the equation below has one solution. what is the solution to the equation?

Systems of Linear Equations

A Linear Equation is an equation for a line.

A linear equation is not always in the class y = 3.5 − 0.5x,

Information technology tin can also be similar y = 0.5(seven − 10)

Or like y + 0.5x = 3.5

Or like y + 0.5x − 3.five = 0 and more.

(Note: those are yet linear equation!)

A System of Linear Equations is when we have 2 or more linear equations working together.

Example: Here are two linear equations:

2x	+	y	=	5
−x	+	y	=	2

Together they are a system of linear equations.

Tin you discover the values of x and y yourself? (Just have a become, play with them a bit.)

Let'south attempt to build and solve a real world instance:

Example: Y'all versus Horse

It's a race!

You tin can run 0.2 km every minute.

The Horse tin run 0.5 km every infinitesimal. But information technology takes 6 minutes to saddle the horse.

How far tin can y'all get before the horse catches you?

Nosotros can make two equations (d=altitude in km, t=time in minutes)

• Yous run at 0.2km every minute, and then d = 0.2t
• The equus caballus runs at 0.5 km per minute, but nosotros take 6 off its fourth dimension: d = 0.v(t−half-dozen)

So we have a system of equations (that are linear):

• d = 0.2t
• d = 0.five(t−6)

We can solve it on a graph:

Do yous see how the horse starts at half-dozen minutes, but then runs faster?

It seems you become caught afterwards ten minutes ... yous just got 2 km abroad.

Run faster adjacent fourth dimension.

Then now you know what a System of Linear Equations is.

Let the states continue to discover out more about them ....

Solving

There can be many ways to solve linear equations!

Let us come across another example:

Case: Solve these two equations:

• ten + y = half-dozen
• −3x + y = 2

The two equations are shown on this graph:

Our task is to notice where the two lines cross.

Well, we can encounter where they cross, then information technology is already solved graphically.

Simply now permit'south solve it using Algebra!

Hmmm ... how to solve this? There tin exist many ways! In this case both equations take "y" so let'southward try subtracting the whole second equation from the first:

x + y − (−3x + y) = 6 − 2

Now permit us simplify information technology:

ten + y + 3x − y = 6 − ii

4x = 4

x = one

So at present we know the lines cantankerous at 10=1.

And nosotros can discover the matching value of y using either of the two original equations (because we know they have the aforementioned value at x=one). Let's utilize the commencement one (yous can try the 2nd one yourself):

ten + y = half dozen

1 + y = 6

y = 5

And the solution is:

x = i and y = 5

And the graph shows us we are right!

Linear Equations

Simply unproblematic variables are allowed in linear equations. No x², y^three, √10, etc:

Linear vs non-linear

Dimensions

Mutual Variables

For the equations to "work together" they share one or more variables:

A System of Equations has two or more equations in 1 or more variables

Many Variables

And then a System of Equations could have many equations and many variables.

Example: 3 equations in 3 variables

2x	+	y	−	2z	=	3
x	−	y	−	z	=	0
10	+	y	+	3z	=	12

In that location can be any combination:

• 2 equations in 3 variables,
• 6 equations in 4 variables,
• nine,000 equations in 567 variables,
• etc.

Solutions

When the number of equations is the aforementioned as the number of variables there is probable to exist a solution. Not guaranteed, but likely.

In fact there are simply iii possible cases:

• No solution
• One solution
• Infinitely many solutions

When there is no solution the equations are called "inconsistent".

One or infinitely many solutions are called "consistent"

Here is a diagram for 2 equations in two variables:

Independent

"Independent" means that each equation gives new data.
Otherwise they are "Dependent".

Also chosen "Linear Independence" and "Linear Dependence"

Example:

• x + y = 3
• 2x + 2y = six

Those equations are "Dependent", because they are really the aforementioned equation, just multiplied by two.

So the 2d equation gave no new information.

Where the Equations are True

The trick is to find where all equations are true at the same time.

True? What does that mean?

Case: You lot versus Horse

The "you" line is true all along its length (but nowhere else).

Anywhere on that line d is equal to 0.2t

• at t=5 and d=1, the equation is true (Is d = 0.2t? Yes, equally 1 = 0.2×five is truthful)
• at t=5 and d=iii, the equation is not truthful (Is d = 0.2t? No, as 3 = 0.2×v is not truthful)

Also the "horse" line is also true all along its length (but nowhere else).

Only only at the bespeak where they cross (at t=ten, d=2) are they both true.

So they have to be true simultaneously ...

... that is why some people call them "Simultaneous Linear Equations"

Solve Using Algebra

Information technology is common to use Algebra to solve them.

Here is the "Horse" case solved using Algebra:

Example: You versus Equus caballus

The organization of equations is:

• d = 0.2t
• d = 0.5(t−half dozen)

In this example it seems easiest to set them equal to each other:

d = 0.2t = 0.v(t−half-dozen)

Start with: 0.2t = 0.5(t − 6)

Expand 0.v(t−6): 0.2t = 0.5t − 3

Subtract 0.5t from both sides: −0.3t = −3

Divide both sides by −0.iii: t = −iii/−0.3 = 10 minutes

At present nosotros know when y'all become defenseless!

