Questions on Algebra: Systems of Linear Equations answered by real tutors!

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tutors have no access to your book. you need to list these problems individually or in groups thanks Mathtut

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You don't say what you are trying to accomplish, though there isn't much you can do with this except to graph it. If developing a graph is your goal, follow this procedure:

Solve the equation for one variable in terms of the other; typically we choose to solve for .









Next, select a value for . Any value will do, but pick something that makes the arithmetic easy, like 0 (zero).

Substitute this value into the equation and do the arithmetic: → when .

That means one of the points on your line is described by the ordered pair: . Plot this point, thus:



Repeat the process with a different selection for . Again, any value will do so long as it is different than the first value you selected. In keeping with the idea of making the arithmetic easy, let's choose 1.

Substitute this value into the equation and do the arithmetic: → when .

That means the second point on your line is described by the ordered pair: . Plot this point, thus:



Finally, draw a straight line all the way across the graph through the two points, thus:

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Hi Fred,
:
what would you like us to do with this equation: graph it? If you looking for the slope and y intercept you need to write it in slope intercept form which is y=mx+b where m is the slope and b is the y intercept. we need more information on what you need.......thanks

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Let x=# of girls and y=# of boys


Since "The number of girls at Sky High School is 60 greater than the number of boys", this tells us that


Because "there are 1250 students all together", this means



Start with the second equation.


Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.


Combine like terms on the left side


Subtract 60 from both sides


Combine like terms on the right side


Divide both sides by 2 to isolate y



Divide




Now that we know that (which means that there are 595 boys), we can plug this into to find (the number of girls)



Start with the first equation.


Plug in


Add


===========================================================================

Answer:


So the answer is and


This means that there are 655 girls and 595 boys

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Please help me solve this equation:
y=3x
x=3y
If you sub for y:
x = 3*3x
1 = 9
There's no solution possible other than x and y are zero.
x = 3*3x
x = 9x
8x = 0
x = 0
y = 0
If you graph these, you get 2 straight lines that intersect at the origin (0,0).

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I think you mean 'solve this system of equations':

Eq. 1:

Eq. 2:

Put each equation in standard form

Eq. 1:




Eq. 2:




Multiply Eq. 2 by 3:

Now add term by term to Eq. 1:







Substitute into either original equation: .

The solution set is

This should have been what you expected because you can see by inspection that the y-intercept for both equations is the point and that since the lines have different slopes, this must be a consistent system.

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If two expressions are equal to the same expression, they are equal to each other.
Both expressions equal y.
Set them equal to each other and solve for x.
Then go back and solve for y.
1.
2.
Then

.
.
.
If you need more help, please post another question.

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x+y=6
x-y=4
Add each colum and you get 2x (+y-y=0)=6+4
2x=10
divide both sides by two and x=5
if x is 5 then y must equal 1
5+1=6
5-1=4

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I assume you want the equation that has the slope of 5 and goes thru the point
(-1,6)
:
y-6=5(x+1)
:
y-6=5x+5
:
y=5x+11....in slope intercept form
:
-5x+y=11...in standard form

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Start with the given system







Plug in into the second equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.


Distribute


Combine like terms on the left side


Subtract 8 from both sides


Combine like terms on the right side


Divide both sides by 5 to isolate y



Divide




Now that we know that , we can plug this into to find



Substitute for each


Add


So our answer is and which also looks like



Notice if we graph the two equations, we can see that their intersection is at . So this verifies our answer.


Graph of (red) and (green)

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Note to the student: Use parentheses to make it clear what you mean. One of the tutors who solved this interpreted your equation to read and the other interpreted it as . You can see that these two equations are not the same thing at all if you look at the results. Given that you had an independent source of the solution set being , then the first interpretation was probably correct, however if you want us to help you properly, in the future you need to be crystal clear about what you want.

