Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!

You can put this solution on YOUR website!
there are a few ways but lets use elimination method which means you want to manipulate the equation so the point where one set of terms are eliminated
:
I assumed that the +0 in eq 1 was really =0
:
3r-4s=0......eq 1
2r+5s=23.....eq 2
:
multiply all terms in eq 1 by 5 and all the terms in eq 2 by 4 and then add the 2 equations together. I will do the math then rewrite the equations next to one another so you can see what is taking place.
:
5(3r-4s=0)---->15r-20s=0 (revised eq 1)
4(2r+5s=23)---->8r+20s=92 (revised eq 2)
:
can you see what happens to the s terms when we add the two equations together. They are eliminated because -20s+20s=0. We are left with 15r+8r=92+0
:
23r=92
:

:
now take r's found value and plug it back into any equation. I am going to use eq 1
:
3(4)-4s=0--->12-4s=0--->4s=12
:

You can put this solution on YOUR website!

Start with the given system of equations:



Multiply the both sides of the first equation by 5.


Distribute and multiply.


Multiply the both sides of the second equation by 4.


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.


Divide both sides by to isolate .


Reduce.


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


Now go back to the first equation.


Plug in .


Multiply.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


So our answer is and .


This means that the system is consistent and independent.

You can put this solution on YOUR website!
divide both sides by -3, keeping in mind that when you multiply or divide and inequality by a negative number you must reverse the inequality
:

:

You can put this solution on YOUR website!
Start with the given inequality.


Divide both sides by to isolate . note: Remember, the inequality sign flips when we divide both sides by a negative number.


Divide.


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

Answer:

So the answer is

You can put this solution on YOUR website!
find the possible values of r in the inequality 5>r-3?
r-3 < 5
Add 3 to both sides to get:
r < 8
======================
Cheers,
Stan H.

You can put this solution on YOUR website!
You could have the following situations:

1. One element in the solution set consisting of an ordered triple. This occurs when all three lines intersect in a single point.

The lines could all lie in the same plane, or

Two of the lines could be in one plane while the third line is in a different plane, in which case the solution set to the system would be an ordered triple representing a point on the line of intersection of the two planes. The two lines that lie in the same plane could be the same line.

All three lines lie in different planes, in which case the point of intersection of the lines and the point of intersection of the planes is the same point.

2. Infinite solutions. All three lines are the same line and any ordered triple that satisfied one of the equations would satisfy the other two.

3. No solution. The three lines do not intersect in the same point or do not intersect at all.

The following situations would have an empty solution set:

Three parallel lines in the same plane.

Three lines possibly, but not necessarily co-planar, such that any pair intersects, but the three possible pairs of lines intersect at different points.

At least one of the three lines, non-coplanar to the other two lines, lies in a plane parallel to a plane containing one or both of the other lines. Such a line is called a 'skew' line.

Two equations representing the same line and a third line parallel or skew to this line.




There may be others, but you get the idea.

You can put this solution on YOUR website!
there are actually more than 3. there is substitution, addition and subtraction commonly called elimination, graphing, matrix, cramers rule, and possibly more. The first two methods are commonly presented in Algebra I and Algebra II
texts. Solving systems using the last two methods is usually first explained
in Linear Algebra. Every method accomplishes the same purpose -- multiple unknowns are found from a
system of simultaneous equations. Different methods are easier to use
depending on the problem. That’s why so many different methods are
shown.
strengths of graphing are you can see it visually and it can be a quick method
strengths of elimination is you generally dont have to manipulate the equations to much and its is pretty quick on most systems.
strengths of substitution- a little easier for some to grasp although it can end up being very tedious with some systems.
:
matrix is one of my favorites because its quick(most of the time) and FUN
:
hope that helps

You can put this solution on YOUR website!
ok let the cost of fix it fast be Cf and the cost of do it right be Cr
let h be the number of hours spent on the job. This problem of course assumes that each plumbing company can do the job in the same amount of time
:
Cf=50h+25
Cr=45h+35
:
now what we are after is when Cf=Cr so lets set their values equal to each other
:
50h+25=45h+35
:
5h=10
:
hours
:
meaning when 2 hours hits the costs will be equal from both companies. Sidenote if greater than 2 hours Cr has the better deal because the initial cost is accounted for and he only charges $45/hr. (45 dollars wow..I can only wish lol....)

