Solving Linear Systems Using SUBSTITUTION | Math10 | jensenmath.ca
Summary
TLDRThis educational video script introduces the method of substitution for solving linear systems, emphasizing the importance of finding exact solutions. It outlines three possible outcomes: a single solution when lines intersect, no solutions when lines are parallel and distinct, and infinite solutions when lines are coincident. The script guides viewers through five steps of the substitution method, including isolating a variable, substituting it into the other equation, solving for the variable, and checking the solution in both original equations. Four examples illustrate these scenarios, providing a clear understanding of solving linear systems algebraically.
Takeaways
- 📚 The lesson focuses on solving linear systems using the substitution method, which provides exact answers compared to graphical estimations.
- 🔍 There are three possible scenarios when solving a linear system: intersection, parallel and distinct lines, and parallel and coincident lines.
- 📈 The first scenario involves lines with different slopes intersecting at a point, leading to a single solution.
- 🚫 The second scenario features parallel but distinct lines that never intersect, resulting in no solutions algebraically shown by an unsolvable equation like 0x = 5.
- ∞ The third scenario describes parallel and coincident lines with the same slope and intercepts, offering infinite solutions, indicated algebraically by an equation like 0x = 0.
- 📝 The substitution method involves five steps: isolating a variable, substituting it into the other equation, solving for the remaining variable, finding the value for the second variable, and checking the solution in both original equations.
- 🔑 The choice of which variable to isolate often depends on finding a coefficient of one or the smallest coefficient for ease of calculation.
- 📉 When substituting, it's crucial to ensure that the x variables in both equations are equal, leading to an equation with a single variable to solve.
- 🔄 The process involves simplifying and rearranging equations to isolate and solve for variables, often starting with distributing and combining like terms.
- 📝 A correct solution must satisfy both original equations, which is verified by substituting the found values back into the equations and checking both sides for equality.
- 📚 The lesson concludes with a preview of the next method for solving linear systems, which is elimination, suggesting a continuation of the algebraic approach to these problems.
Q & A
What is the main goal of solving a linear system?
-The main goal of solving a linear system is to find the values of the variables that satisfy all of the equations in the system, which can be graphically represented as the point of intersection of the lines.
What are the three scenarios that can occur when solving a linear system?
-The three scenarios are: 1) The lines intersect, meaning there is a single solution. 2) The lines are parallel and distinct, meaning there are no solutions. 3) The lines are parallel and coincident, meaning there are an infinite number of solutions.
What is the algebraic method called that is taught in this lesson?
-The algebraic method taught in this lesson is called substitution.
How many steps are there in the substitution method for solving linear systems?
-There are five steps in the substitution method: 1) Rearrange one of the original equations to isolate a variable. 2) Substitute the isolated variable into the other equation. 3) Solve the new equation for the remaining variable. 4) Plug the answer back into one of the original equations to solve for the other variable. 5) Check the solution in both equations.
Why is it easier to isolate a variable with a coefficient of one?
-It is easier to isolate a variable with a coefficient of one because it simplifies the process of rearranging the equation, as you do not need to divide or multiply through by the coefficient.
What does it mean if you end up with an equation like 0x = 5 during the substitution process?
-If you end up with an equation like 0x = 5 during the substitution process, it indicates that there are no solutions to the linear system because there is no value for x that can satisfy the equation.
How can you identify when the lines represented by the equations are parallel and coincident?
-You can identify when the lines are parallel and coincident if, during the substitution process, you end up with an equation that is true for all values of the variable, such as 0x = 0.
What is the purpose of checking the solution in both original equations?
-The purpose of checking the solution in both original equations is to verify that the found values of x and y indeed satisfy both equations, ensuring the accuracy of the solution.
Can you provide an example of a situation where the substitution method would yield infinitely many solutions?
-The substitution method would yield infinitely many solutions when the lines represented by the equations are parallel and coincident, as in the case where the equations have the same slope and intercepts, and the substitution process results in an equation like 0y = 0.
What is the next algebraic method for solving linear systems that will be taught after substitution?
-The next algebraic method for solving linear systems that will be taught after substitution is called elimination.
Outlines
📚 Introduction to Solving Linear Systems by Substitution
This paragraph introduces the concept of solving linear systems using the substitution method. It explains the goal of finding values for variables that satisfy all equations in the system. The instructor outlines three possible scenarios when solving a linear system: intersection of lines with different slopes, parallel lines that never intersect, and coincident lines that overlap completely. The paragraph also outlines the algebraic implications of each scenario, such as obtaining a single solution, no solution, or infinitely many solutions, and introduces the five-step process for substitution that includes isolating a variable, substituting it into the other equation, solving for the remaining variable, and checking the solution in both original equations.