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To solve this equation group all x on one side and rest on the other side of = sign. Remember that whatever you do to left side of the equation the same you have to do to the right, and opposit.
x+1=2/3x subtract 2/3x from both sides so you will not have any x on the right side
x-2/3x-1=2/3x-2/3x
x-2/3x-1=0 add 1 to both sides to remove number from rigt side (we want only x there)
x-2/3x-1+1=-1
x-2/3x=-1
3/3x-2/3x=-1
1/3x=-1 /3 multiply by 3 to get just x on the left side
x=-3

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The difficulty in the question is that there is a variable in the denominator of the right hand side of the equation. Also note that something must be incorrect in your question as x = -3 is not a valid solution (can be seen by simple substitution). The first step is to multiply both sides of the equation by 3x so that there are no more "fractions" in the question. This yields:
3(x^2) + 3x = 2
So, converting to standard form of a quadratic, we obtain
3(x^2) + 3x - 2 = 0
Now you can use whichever method you like, such as the quadratic formula to get
x = (-3 + sqrt(33))/6
and
x = (-3 - sqrt(33))/6

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(x - y = 4) & (x + y = 2).
Rewrite each equation to slope y-intercept form
x-y=4 Can be rewritten as y=x-4 (Equation #1)
x+y=2 Can be rewritten as y=2-x (Equation #2)
REplace equation #2 into Equation #1
so 2-x=x-4
Gather all like terms on the same side of the equation
So 2x=6
Solve for x
x=3
Substitute x=3 into Equation #2
y=2-3=-1
So the solution is (3,-1)
I hope that this helps. This is a general approach that should work with all similar problems.

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Can you please help me by WALKING me through how we determine that the intesect points for these two equations is (3,-1): (x - y = 4) & (x + y = 2). I need to see each step in the process of coming up with (3, -1). I know that these two points solves both equations. Thanks, Charlie
---------------------
You determine the "solution" for the system of the 2 eqns. The solution is the point, or the values that satisfy both eqns. In graphical terms, it's where the 2 lines cross (these are linear, so they're straight lines).
x - y = 4
x + y = 2
Solve by elimination and substitution, since the coeffs are all 1's.
Add the 2 eqns
2x + 0y = 6
x = 3
---------
Sub x into either eqn to find y, I'll use eqn 1.
3 - y = 4
y = -1
---------
Those 2 values are the point (3,-1)

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Rewrite each equation in slope & y-intercept form
y-x=1 Equation #1
y+x=3 Equation #2
Equation #1 in slope & y-intercept form
y=x+1
Equation #2 in slope & y-intercept form
y=-x+3
The graphs of the two lines are
From the graph it appears that the solution (point of intersection) is (1,2)
It is always a good idea to check your answer. Substitute the solution (1,2) into the equations to see if equality still is the case.

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I need help with the following
- Find the shortest distance from the point (3,1) to the line 3x+y=-2 to the nearest tenth.
-----------------------
Draw the picture.
Find the equation of the line perpendicular to the given line and passing
thru (3,1)
---------------
slope of the given line: -3
slope of the perpendicular line: 1/3
intercept for this line: 1 = (1/3)3 + b
b = 0
Equation of the perpendicular line: y = (1/3)x
---------------------------------
intercept of the given line and the perpendicular line:
y = (1/3)x
y = -3x-2
---------------
Substitute to solve for "x": (1/3)x = -3x-2 ; (8/3)x=-2;x =-3/4
Since y = (1/3)x y = (1/3)(-3/4) = -1/4
------
So the intercept point is (-3/4,1/4)
-------------------
Find the distance from (-3/4,-1/4) t0 (3,1)
distance = sqrt[(1--1/4)^2 + (3--3/4)^2] = sqrt[(25/16) + (225/16)]
= sqrt[250/16] = (5/4)sqrt(10)
============================================
Cheers,
Stan H.

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Start with the given system of equations:



Multiply the both sides of the first equation by 2.


Distribute and multiply.


So we have the new system of equations:



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:





Group like terms.


Combine like terms. Notice how the y terms cancel out.


Simplify.


Divide both sides by to isolate .


------------------------------------------------------------------


Now go back to the first equation.


Plug in .


Multiply.


Multiply both sides by the LCD to clear any fractions.


Distribute and multiply.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


========================================================

Answer:


So the solutions are and


which forms the ordered pair

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What are the basic steps to solving a linear equation?
How do you solve 80-9y=6y?
I don't understand it at all!
-----------------
Like most things, it's easy IF you know how.
The object is to get y on one side of the equal sign by itself.
80-9y=6y
Add 9y to both sides
--> 80 = 15y
Then divide by 15
80/15 = y
Then reduce (not always possible or necessary)
y = 16/3
That's it. Take a break.

