Class 10 Exam > Class 10 Videos > Substitution Method - Pair of Linear Equations in Two Variables
Substitution Method - Pair of Linear Equations in Two Variables Video Lecture - Class 10
FAQs on Substitution Method - Pair of Linear Equations in Two Variables Video Lecture - Class 10
| 1. What is the substitution method in solving a pair of linear equations in two variables? |  |
Ans. The substitution method is a technique used to solve a pair of linear equations in two variables. In this method, we solve one equation for one variable and substitute that value into the other equation. By doing this, we can find the values of both variables and determine the solution to the system of equations.
| 2. How do you solve a pair of linear equations using the substitution method? | |
Ans. To solve a pair of linear equations using the substitution method, follow these steps:
1. Solve one equation for one variable in terms of the other variable.
2. Substitute the expression obtained into the other equation.
3. Solve the resulting equation to find the value of the variable.
4. Substitute this value back into any of the original equations to find the value of the other variable.
5. Verify the solution by substituting the obtained values into both equations.
| 3. Can the substitution method be used for any pair of linear equations? | |
Ans. Yes, the substitution method can be used for any pair of linear equations. It is a general method that can be applied to any system of linear equations to find their solution. However, it may not always be the most efficient method, especially when dealing with complex equations or large systems.
| 4. What are the advantages of using the substitution method in solving linear equations? | |
Ans. The substitution method has several advantages when solving linear equations:
1. It is a straightforward and systematic method.
2. It can be used for any pair of linear equations.
3. It always guarantees a solution, either a unique solution or infinitely many solutions.
4. It allows for easy verification of the solution by substituting the values back into the original equations.
| 5. Are there any limitations or drawbacks to using the substitution method? | |
Ans. While the substitution method is a useful technique, it does have some limitations:
1. It may involve complex algebraic manipulations, especially when dealing with equations involving fractions or radicals.
2. It may not be the most efficient method for large systems of equations, as it requires multiple substitutions.
3. It relies on the ability to solve equations for a single variable, which may not always be possible or straightforward.
4. It may not provide an intuitive understanding of the geometric interpretation of the solution in terms of lines or planes in coordinate space.
Text Transcript from Video
In the last video, we saw what
a system of equations is.
And in this video, I'm going
to show you one algebraic
technique for solving systems of
equations, where you don't
have to graph the two lines and
try to figure out exactly
where they intersect.
This will give you an exact
algebraic answer.
And in future videos,
we'll see more
methods of doing this.
So let's say you had
two equations.
One is x plus 2y is equal to 9,
and the other equation is
3x plus 5y is equal to 20.
Now, if we did what we did in
the last video, we could graph
each of these.
These are lines.
You could put them in either
slope-intercept form or
point-slope form.
They're in standard
form right now.
And then you could graph each
of these lines, figure out
where they intersect, and that
would be a solution to that.
But it's sometimes hard to
find, to just by looking,
figure out exactly where
they intersect.
So let's figure out a way to
algebraically do this.
And what I'm going to do is
the substitution method.
I'm going to use one of the
equations to solve for one of
the variables, and then I'm
going to substitute back in
for that variable over here.
So let me show you what
I'm talking about.
So let me solve for x using
this top equation.
So the top equation says x
plus 2y is equal to 9.
I want to solve for x, so let's
subtract 2y from both
sides of this equation.
So I'm left with x is
equal to 9 minus 2y.
This is what this first equation
is telling me.
I just rearranged
it a little bit.
The first equation
is saying that.
So in order to satisfy both of
these equations, x has to
satisfy this constraint
right here.
So I can substitute this
back in for x.
We're saying, this top equation
says, x has to be
equal to this.
Well, if x has to be equal
to that, let's
substitute this in for x.
So this second equation
will become 3 times x.
And instead of an x, I'll write
this thing, 9 minus 2y.
3 times 9 minus 2y, plus
5y is equal to 20.
That's why it's called the
substitution method.
I just substituted for x.
And the reason why that's
useful is now I have one
equation with one unknown,
and I can solve for y.
So let's do that
3 times 9 is 27.
3 times negative 2 is negative
6y, plus 5y is equal to 20.
Add the negative 6y plus the 5y,
add those two terms. You
have 27-- let's see, this will
be-- minus y is equal to 20.
Let's subtract 27
from both sides.
And you get-- let me
write it out here.
So let's subtract 27
from both sides.
The left-hand side, the 27's
cancel each other out.
And you're left with negative y
is equal to 20 minus 27, is
negative 7.
And then we can multiply both
sides of this equation by
negative 1, and we get
y is equal to 7.
So we found the y value of the
point of intersection of these
two lines. y is equal to 7.
Let me write over here, so I
don't have to keep scrolling
down and back up.
y is equal to 7.
Well, if we know y, we can
now solve for x. x is
equal to 9 minus 2y.
So let's do that.
x is equal to 9 minus 2
times y, 2 times 7.