🔍 Detailed Walkthrough of the Substitution Method
The second paragraph provides a step-by-step guide on how to apply the substitution method to solve linear systems. It begins by demonstrating how to isolate a variable with a coefficient of one for simplicity. The example shows substituting the isolated variable into the other equation, solving for the remaining variable, and then substituting back to find the value of the first variable. The paragraph emphasizes the importance of checking the solution in both original equations to ensure accuracy. It also touches on the process of solving equations when coefficients are not one, showcasing how to rearrange and simplify equations to find the solution.
📉 Exploring Different Scenarios in Linear Systems
This paragraph delves into the different scenarios that can arise when solving linear systems: lines intersecting, lines being parallel and distinct, and lines being parallel and coincident. It explains the algebraic outcomes of these scenarios, such as obtaining no solution or infinitely many solutions, and how they relate to the geometric representation of lines. The instructor uses examples to illustrate each scenario, demonstrating how to identify when lines do not intersect or are coincident through the algebraic process, and emphasizes the importance of checking solutions to confirm they satisfy both original equations.
Mindmap
Keywords
💡Linear Systems
💡Substitution
💡Variables
💡Equations
💡Point of Intersection
💡Slope
💡Parallel Lines
💡Algebraic Solution
💡Coefficient
💡Isolate
💡Check Solution
Highlights
Lesson focuses on solving linear systems by substitution, providing an algebraic method for exact answers.
Three scenarios when solving a linear system: intersection, parallel and distinct lines, and parallel and coincident lines.
Algebraic solutions for each scenario: single solution, no solution, and infinite solutions.
Graphical representation of solutions as the point of intersection for intersecting lines.
Steps for substitution method: isolate a variable, substitute, solve for one variable, solve for the other, and check the solution.
Example of solving a system with a single solution using substitution.
Demonstration of checking the solution in both original equations for accuracy.
Explanation of no solution scenario when lines are parallel and distinct, leading to an unsolvable equation.
Illustration of an infinite solution scenario when lines are parallel and coincident, sharing all x-y points.
Technique of rearranging equations for easier substitution of isolated variables.
Process of solving for the remaining variable after substitution.
Importance of checking the solution in both equations to ensure it lies on both lines.
Approach to solving when coefficients of variables are not one, by isolating the variable with the smallest coefficient.
Example of a system with no solution due to the resulting equation being impossible to satisfy.
Final example leading to infinitely many solutions, indicating the lines are coincident.
Conclusion summarizing the substitution method and previewing the next lesson on elimination.
Transcripts
[Music]
here is our second lesson on solving
linear systems solving linear systems by
substitution so remember that solving a
linear system
means to find the values of the
variables that satisfy all of the
equations in the system
we're going to be looking at four
different systems of equations one two
three four notice each of those systems
has two linear equations so our job for
each of those is to figure out an
ordered pair x y that satisfies both of
the equations so graphically speaking
what we're doing is we're going to be
finding the point of intersection the
point where the lines intersect in the
first lesson of this unit we learned how
to do that graphically
but sometimes we had to estimate the
solutions so in this lesson we're going
to learn an algebraic method called
substitution where we'll make sure we
get exact answers for each linear system
let me take you through again the three
scenarios that can happen when solving a
linear system and then talk about when
we're doing this algebraic solution
what's going to happen for each of these
scenarios
so the first scenario that could happen
is that your two lines intersect so
there is a point of intersection
if that happens the two lines are going
to have different slopes
and usually different x and y intercepts
unless they intersect on the x or y axis
and you'll get one solution if the lines
intersect
and what's going to happen algebraically
you'll get a single answer for both x
and y that satisfy both equations
the second scenario is what if your
lines are parallel and distinct
notice in the diagram the two lines run
parallel to each other but never
actually intersect
if that's the case the equations of the
lines are going to have the exact same
slope
but their y and x intercepts are going
to be different
you will get no solutions when trying to
solve this system of equations and
what's going to happen algebraically
when you're trying to solve this system
is that you'll get an equation that
isn't true for any value of the variable
so for example in your solving process
you may get to a point where you get an
equation that needs to be solved like 0x
equals 5. now there's no value you could
plug in for x to make that equation true
because 0 times anything is 0. there's
no way it could be 5.