Knowing t we can calculate d: d = 0.2t = 0.2×10 = 2 km

And our solution is:

t = ten minutes and d = 2 km

Algebra vs Graphs

Why use Algebra when graphs are then like shooting fish in a barrel? Because:

More than than 2 variables tin't exist solved past a simple graph.

So Algebra comes to the rescue with two popular methods:

• Solving By Substitution
• Solving By Elimination

Nosotros volition come across each one, with examples in 2 variables, and in 3 variables. Here goes ...

Solving Past Substitution

These are the steps:

• Write one of the equations so information technology is in the style "variable = ..."
• Supercede (i.e. substitute) that variable in the other equation(s).
• Solve the other equation(s)
• (Echo as necessary)

Here is an instance with 2 equations in ii variables:

Example:

• 3x + 2y = 19
• 10 + y = eight

We tin can start with whatsoever equation and whatsoever variable.

Permit's use the second equation and the variable "y" (it looks the simplest equation).

Write one of the equations so information technology is in the way "variable = ...":

Nosotros can subtract x from both sides of x + y = viii to get y = eight − x. Now our equations wait like this:

• 3x + 2y = 19
• y = 8 − 10

At present supplant "y" with "8 − x" in the other equation:

• 3x + 2(8 − x) = 19
• y = 8 − x

Solve using the usual algebra methods:

Expand 2(eight−10):

• 3x + 16 − 2x = nineteen
• y = 8 − x

And so 3x−2x = 10:

• x + 16 = xix
• y = 8 − x

And lastly xix−16=3

• ten = iii
• y = 8 − x

Now we know what x is, nosotros tin put information technology in the y = 8 − x equation:

• x = 3
• y = 8 − iii = 5

And the respond is:

10 = 3
y = 5

Annotation: because in that location is a solution the equations are "consequent"

Bank check: why don't you bank check to encounter if ten = iii and y = 5 works in both equations?

Solving By Substitution: 3 equations in iii variables

OK! Allow's move to a longer example: 3 equations in 3 variables.

This is non hard to do... information technology just takes a long time!

Example:

• x + z = 6
• z − 3y = vii
• 2x + y + 3z = 15

We should line up the variables neatly, or we may lose rail of what we are doing:

x			+	z	=	6
	−	3y	+	z	=	7
2x	+	y	+	3z	=	15

WeI can start with any equation and whatsoever variable. Let'due south use the first equation and the variable "x".

Write 1 of the equations so information technology is in the fashion "variable = ...":

ten					=	6 − z
	−	3y	+	z	=	vii
2x	+	y	+	3z	=	15

Now replace "ten" with "6 − z" in the other equations:

(Luckily there is only one other equation with 10 in it)

	x					=	6 − z
		−	3y	+	z	=	7
two(6−z)		+	y	+	3z	=	fifteen

Solve using the usual algebra methods:

ii(six−z) + y + 3z = xv simplifies to y + z = 3:

ten					=	6 − z
	−	3y	+	z	=	7
		y	+	z	=	3

Skilful. Nosotros have made some progress, but not in that location even so.

Now repeat the process, but just for the last 2 equations.

Write i of the equations so it is in the style "variable = ...":

Let'southward choose the last equation and the variable z:

x					=	6 − z
	−	3y	+	z	=	seven
				z	=	three − y

At present replace "z" with "iii − y" in the other equation:

ten					=	6 − z
	−	3y	+	iii − y	=	7
				z	=	3 − y

Solve using the usual algebra methods:

−3y + (3−y) = seven simplifies to −4y = four, or in other words y = −ane

x					=	6 − z
		y			=	−i
				z	=	3 − y

Almost Done!

Knowing that y = −1 nosotros tin summate that z = 3−y = 4:

x					=	half dozen − z
		y			=	−1
				z	=	4

And knowing that z = 4 we can calculate that x = six−z = ii:

x					=	2
		y			=	−1
				z	=	4

And the respond is:

x = 2
y = −1
z = four

Check: please check this yourself.

Nosotros tin utilise this method for 4 or more equations and variables... just practice the same steps again and again until it is solved.

Conclusion: Commutation works nicely, but does take a long time to do.

Solving By Elimination

Emptying tin be faster ... but needs to be kept dandy.

"Eliminate" ways to remove: this method works by removing variables until in that location is just one left.

The thought is that we can safely:

• multiply an equation past a constant (except naught),
• add together (or subtract) an equation on to another equation

Like in these examples:

WHY can we add together equations to each other?

Imagine two actually simple equations:

x − 5 = three
five = 5

We can add the "5 = 5" to "x − v = iii":

ten − v + 5 = three + v
ten = 8

Endeavor that yourself only use 5 = 3+ii equally the 2nd equation

Information technology volition still work just fine, because both sides are equal (that is what the = is for!)

