24 eight times.
8 times 3 is 24.
No remainder.
So x is equal to 0.48.
So there you have it.
We figured out, using
elimination, that the cost of
a candy bar is equal to $0.48,
and that the cost of a Fruit
Roll-Up is equal to $0.35.
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