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A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car traveled at 100 km/h. For the rest of the trip it traveled at 90 km/h. How far did the car travel at each speed?
--------------------------
100 kph section DATA:
Distance = x km ; rate = 100 kph ; time = d/r = x/100 hrs.
--------------------
90 kph section DATA:
Distance = 470 - x km ; rate = 90 kph ; time = d/r = (470-x)/90 hrs
==========================
EQUATION:
time + time = 5 hrs
x/100 + (470-x)/90 = 5
90x + 100(470-x) = 450*100
90x + 100*470 - 100x = 450*100
-10x = -20*100
x = 200 km (distance traveled at 100 kph)
470-x = 270 km (distance traveled at 90 kph)
================================================
Cheers,
Stan H.

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Hi, Hope I can help,
.
I need help!!
- A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
We will start fresh
.
First we need to know a formula
.
, or with abbreviations,
.
A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
This problem is a little tricky, since the car is going two speeds, instead of just one, we also have two different times.
.
This will mean we will have two unknowns, we can find the other unknowns with the information we have
.
A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
Remember that , we don't know the time or distance for either one of the speeds, but we know the total distance and total time
.
We can say that the time it took when the car went 90 km/h, the time it took would be
.
We can say that how long it took when the car went 100 km/h, the time it took would be , ( if you subtract the other time, from the total time, you will find the second unknown time )
.
It is the same concept with the distances
.
one distance would be
.
the second unknown distance would be
.
We can now find equations to this problem
.

.
Equation 1
.
if the car went 100 km/h ( rate )
.
if the car took hours ( time ), ( doesn't matter which time we could use "x" if we wanted, but we will use this one )
.
These two multiplied together will equal a distance , ( once again it doen't matter which distance, we could use "y" if we wanted )
.
Now we would put the words into a formula,
.
= , this is the first equation
.
Now we will use the other distance, time, and rate, to find the second equation
.
If the car went 90 km/h ( rate )
.
If the car took hours ( time )
.
If the car drove km ( distance )
.
Now replace "r", "t" and "d" with the other veriables
.
= , this is the second equation
.
Now we can put these two equations side by side
.

.
To make it easier, we will simplify the two equations
.
Equation 1, , we can use the distribution to simplify this equation
.
= =
(remember the signs
.
= , we don't have to move anything right now, so this is the simplified form
.
Equation 2, we will just multiply the "90" and "x"
.
= , this is simplified
.
Now we can start solving for the two unknowns
.

.
This is a system of equations, the easiest way to solve this system is by substitution
.
Since , we can just simply replace "y" in the first equation with "90x"
.
=
.
= , now we can just solve for "x", we will move (-100x) to the right side
.
= = , now we will move "470" to the left side
.
= = , to find "x" we will divide each side by "10"
.
= = = = ,
.
(remember "x" is the time in hours), we can replace "x" with "3" in the second equation, to find "y"
.
= = , , we can check our two answers by replacing "x" and "y" in the first equation
.
x = 3
.
y = 270
.
= = = (True)
.
Here are the simplified equations with the numbers
.

.

.
If we look at the formula,
.

.
This means that the car went 200 km at 100 km/h, and it went 100 km/h for 2 hours
.
The car went 270 km at 90 km/h, and it went 90 km/h for 3 hours
.
The distances add up to the total distance, as well as the time with the total time
.
The car went 200 km at 100 km/h
.
The car went 270 km at 90 km/h
.
Hope I helped, Levi

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How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are needed?
------------------------
active + active = active
0.40x + 0.80(20-x) = 0.50*20
40x + 1600 - 80x = 1000
-40x = -600
x = 15 liters (amount of 40% antifreeze needed in the mixture)
-------------
20-x = 5 liters (amount of 80% antifreeze needed in the mixture)
=========================
Cheers,
Stan H.