You can put this solution on YOUR website!
1/400=.0025(100)=.25%

You can put this solution on YOUR website!
let x = what dennis has now.
let y = what julia has now.
---
dennis savings is 2/5 more than what julia has.
x = y + 2y/5 = 7y/5
---
after dennis gives 500 to julia, he has 1/5 of what julia has after she receives the 500.
x - 500 = (y + 500)/5
---
you have 2 equations that you need to solve simultaneously.
they are:
x = 7y/5 (first equation)
x-500 = (y+500)/5 (second equation)
---
since you already have x = 7y/5, you can solve by substituting 7y/5 for x in the second equation.
you get:
7y/5 - 500 = (y+500)/5
multiply each side of the equation by 5 to clear the fractions and get:
7y - 2500 = y + 500
subtract y from both sides of the equation to get:
6y - 2500 = 500
add 2500 to both sides of the equation to get:
6y = 3000
divide both sides of the equation by 6 to get:
y = 500
---
substitute 500 for y in the first equation to get:
x = 7*(500)/5 = 7*100 = 700
---
you have:
x = 700
y = 500
---
this means that:
dennis has 700
julia has 500
---
first part of problem states that dennis had 2/5 more than what julia had.
700 = 500 + 2/5 * 500 = 500 + 200 = 700
this part is good.
---
second part of problem states that dennis has 1/5 what julia has after he gives her 500.
700 - 500 = 200 = 1/5 * (500 + 500) = 1/5 * (1000) = 200
this part is good.
---
answer has been checked and is good.
dennis had 700 and julia had 500 before dennis gave her the 500.
dennis had 200 and julia had 1000 after dennis gave her the 500.
---

You can put this solution on YOUR website!
It says classify each system and give the # of solutions but i cant get it. The problem is 3x=-y-2 and 2y+4=-6x
----------------------
Put each equation in slope-intercept form.
y = 3x-2
y = -3x -2
-----------------
The slopes are 3 and -3. Since they a different the
lines must intersect at a single point.
----------
The system is said to be consistent and it has one
Real.
-------------
Note:
If the slopes are the same and the intercepts are the
same, the system is dependent and has an infinite number
of solutions.
If the slopes are the same and the intercepts are different,
the system is inconsistent and has no solution.
==========================
Cheers,
Stan H.

You can put this solution on YOUR website!
lets call the number of small vans and large vans purchased ,be S and L, respectively
:
...eq 1 (for purchases)
:
...........eq 2(for maintanence)
:
divide equation 1 by 10000
:
revised eq 1
revised eq 2
:
take s's value in eq 2 and plug it into eq 1
:

:
multiply all terms by 4
:

:

:

:

:
so if we purchase 4 Large vans and 2 small vans we can carry
:
4(15)+2(7)=74 passengers

You can put this solution on YOUR website!
Define linear programming.

You can put this solution on YOUR website!
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 solutions).

You can put this solution on YOUR website!
-2y=6x-2--->y=-3x+1.....(after dividing by -2)
:
y+3x=1--->y=-3x+1
:
this is the same line therefore there are infinite solutions

You can put this solution on YOUR website!
what is the question?????
:
you have two equations with no instructions on what to do with them

You can put this solution on YOUR website!
I need help understanding the question as well as answering it.
x^5+y^3=3
x^2-y^12=2
-------------
What is the question?

You can put this solution on YOUR website!
Set up the simplex matrix used to solve this linear programming problem. all variables are nonnegative.
Maximize subject to

Introduce slack variables , , and ,
to "take up the slack" between the left sides
and the right sides of the first three and the
objective function, the one to maximize, at the bottom: 



Rearrange the terms in the bottom equation
and put in 0 and 1 coefficients of all the 
variables:

   

Then we set up the matrix with petitions as follows:


  
That matrix is called "the initial tableau".

That's all you were asked for, but I thought
I'd show you how to solve it.

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

The numbers on the bottom left, 
all but the last one, are called "indicators",
Pick the indicator that is the most negative.  
That is, the . It is in column 3, so 
we call column 3 the "pivot column".

Next divide each of the three numbers above it into the number
at the far right of the row it's in. That is we do these
three divisions:

  15    48        48
4)60, 1)48, and 1)48 

The smallest of these is 15, obtained from row 1,
So we call row 1 the "pivot row".

And we call the number in both the pivot row and the
pivot column the "pivot element", which is 4.

Now we divide the pivot row by that number, 4, to make 
the pivot element become 1.



Now we use the pivot row to get 0's everywhere
else in the pivot column.