Or x is equal to 9 minus 14, or
x is equal to negative 5.
So we've just, using
substitution, we've been able
to find a pair of x
and y points that
satisfy these equations.
The point x is equal to negative
5, y is equal to 7,
satisfy both of these.
And you can try it out.
Negative 5 plus 2 times 7,
that's negative 5 plus 14,
that is indeed 9.
You do this equation.
3 times negative 5 is negative
15, plus 5 times y,
plus 5 times 7.
So negative 15 plus
35 is indeed 20.
So this satisfies
both equations.
If you were to graph both of
these equations, they would
intersect at the point
negative 5 comma 7.
Now let's use our newly
found skill to do
an actual word problem.
Let's say that they tell
us that the sum of
two numbers is 70.
And they differ-- or maybe we
could say their difference--
they differ by 11.
What are the numbers?
So let's do this word problem.
So let's define some
variables.
Let's let x be the larger
number, and let y be the
smaller number.
I'm just arbitrarily creating
these variables.
One of them is larger
than the other.
They differ by 11.
Now, this first statement, the
sum of the two numbers is 70.
That tells us that x plus
y must be equal to 70.
That second statement, that
they differ by 11.
That means the larger number
minus the smaller
number must be 11.
That tells us that x minus
y must be equal to 11.
So there we have it.
We have two equations
and two unknowns.
We have a system of
two equations.
We can now solve it using
the substitution method.
So let's solve for x on this
equation right here.
So if you add y to both
sides of this
equation, what do you get?
On the left-hand side, you just
get an x, because these
cancel out.
And then on the right-hand side,
you get x is equal to 11
plus y, or y plus 11.
So we get x is equal
to 11 plus y
using the second equation.
And then we can substitute it
back into this top equation.
So instead of writing x plus
y is equal to 70, we can
substitute this in for x.
We've already used the second
equation, the magenta one, now
we have to use the
top constraint.
So if we substitute this in, we
get y plus 11-- remember,
this is what x was, we're
substituting that in for x--
plus y is equal to 70.
This is x.
And that constraint was given to
us by this second equation,
or by this second statement.
I just substituted this x with
y plus 11, and I was able to
do that because that's the
constraint the second
equation gave us.
So now let's just solve for y.
We get y plus 11, plus
y is equal to 70.
That's 2y plus 11
is equal to 70.
And then if we subtract 11 from
both sides, we get 2y is
equal to-- what is that?
59?
You subtract 10 from
70, you get 60, so
it's going to be 59.
So y is equal to 59 over 2.
Or another way to write it,
you could write that as 59
over 2 is the same thing as--
let's see-- 25-- 29.5.
y is equal to 29.5.
Now, what is x going
to be equal to?
Well, we already figured out
x is equal to y plus 11.
So x is going to be equal to
29.5-- that's what y is, we
just figured that out-- plus 11,
which is equal to-- so you
add 10, you get 39.5.
You add another 1,
you get 40.5.
And we're done.
If you wanted to find the
intersection of these two
lines, it would intersect at
the point 40.5 comma 29.5.
And you could have used this
equation to solve for x and
then substituted in this one.
You could have used this
equation to solve for y and
then substituted in this one.
You could use this equation
to solve for y and then
substitute into that equation.
The important thing is, is
you use both constraints.
Now let's just verify that
this actually works out.
What's the sum of these
two numbers?
40.5 plus 29.5, that
indeed is 70.
And the difference between
the two is indeed 11.
They're exactly 11 apart.
Anyway, hopefully you
found that useful.
a system of equations is.
And in this video, I'm going
to show you one algebraic
technique for solving systems of
equations, where you don't
have to graph the two lines and
try to figure out exactly
where they intersect.
This will give you an exact
algebraic answer.
And in future videos,
we'll see more
methods of doing this.
So let's say you had
two equations.
One is x plus 2y is equal to 9,
and the other equation is
3x plus 5y is equal to 20.
Now, if we did what we did in
the last video, we could graph
each of these.
These are lines.
You could put them in either
slope-intercept form or
point-slope form.
They're in standard
form right now.
And then you could graph each
of these lines, figure out
where they intersect, and that
would be a solution to that.
But it's sometimes hard to
find, to just by looking,
figure out exactly where
they intersect.
So let's figure out a way to
algebraically do this.
And what I'm going to do is
the substitution method.
I'm going to use one of the
equations to solve for one of
the variables, and then I'm
going to substitute back in
for that variable over here.
So let me show you what
I'm talking about.
So let me solve for x using
this top equation.
So the top equation says x
plus 2y is equal to 9.
I want to solve for x, so let's
subtract 2y from both
sides of this equation.
So I'm left with x is
equal to 9 minus 2y.
This is what this first equation
is telling me.
I just rearranged
it a little bit.
The first equation
is saying that.
So in order to satisfy both of
these equations, x has to
satisfy this constraint
right here.