so if that happens to you you know what
you have is a system of equations where
the two lines are parallel and distinct
meaning there are no solutions
the other option is that the two lines
are parallel and coincident which means
the lines are actually right on top of
each other so i know it looks like you
only see one line there but the red and
blue line are actually right on top of
each other in that graph so when that
happens
the equations of the lines will have the
exact same slope
and their x and y intercepts are going
to be the exact same
which means that there are an infinite
number of solutions to this system of
equations
because they share every x y point so
there are an infinite number of x y
points that satisfy both equations
what's going to happen algebraically
that tell you that the lines are
parallel and coincident you're going to
get an equation in your solving process
that's true for all values of the
variable
so for example you may get an equation
that says 0x
equals 0.
if that happens well you could plug in
anything for x and that equation is true
right zero times anything is zero so if
that happens to you in your solving
process then you know what you have are
two lines that are parallel and
coincident meaning there are an infinite
number of solutions to the system so
let's go through four examples and
within those four practice examples we
will encounter each of these three
different scenarios when we're using the
method of substitution we're going to
follow these five steps we'll start by
rearranging either of the original
equations to isolate a variable either x
or y doesn't matter we'll choose
whatever is easiest
we then substitute what the isolated
variable is equal to into the other
equation that's why this is called
substitution because the second step we
have to do a substitution
step three we solve the new equation for
the one variable that is in that
equation and then plug that answer for
the variable back into either of the
original equations to solve for the
other variable and of course step five
we should always check our answer in
both equations to make sure it's right
i'll reference those five steps while we
solve these systems
so when doing substitution i like to set
it up so that i have the equations
beside each other
so i'll have
line one i'll rewrite here
and i'll write line two right beside it
so keep in mind our goal is to find what
value of x and y satisfies both
equations step one tells us we have to
pick one of the variables in either of
these equations to isolate
i like to look for one that has a
coefficient of one that's usually
easiest to isolate so i notice the x in
line 1 has a coefficient of 1. so let's
isolate that
so i'll rearrange this by subtracting
the 4y to the other side and it would be
x equals 6 minus 4y
that's step one done step two says we
need to now take
what the isolated variable is equal to
six minus four y
and substitute it in for x into the
other equation
so we'll replace the x in line two with
six minus four y
so line two equals two
times six minus four y
minus three y
equals one
so really what we're doing is we're
forcing their x variables to be equal
now we're solving for what value of y
we'll make that true so we have an
equation that only has one variable it
only has the variable y we can solve
that equation that's what step three
tells us to do so to solve this equation
i think the easiest thing to do is start
by distributing this 2 into the brackets
so 2 times 6 is 12
minus 2 times 4y is 8y
i'll collect my like terms i have 12
minus 11y equals 1.
and now i need to isolate y i think i'll
move the negative 11y to the right and
bring the constant of 1 to the left it
changes both of the signs of those terms
i have 11 equals 11y so therefore y is
1.
so i have my answer for y that's step 3
done step 4 says i now need to sub that
answer for y back into either original
equation to solve for x
usually the easiest way to solve for the
other variable is to use this rearranged
version of the other equation that
already has the other variable isolated
right just take our answer for y
sub it in for y and evaluate
so i have x equals 6 minus 4 times we
just solved for y to be one so sub n one
we get x equals six minus four which is
two so our solution to the linear system
is x equals two and y equals one those
are the values of the variables that
make both of the original equations true
now we should check to make sure that
that's the correct answer
so i'm going to check the solution x
equals to y equals one
i need to check it in both equations
i'll check it in line one first
by doing a left side right side check
the left side is x plus 4y the right
side is 6. if i plug in my solution for
x and y so 2 plus 4 times 1
the left side is equal to the right side
good so
the solution to one satisfies the first
equation
but does it also satisfy the second
equation right all we've proven by
plugging into one equation is proving
that the point to one is on line one is
it also on line two let's have a look
so i'll plug in my 0.21 for x and y
i get 4 minus 3 which is 1 left side and
right side equal okay so we've proven
that the 0.21 is on both lines therefore
that's where the lines intersect so this
is the correct solution to our system so
we should write our final answer
the solution is
and we can either write it as an x y
point or we could just list the answers
for x and y
i'll write it like that the solution is
x equals 2 y equals 1. but like i said
if you want to write it as an x y point
2 1 that's fine as well
let's try another one
i'm going to set it up the same way
write the equations beside each other
and now i need to isolate one of the
variables in either equation like i said
in the last example i try and look for a
variable that has a coefficient of one
and i notice the y in the second
equation has a coefficient of one so
it'll be easy to isolate that variable
i'll just move the 7x to the other side
and i get y equals negative 7x now i
need to substitute
what this equation tells me y is equal
to negative 7x into the other equation
for y
so i'll rewrite line 1 with negative 7x
plugged in for y
so i have 5x minus 3 times
negative 7x
minus 2 equals 0.