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slope = which is the change in y values of two points over the change in the x values of those same two point. Direction is important but it doesnt matter which point you start with and which you end with as long as your consistent in your direction.
example take points (2,5) and point (-3,4)
to find the slope you can start with the first point or the second... I will start with the second point using the y and x values of that point FIRST.
:
(4-5)/(-3-2)=slope=-1/-5=1/5 or you could start with first point
:
(5-4)/(2-(-3))=1/5
either way you end up with the slope but you cannot ever start with the first point on the y such as (5-4) and then change direction such as using the second point first (-3-2)......as you can see that changes the sign to
-1/5...DIRECTION is VERY important

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At first I thought A. B and C were choices of answers.........
:
they want you to find m such that it satisfys the following conditions. The problem was written very poorly.
:
A one solution
B.infinite solutions
C no solutions
:
:
A) m could be almost anything except 1^2.......so let m=-1 then we have y=1^2x-1 and y=-1x-1 and those lines will cross in one place as they are perpendicular to each other
B.) in order for this system to have infinite solutions it has to be the same line so m=1^2
C) There is no value of m that will satisfy this condition because the slope must be the same(parallel lines) to have no solutions. The problem is the y intercept would also have to change to be a parallel line because if m is equal to 1^2 we would have the same answer as B( same line-infinite

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I got a question I hardly understand let alone solve.
What value of m gives a system
y=1^2 x-2
y=mx-1 with
a) one solution
b) an infinitely many solutions
c) no solution
There are three possibilities when solving linear equations (that one you gave is linear): one solution, infinitely many, or no solution
One solution exists when the graph is intersecting lines.
No solution exists when the graph is parallel lines.
Infinite solutions exists when the graph is coincident lines.
Let's go back.
y=1^2 x-2 >>>I'm confused. I'll assume it's
Since the equation is in the slope intercept form (y=mx+b), the slope of this line is 1/2.
For the line to intersect but not coincide, the slope of the second line must not be 1/2.
Therefore, the equations have one solution when, m is not equal to 1/2.Ans.
b) Two lines are coincident when they have the same slope and y-intercepts. The y-intercepts of you given lines are -2 and -1, which can never be equal.
Therefore, no value of m would make the system have infinite solutions.
c) Two equations have no solutions when their graphs are parallel, i.e, same slope but diff. y-intercepts.
Since their intercepts are -2 and -1. This system is possible.
Also, the lines must have the same slope.
Therefore, m=1/2 will make the system have no solution.

Power up,
HyperBrain!

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Start with the first equation.


Distribute


Combine like terms.


Subtract 6 from both sides.


Combine like terms.


---------------------------------------------------


Move onto the second equation.


Distribute


Combine like terms.


Subtract 5 from both sides.


Combine like terms.



==================================================================






So we have the system of equations:




Let's solve this system by substitution


Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.


So let's isolate y in the first equation

Start with the first equation


Subtract from both sides


Rearrange the equation


Divide both sides by


Break up the fraction


Reduce



---------------------

Since , we can now replace each in the second equation with to solve for



Plug in into the second equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.



Distribute to


Multiply


Combine like terms on the left side


Add 36 to both sides


Combine like terms on the right side


Divide both sides by 11 to isolate x



Divide





-----------------First Answer------------------------------


So the first part of our answer is:









Since we know that we can plug it into the equation (remember we previously solved for in the first equation).



Start with the equation where was previously isolated.


Plug in


Multiply


Combine like terms



-----------------Second Answer------------------------------


So the second part of our answer is:









-----------------Summary------------------------------

So our answers are:

and

which form the ordered pair


This means that the system is consistent and independent.








Now let's graph the two equations (if you need help with graphing, check out this solver)


From the graph, we can see that the two equations intersect at . This visually verifies our answer.




graph of (red) and (green) and the intersection of the lines (blue circle).

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Make a scatter plot of the data below.
x: -5,-3,-3,0,1,2,5,6
y: -4,12,10,-6,8,0,3,-9
Describe the correlation of the data. If possible, fit a line to the data and write an equation of the line.

The linear correlation is bad, 
because the points don't appear to
fall in a straight line.

The correlation coefficient is
merely -.3839067252 which in
absolute value is not a good
fit, for it isn't close to 1
at all.  The best fitting line,
called the "regression line"
has this equation:

y = -.7624565469x+2.035921205

On the graph that line is:



I found this line on the TI-84.  If you want to know how,
post again.

Edwin

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A system of equations that produces the situation "there are no solutions" means that the graphs of the lines represented by the linear equations never intersect, the lines are parallel.
A system of equations that produces the situation "there are infinitely many solutions" means that the two are equations are really one and the same equation. The mathematical way of saying this is that the system is said to be dependent. One equation is a multiple of the other equation.

