All language subtitles for English (auto-generated)_en_0_Solving Systems of Two Equations and Two Unknowns_ Graphing_ Substitution_ and Elimination(1080P_HD)

it's professor Dave let's solve some

linear systems by now we have gotten

pretty good at finding solutions to

equations whether they are linear

quadratic or beyond but what about when

we have a system of two different linear

equations can there be a solution to

this set can there be a value or set of

values that works for both equations

usually there will be so let's find out

some different ways that we can get to

this solution first of all since we are

talking about lines we can graph them

say we have y equals x and y equals

negative x plus 4 if we place both of

these lines on the coordinate plane we

can see that they do indeed cross if one

line represents all of the pair's and x

and y values that are solutions to one

equation and the other line represents

all the pairs of x and y values that are

solutions to the other equation then it

must necessarily be the case that the

point where they cross must be a

solution to both equations because this

point is on both lines we can find this

solution graphically if our graphs are

very neat and tidy or if we are using a

graphing calculator because as long as

the graph is accurate we can just

observe that the lines cross at the

point 2 2 and therefore x equals 2 and y

equals 2 represents the solution to this

system of linear equations most of the

time two lines will cross precisely once

and there will therefore be one unique

solution to the system however other

situations include parallel lines which

never cross so there will be no solution

to the system and we could also have two

different equations that describe

precisely the same line in which case

there are infinitely many solutions as

every point on the line will be a

solution for both equations

but as we said we are typically

concerned with situations where two

lines cross once and we want to be able

to get that one solution when graphing

isn't so easy for one reason or another

there are other methods we can use to

find the solution the first technique we

will look at is called substitution say

we have these two equations 2x plus y

equals 4 and X plus 2y equals negative 1

what we could do is solve for one of

these variables in terms of the other

for example let's solve for y in the

first equation subtracting 2x from both

sides gives us y equals 4 minus 2x now

what we can do is substitute this value

of y for the Y in the other equation

that means instead of Y we insert the

quantity 4 minus 2x now we have a new

equation with only X which will make it

possible to solve for X we distribute

the 2 to get 8 minus 4x combine like

terms bring the 8 to the other side and

we get negative 3x equals negative 9 or

x equals 3 that means 3 is the

x-coordinate for the point where these

lines cross to get the Y value we just

plug this x value into either of these

equations using the top 1/2 times 3 is 6

and if 6 plus y equals 4 we can subtract

6 from both sides to get y equals

negative 2 so x equals 3 and y equals

negative 2 or the point 3 negative 2 is

the unique solution to this system

substitution isn't the only technique we

can use we can use another technique

called elimination say we have another

set of linear equations negative 3x plus

2y equals 11 and 5x minus 2y equals 5

once again we absolutely can use the

substitution method but it will

sometimes be easier to go another route

we can combine these equations

in such a way so as to get a new one

where one of the variables has

disappeared to do this in this

particular case we just add the

equations together this is valid because

if a equals B and C equals D then a plus

C must equal B plus D so we are just

exploiting this logical truth to solve a

problem negative 3x plus 5x gives us 2x

then 2y plus negative 2y gives us 0 the

y term goes away and 11 plus 5 gives us

16 so dividing by 2x equals 8 then we

can plug 8 into one of these like the

bottom one we get 40 minus 2y equals 5

negative 2y equals negative 35 and y

equals 35 halves so this was a pretty

fast way to solve this system of course

it won't always be as easy as just

adding them together sometimes we will

have to subtract if both of the Y terms

were positive we would subtract one from

the other

to get the Y terms to go away and we

would then have to be careful with our

arithmetic making sure we get the right

sign on all the values in the answer

furthermore we might have to multiply

one or both of the equations by some

constant to be able to do elimination

say we have 11x plus 2y equals 13 and 7x

minus 3y equals 9 the Y values are

pretty close but we can't add them

together as is and have them cancel

instead we have to realize that the

least common multiple of 2 & 3 is 6 so

we multiply the top equation by 3 and

the bottom equation by 2 that gives us

30 3x plus 6y equals 39 and 14x minus 6y

equals 18 now we are ready to combine

them because positive 6y and negative 6y

will indeed cancel each other out that

gives us 47 x equals 57 we can then

solve for X and plug back into one of

the equations to get the Y value with

lots of multiplying the solutions can

get messy but that's just how they go

sometimes so in summary we can see that

we have three options available to solve

for a system with two equations and two

unknowns we can either use substitution

by solving for one variable in terms of

the other and plugging that into the

other equation we can use elimination by

manipulating one or both of the

equations in order to combine them in

some manner so as to get one variable to

go away or we can just graph everything

and see where the two lines cross if

they cross once there is one solution if

they don't cross there are no solutions

if they are the same line there are

infinitely many solutions so why is this

useful let's look at a real world

example say that Penelope and Artemis

worked a total of 70 hours this week

however Penelope worked 2 hours less

than twice the number of hours that

Artemis worked how many hours did each

of them work while this example is

relatively simple and if we did some

guessing and checking we would

eventually find the answer that wouldn't

be doing algebra and if we learn the

algebra we can apply it to much more

complicated systems where guess and

check just isn't an option first let's

write some equations the two of them

worked a total of 70 hours so let's

write P plus a equals 70 then we know

that Penelope worked 2 hours less than

twice what Artemis did so P must be

equal to 2a minus 2 there's our two

equations with two unknowns

how shall we solve in this case

substitution seems pretty easy as we've

already solved for P in the bottom

equation let's put 2a minus 2 in the

place of P in the top equation that

gives us 2a minus 2

plus a or three a minus two on the Left

we add two to both sides divide by three

and we get a equals 24 then we can use

either of these equations to find p2

times 24 is 48 minus 2 is 46

so Penelope and Artemis worked 24 and 46

hours this week let's check

comprehension thanks for watching guys

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free to email me professor Dave explains

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