now i have an equation where the only
unknown is an x we can solve that
equation so to solve this i'm going to
do a multiplication first i've got
negative 3 times negative 7x that's
positive 21x
i'll collect my like terms 5x plus 21x
is
26x i'll move the negative 2 over and
when i isolate x i'd have to divide both
sides by 26 which would give me x equals
2 over 26
which we could of course reduce to 1
over 13.
so that's my solution for x
we now need to take that solution for x
and plug it back into either original
equation but like i told you last time
it's almost always easiest sub it back
into the rearranged version of the other
equation that already has y isolated
i can just sub in 1 over 13 for x right
there and easily solve for y so i have y
equals negative 7 times 1 over 13
which gives me negative 7 over 13.
so my solution
is x equals 1 over 13
y equals
negative 7 over 13.
and just like the last one we should do
a left side right side check plug it
back into both original equations and
prove that it satisfies both equations
so i'll just do that quickly just so you
can see that it actually works
[Music]
there we go so you can see that the
point 1 over 13 comma negative 7 over 13
satisfies both equations which means the
point is on both of the lines which
means that's where the lines intersect
so we have the correct answer
part c
so in both of the first two examples we
were able to get a solution so the lines
were not parallel to each other they had
a point of intersection
let's look at this one this one's going
to be a little bit different i'll set it
up the same way i'll write equation one
and two beside each other
so the first step for solving by
substitution is to isolate a variable
i'll isolate the x in the second
equation so i have x equals 6 minus y
and now i need to take what that
equation tells me x is equal to 6 minus
y and replace the x in the other
equation with that value
so when i do that i'll have 2 times 6
minus y because that's what the other
equation tells me x is equal to
plus 2i equals 7. let's try and solve
this i'll distribute the 2 into the
brackets and i would have 12 minus 2y
plus 2y
equals 7. i'll move all the constant
terms to the right side of the equation
and leave the variable terms on the left
so i have negative 2y
plus 2y equals 7 minus 12. well negative
2y plus 2y that's 0y
and 7 minus 12 is negative 5. what i see
here is an equation
where there are no possible values i
could plug in for y and get an answer
zero times anything is zero there's no
way i could get negative five right zero
times y is zero
and zero clearly doesn't equal negative
five so when you get something like this
that tells you there are no solutions to
the linear system what do we know
therefore about the two lines the two
lines must be parallel and distinct
they have the same slope but different
intercepts
let's do our last example
once again let's set up the equations
beside each other
i'm going to have to isolate a variable
now none of these variables have a
coefficient of 1. so i'll just pick the
one with the smallest coefficient i'll
pick the x in the first equation let's
isolate that so i'll start by isolating
3x 3x equals 2 minus 4y i'll divide both
sides of this equation by 3 i'd have x
equals
2 over 3
minus 4 over 3 y
now i need to take what x is equal to
and sub it into the other equation for x
when i do that i'll have nine
times two over three minus four over
three y
plus 12y
equals 6. let's see what happens when we
try and solve this
well i'll distribute the 9 into the
brackets
and there's ways we could simplify this
first but i'll just expand it and
simplify after so i'd have 18 over 3
right 9 times 2 is 18. so 18 over 3
minus 9 times 4 is 36
over 3 y
plus 12 y equals 6.
18 divided by 3 is 6
and 36 divided by 3 is 12. so i've got 6
minus 12y
plus 12y
equals 6. i'll move the constant term 6
to the right side to be with the other
constant term so i have negative 12y
plus 12y equals 6 minus 6.
that gives me 0y
equals 0. well for what values of y is
this equation true
for any value of y you could plug in
anything you want for y and it makes
that equation true right zero times
anything is zero so this equation has
infinitely many solutions which means
the linear system has infinitely many
solutions therefore
the system has infinite
solutions that happens when the two
lines are right on top of each other
they're parallel and coincident they
have the same slope and the same
intercepts
all right i hope those examples helped
you with the method of substitution stay
tuned for the next lesson where we'll
learn another algebraic method for
solving linear systems called
elimination
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