So I can substitute this
back in for x.
We're saying, this top equation
says, x has to be
equal to this.
Well, if x has to be equal
to that, let's
substitute this in for x.
So this second equation
will become 3 times x.
And instead of an x, I'll write
this thing, 9 minus 2y.
3 times 9 minus 2y, plus
5y is equal to 20.
That's why it's called the
substitution method.
I just substituted for x.
And the reason why that's
useful is now I have one
equation with one unknown,
and I can solve for y.
So let's do that
3 times 9 is 27.
3 times negative 2 is negative
6y, plus 5y is equal to 20.
Add the negative 6y plus the 5y,
add those two terms. You
have 27-- let's see, this will
be-- minus y is equal to 20.
Let's subtract 27
from both sides.
And you get-- let me
write it out here.
So let's subtract 27
from both sides.
The left-hand side, the 27's
cancel each other out.
And you're left with negative y
is equal to 20 minus 27, is
negative 7.
And then we can multiply both
sides of this equation by
negative 1, and we get
y is equal to 7.
So we found the y value of the
point of intersection of these
two lines. y is equal to 7.
Let me write over here, so I
don't have to keep scrolling
down and back up.
y is equal to 7.
Well, if we know y, we can
now solve for x. x is
equal to 9 minus 2y.
So let's do that.
x is equal to 9 minus 2
times y, 2 times 7.
Or x is equal to 9 minus 14, or
x is equal to negative 5.
So we've just, using
substitution, we've been able
to find a pair of x
and y points that
satisfy these equations.
The point x is equal to negative
5, y is equal to 7,
satisfy both of these.
And you can try it out.
Negative 5 plus 2 times 7,
that's negative 5 plus 14,
that is indeed 9.
You do this equation.
3 times negative 5 is negative
15, plus 5 times y,
plus 5 times 7.
So negative 15 plus
35 is indeed 20.
So this satisfies
both equations.
If you were to graph both of
these equations, they would
intersect at the point
negative 5 comma 7.
Now let's use our newly
found skill to do
an actual word problem.
Let's say that they tell
us that the sum of
two numbers is 70.
And they differ-- or maybe we
could say their difference--
they differ by 11.
What are the numbers?
So let's do this word problem.
So let's define some
variables.
Let's let x be the larger
number, and let y be the
smaller number.
I'm just arbitrarily creating
these variables.
One of them is larger
than the other.
They differ by 11.
Now, this first statement, the
sum of the two numbers is 70.
That tells us that x plus
y must be equal to 70.
That second statement, that
they differ by 11.
That means the larger number
minus the smaller
number must be 11.
That tells us that x minus
y must be equal to 11.
So there we have it.
We have two equations
and two unknowns.
We have a system of
two equations.
We can now solve it using
the substitution method.
So let's solve for x on this
equation right here.
So if you add y to both
sides of this
equation, what do you get?
On the left-hand side, you just
get an x, because these
cancel out.
And then on the right-hand side,
you get x is equal to 11
plus y, or y plus 11.
So we get x is equal
to 11 plus y
using the second equation.
And then we can substitute it
back into this top equation.
So instead of writing x plus
y is equal to 70, we can
substitute this in for x.
We've already used the second
equation, the magenta one, now
we have to use the
top constraint.
So if we substitute this in, we
get y plus 11-- remember,
this is what x was, we're
substituting that in for x--
plus y is equal to 70.
This is x.
And that constraint was given to
us by this second equation,
or by this second statement.
I just substituted this x with
y plus 11, and I was able to
do that because that's the
constraint the second
equation gave us.
So now let's just solve for y.
We get y plus 11, plus
y is equal to 70.
That's 2y plus 11
is equal to 70.
And then if we subtract 11 from
both sides, we get 2y is
equal to-- what is that?
59?
You subtract 10 from
70, you get 60, so
it's going to be 59.
So y is equal to 59 over 2.
Or another way to write it,
you could write that as 59
over 2 is the same thing as--
let's see-- 25-- 29.5.
y is equal to 29.5.
Now, what is x going
to be equal to?
Well, we already figured out
x is equal to y plus 11.
So x is going to be equal to
29.5-- that's what y is, we
just figured that out-- plus 11,
which is equal to-- so you
add 10, you get 39.5.
You add another 1,
you get 40.5.
And we're done.
If you wanted to find the
intersection of these two
lines, it would intersect at
the point 40.5 comma 29.5.
And you could have used this
equation to solve for x and
then substituted in this one.
You could have used this
equation to solve for y and
then substituted in this one.
You could use this equation
to solve for y and then
substitute into that equation.
The important thing is, is
you use both constraints.
Now let's just verify that
this actually works out.
What's the sum of these
two numbers?
40.5 plus 29.5, that
indeed is 70.
And the difference between
the two is indeed 11.
They're exactly 11 apart.
Anyway, hopefully you
found that useful.
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