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I did this one already. You can see how it's done and do the other 2.
---------------------------------
Solve for X, Y, and Z in the following equations:
22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14
----------------
To use elimination methods, you have to pick one of the 3 variables to be eliminated that will give you 2 eqns in 2 unknowns. In these 3, y has the smallest coefficients, so I'd pick it and use the 3rd eqn to subtract from the other two, since its coeff is 2.
---------------------
Multiply eqn 3 by 5 and eqn 1 by 2
44x + 10y + 14z = 24 from eqn 1
45x + 10y + 60z = 70 from eqn 3. Subtract 1 from 3
x + 46z = 46 That's one.
--------------
Now multiply eqn 3 x3 and eqn 2 x2
20x + 6y + 4z = 10 from eqn 2
27x + 6y + 36z = 42 from eqn 3 Subtract 2 from 3 (to give pos coeffs)
7x + 32z = 32 That's the 2nd eqn in x and z only.
------------------
7x + 32z = 32 Multiply the other one by 7
7x + 322z = 322
Subtract
-290z = -290
z = 1
-----------
Sub into either of the 2 variable eqns
7x + 32*1 = 32
7x = 0
x = 0
----------
Now sub into the original eqn 3
9X + 2Y + 12Z = 14
9*0 + 2y + 12*1 = 14
2y = 2
y = 1
-------
So it's 0, 1, 1 for x, y, z
The important thing to remember is to eliminate one variable first, then another, etc.

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c. 4x + y =17 or y=-4x+17 (red line)
2x + 3y=21 or 3y=-2x+21 or y=-2x/3+21/3 or y=-2x/3+7 (green line).
(graph 300x200 pixels, x from -6 to 5, y from -10 to 10, of TWO functions -4x +17 and -2x/3 +7).
ANSWER (3,5)
d. 18x + y=42 or y=-18x+42 (red line)
6x-3y=-6 or -3y=-6x-6 or y=-6x/-3-6/-3 or y=2x+2 (green line).
(graph 300x200 pixels, x from -6 to 5, y from -10 to 50, of TWO functions -18x +42 and 2x +2).
ANSWER (2,6)

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Notice the 3Y in the first equation and the 9Y
in the second equation.  If you multiply the
first equation by -3, the 3Y will become -9Y and
then if you add the two equations, the Y's will
cancel.  So let's multiply the first equation by
-3:



Simplifying:



Now we add corresponding terms.
-21X added to 5X gives -16X,
-9Y added to 9Y gives 0,
-42 added to 10 gives -32, so
by adding vertically term by 
term we have the equation:



Divide both sides by -16:



Now pick either one of the original
equations and substitute 2 for X:



-----



Notice the 4X in the first equation and the 2X
in the second equation.  If you multiply the
second equation by -2, the 2X will become -4X and
then if you add the two equations, the X's will
cancel.  So let's multiply the second equation by
-2:



Simplifying:



Now we add corresponding terms.
4X added to -4X gives 0,
Y added to -6Y gives -5Y,
16 added to -48 gives -32, so
by adding vertically term by 
term we have the equation:



Divide both sides by -5:




We could substitute this fraction in
for Y as we did in the previous problem.
However, unlike the previous problem,
 is difficult to subtitute.
So instead we will start over:



and this time we'll eliminate Y:

Notice the Y in the first equation and the 3Y
in the second equation.  If you multiply the
first equation by -3, the Y will become -3Y and
then if you add the two equations, the Y's will
cancel.  So let's multiply the first equation by
-3:



Simplifying:



Now we add corresponding terms.
-12X added to 2X gives -10X,
-3Y added to 3Y gives 0,
-48 added to 24 gives -24, so
by adding vertically term by 
term we have the equation:



Divide both sides by -16:

 



---



You do this one by yourself.  It's just like the first one.

Answer: , 

Edwin

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Start with the given system of equations:



Multiply the both sides of the first equation by 3.


Distribute and multiply.


So we have the new system of equations:



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:





Group like terms.


Combine like terms.


Simplify.


Since is NEVER true, this means that there are no solutions. So the system is inconsistent.

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6x + 2y = 2
3x + 5y = 5 multiply this equation by -2 & then add them.
-6x-10y=-10
6x+2y=2
-----------------
-8y=-8
y=-8/-8
y=1 ans.
6x+2*1=2
6x+2=2
6x=2-2
6x=0
x=0 ans.
Proof:
3*0+5*1=5
5=5

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I'll do the first two to get you started


a)


Start with the given system of equations:




Let's solve the system by substitution


Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.