We tin also swap equations around, so the 1st could become the 2nd, etc, if that helps.

OK, time for a full example. Let's employ the 2 equations in 2 variables example from earlier:

Example:

• 3x + 2y = 19
• 10 + y = 8

Very important to keep things corking:

3x	+	2y	=	19
ten	+	y	=	viii

Now ... our aim is to eliminate a variable from an equation.

First nosotros run across there is a "2y" and a "y", and so allow's piece of work on that.

Multiply the second equation past two:

3x	+	2y	=	19
2x	+	twoy	=	16

Subtract the second equation from the showtime equation:

x			=	3
2x	+	2y	=	16

Yay! Now nosotros know what 10 is!

Adjacent we see the 2d equation has "2x", so let's halve information technology, and then subtract "x":

Multiply the 2nd equation by ½ (i.e. divide past two):

x			=	iii
10	+	y	=	8

Subtract the start equation from the 2d equation:

x			=	3
		y	=	v

Done!

And the answer is:

x = 3 and y = v

And here is the graph:

The blue line is where 3x + 2y = 19 is true

The carmine line is where 10 + y = 8 is truthful

At 10=three, y=v (where the lines cross) they are both true. That is the reply.

Here is another case:

Example:

• 2x − y = 4
• 6x − 3y = 3

Lay information technology out neatly:

2x	−	y	=	4
6x	−	3y	=	3

Multiply the beginning equation by 3:

6x	−	3y	=	12
6x	−	3y	=	3

Subtract the second equation from the kickoff equation:

0	−	0	=	9
6x	−	3y	=	3

0 − 0 = 9 ???

What is going on here?

Quite only, in that location is no solution.

They are actually parallel lines:

And lastly:

Case:

• 2x − y = 4
• 6x − 3y = 12

Neatly:

2x	−	y	=	4
6x	−	3y	=	12

Multiply the first equation by three:

6x	−	3y	=	12
6x	−	3y	=	12

Subtract the second equation from the first equation:

0	−	0	=	0
6x	−	3y	=	3

0 − 0 = 0

Well, that is actually TRUE! Nothing does equal zero ...

... that is because they are actually the same equation ...

... and then there are an Infinite Number of Solutions

They are the same line:

And then at present we accept seen an example of each of the iii possible cases:

• No solution
• One solution
• Infinitely many solutions

Solving By Elimination: three equations in 3 variables

Before we start on the next example, let's look at an improved way to practice things.

Follow this method and we are less likely to make a mistake.

Outset of all, eliminate the variables in gild:

• Eliminate xs commencement (from equation two and three, in gild)
• then eliminate y (from equation 3)

So this is how nosotros eliminate them:

Nosotros and so have this "triangle shape":

At present first at the lesser and work back up (called "Back-Substitution")
(put in z to find y, then z and y to find ten):

And we are solved:

Likewise, we will notice it is easier to do some of the calculations in our head, or on scratch paper, rather than always working within the fix of equations:

Example:

• 10 + y + z = 6
• 2y + 5z = −iv
• 2x + 5y − z = 27

Written neatly:

x	+	y	+	z	=	6
		2y	+	5z	=	−4
2x	+	5y	−	z	=	27

First, eliminate x from second and tertiary equation.

There is no 10 in the 2nd equation ... movement on to the 3rd equation:

Decrease 2 times the 1st equation from the 3rd equation (simply do this in your head or on scratch newspaper):

And we get:

x	+	y	+	z	=	6
		2y	+	5z	=	−4
		3y	−	3z	=	xv

Next, eliminate y from 3rd equation.

Nosotros could subtract one½ times the 2nd equation from the 3rd equation (because ane½ times two is 3) ...

... but we can avoid fractions if we:

• multiply the tertiary equation past 2 and
• multiply the 2d equation by iii

and then do the subtraction ... like this:

And we end up with:

x	+	y	+	z	=	half-dozen
		2y	+	5z	=	−4
				z	=	−2

We now take that "triangle shape"!

At present go back up again "dorsum-substituting":

We know z, so 2y+5z=−4 becomes 2y−x=−four, then 2y=six, then y=3:

x	+	y	+	z	=	6
		y			=	3
				z	=	−ii

Then x+y+z=half-dozen becomes x+three−2=6, so x=6−3+two=five

x					=	5
		y			=	3
				z	=	−2

And the respond is:

x = v
y = 3
z = −2

Check: delight check for yourself.

General Communication

In one case you go used to the Emptying Method it becomes easier than Substitution, considering you just follow the steps and the answers appear.

But sometimes Exchange tin can give a quicker result.

• Substitution is often easier for small cases (like 2 equations, or sometimes three equations)
• Elimination is easier for larger cases

And it always pays to look over the equations first, to see if there is an easy shortcut ... and then experience helps.

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Source: https://www.mathsisfun.com/algebra/systems-linear-equations.html

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