Solving Linear Systems Using SUBSTITUTION | Math10 | jensenmath.ca

Summary

This 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

00:00

📚 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.

05:02

🔍 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.

10:03

📉 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.

Keywords

💡Linear Systems

Linear Systems refer to a set of two or more linear equations with the same set of variables. In the context of the video, the main theme revolves around finding the values of variables that satisfy all equations in the system, which is essential for understanding the method of substitution being taught.

💡Substitution

Substitution is an algebraic method used to solve linear systems by isolating one variable in one equation and replacing it with its equivalent expression in the other equation. The video script provides a step-by-step guide on how to apply this method to find exact solutions for linear systems.

💡Variables

Variables in the context of the video are the unknowns, typically represented by x and y, which are part of the linear equations in the system. The process of solving a linear system involves finding the values of these variables that satisfy all equations.

💡Equations

Equations in the script are the mathematical statements that represent the linear relationships between variables. The video focuses on systems of two linear equations, each with two variables, and how to solve them using substitution.

💡Point of Intersection

A point of intersection is the solution to a system of linear equations, graphically represented as the point where two lines meet. The video script explains that finding this point algebraically ensures exact solutions rather than estimations.

💡Slope

Slope in the video refers to the steepness or gradient of a line, which is a key characteristic of a line's equation. The script discusses how different slopes indicate intersecting lines, while the same slope can indicate parallel lines.

💡Parallel Lines

Parallel lines are lines that run alongside each other without ever intersecting. The script explains two scenarios involving parallel lines: when they are distinct (no solutions) and when they coincident (infinite solutions), both of which are important in understanding the outcomes of solving linear systems.

💡Algebraic Solution

An algebraic solution is the process of finding the values of variables using mathematical operations and equations. The video script emphasizes the importance of algebraic methods like substitution for obtaining precise answers to linear systems.

💡Coefficient

Coefficients are the numerical factors that multiply the variables in an equation. The script mentions looking for variables with coefficients of one for easier isolation during the substitution process.

💡Isolate

To isolate, in the context of the video, means to manipulate an equation to express one variable in terms of the other or in terms of a constant. This is a fundamental step in the substitution method, as it allows for the replacement of variables in the other equation.

💡Check Solution

Checking the solution involves verifying that the found values of variables satisfy all original equations in the system. The script demonstrates the importance of this step to ensure the accuracy of the solution obtained through the substitution method.

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