So let's isolate y in the first equation

Start with the first equation


Subtract from both sides


Rearrange the equation


Divide both sides by


Break up the fraction


Reduce



---------------------

Since , we can now replace each in the second equation with to solve for



Plug in into the second equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.



Combine like terms on the left side


Subtract 6 from both sides


Combine like terms on the right side





-----------------First Answer------------------------------


So the first part of our answer is:









Since we know that we can plug it into the equation (remember we previously solved for in the first equation).



Start with the equation where was previously isolated.


Plug in


Multiply


Combine like terms



-----------------Second Answer------------------------------


So the second part of our answer is:









-----------------Summary------------------------------

So our answers are:

and

which form the point








Now let's graph the two equations (if you need help with graphing, check out this solver)


From the graph, we can see that the two equations intersect at . This visually verifies our answer.




graph of (red) and (green) and the intersection of the lines (blue circle).

---

b)


Start with the given system of equations:



Multiply the both sides of the first equation by -3.


Distribute and multiply.


So we have the new system of equations:



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:





Group like terms.


Combine like terms. Notice how the y terms cancel out.


Simplify.


Divide both sides by to isolate .


Reduce.


------------------------------------------------------------------


Now go back to the first equation.


Plug in .


Multiply.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


So our answer is and .


Which form the ordered pair .


This means that the system is consistent and independent.


Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer.


Graph of (red) and (green)

You can put this solution on YOUR website!
Let x = the number

1x -5 = -3
x-5+5 = -3+5
x= 2

R^2

You can put this solution on YOUR website!
The value V, in dollars, of a shopkeeper's inventory software program is given by V = -40t + 350, where t is the number of years since the shopkeeper first bought the program. Find the value of the software after 0 yr., 4 yr., and 6 yr.
All you have to do is replace t with each year given and then simplify.
In other words, let t = 0 and simplify.
Then, let t = 4 and simplify.
Then, let t = 6 and simplify.
I'll do the first one.
V = -40(0) + 350
V = 0 + 350
V = 350
So, after 0 years, the software has a value of $350.
Do you understand?
Now, do the same using 4 years and then 6 years.

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Since , this means that EVERY point on the line has an x-coordinate of -1. So we have the points


(-1,-2),(-1,-1),(-1,0),(-1,1),(-1,2)(-1,3), etc...


Notice how EVERY point has an x-coordinate of -1


Now let's plot the points






Now draw a line through the points





So we can see that the graph of is a vertical line.



Note: ANY equation in the form where "k" is any number will graph a vertical line.

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Start with the second equation.


Divide both sides by 4 to isolate y.


Move onto the first equation


Plug in


Distribute


Multiply EVERY term by the LCD 4 to clear out the fraction.


Distribute.


Combine like terms on the left side.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce. So this is the first answer.


-----------------------------------------------


Go back to the isolated equation


Plug in


Multiply


Add


Reduce. So this is the second answer.


========================================================

Answer:

So the solutions are and which form the ordered pair (5,8)

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a) Write and equation that models the path of the ordered pairs of (-4,0) and (0,8).
slope = (8-0)/(0--4) = 2
intercept: Since f(0) = 8,8 is the intercept
Equation:
y = 2x + 8
-----------------------------------------

b) Write and equation that models the path of the ordered pairs of (0,8) and (-4,16).
slope = (16-8)/(-4-0) = 8/-4 = -2
intercept: since f(0) = 8, the intercept is .
Equation:
y = -2x+8
-----------------------------------------
c) Does the path of the puck form a right angle? Explain.
No, because the product of the slopes is not -1.
====================================
Cheers,
Stan H.

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Rewriting each of the given equation into the "point-slope" form of
y = mx + b
.
Line a: y = 3x + 6
.
Line b: 3x + y = 6
y = -3x + 6
.
Line c: 3y = 2x + 18
y = (2/3)x + 6
.
Since NO two lines have the same slope -- no lines are parallel.
And, NO two lines have slopes that are "negative reciprocals"
.
Therefore, NONE of the lines are parallel OR perpendicular to each other.

You can put this solution on YOUR website!
Rewriting each of the given equation into the "point-slope" form of
y = mx + b
.
Line a: -2x+y=4
y = 2x+4
.
Line b: 2