Then we get a 0 where the 1 is in row 2 by multiplying
-1 times the pivot row to row 2. 
 
Then we get a 0 where the 1 is in row 2 by multiplying
-1 times the pivot row to row 3.

Then we get a 0 where the -11 is in the bottom row by 
multiplying 11 times the pivot row to the bottom row.

Now we have:



Usually some of the indicators on the bottom
row will still be negative, and we would have 
to repeat the operation until there were no
negative indictors on the bottom row.  However 
this particular problem became solved with just 
one operation, so we convert this into a system 
of equations:






Now look at the bottom equation:



Solve for 



Since none of the variables can be negative,
f will be the largest value it can possibly be
when nothing is subtracted from the 165, so 
that maximum is reached when ,, and  are 
all taken to be . So the system now becomes



or



So  has a maximum value of  when
, , and 

Edwin

You can put this solution on YOUR website!
solve each system..
2. y=2x+8
y=3x-1

I will only do this one and you must do the rest.
We have two equations where y is written in terms of x.
Equate both equations to find x.
2x + 8 = 3x - 1
2x - 3x = -8 - 1
-x = -9
x = -9/-1
x = 9/1
x = 9
Now that we know x = 9, plug that value for x into EITHER of the original equations given to find y.
I will use y = 2x + 8 but you can use the other equation as well.
y = 2(9) + 8
y = 18 + 8
y = 26
The solution for this system of equations in two variables is the point
(9, 26). BOTH equations meet at the point (9, 26) and so this is why that point is the solution.
You can also write the answer as x = 9 and y = 26

You can put this solution on YOUR website!

6.

----->





thank you,
Jojo

You can put this solution on YOUR website!
let a = number of adult tickets sold.
let s = number of student tickets sold.
-----
a + s = 198 since the total number of tickets sold is 198.
-----
5*a + 3*s = 878 since the total amount of money made is 878 dollars and each adult ticket brought in 5 dollars and each student ticket brought in 3 dollars.
-----
you have 2 equations that need to be solved simultaneously. this means the same value for a and the same value for s applies to both equations.
-----
the 2 equations are:
a + s = 198
5*a + 3*s = 878
-----
to solve by substitution, take one of the equations and solve for one variable in terms of the other.
take a+s = 198 and solve for a.
a+s = 198
subtract s from both sides:
a = 198 - s
-----
substitute 198-s for a in the second equation:
5*a + 3*s = 878
5*(198-s) + 3*s = 878
990 -5*s + 3*s = 878
combine like terms:
990 - 2*s = 878
subtract 878 and add 2*s to both sides of the equation:
990 - 878 = 2*s
combine like terms:
112 = 2*s
divide both sides by 2:
56 = s
-----
you have s = 56.
take your first equation and solve for a.
a + s = 198
a + 56 = 198
a = 198-56
a = 142
-----
you have:
a = 142
s = 56
-----
takes these numbers and plug into your second equation:
5*a + 3*s = 878
5*142 + 3*56 = 878
710 + 168 = 878
878 = 878
equation is true.
values for a and s are good.
your answer is:
142 adult tickets were sold and 56 student tickets were sold.

You can put this solution on YOUR website!

Here's the correct solution.  Find your
errors and why you made them. Lots of
times things don't divide evenly, so gah,
you just get a fraction, that's all:

 6x - 24 + 4x = -2x + 6
10x - 24      = -2x + 6
12x - 24      =       6
    + 24            +24
12x           =      30
          12x = 30
            x = 
            x = 

Check:



Edwin

You can put this solution on YOUR website!
Your steps are correct, you are not watching your signs;
6x-24+4x= -2x +6
you are going to add the 6x + 4x; the minus sign goes with the 24, not the 4. The 4 is positive.
10x-24= -2x+ 6
Then you should have added the 2x to both sides, it is a negative 2, so you need to add to cancel it out;
12x-24=6
+24 +24
you would add 24 to both sides since it is negative;
12x = 30
The answer is a mixed fraction;
x =
x =
CHECK:
6x-24+4x= -2x +6





30 = 30
:)

You can put this solution on YOUR website!
These problems are always easier if you put letters in place
of the things you're looking for:
Let = the pounds of peanuts needed
Let = the pounds of cashews needed
Let = the pounds of raisins needed
The 1st sentence says it's a 9-pound mixture, so
(1)
Peanuts cost $1.50 per pound, so the cost of the peanuts will be

Cashews cost $2.00 per pound, so the cost of the cashews will be

Raisins cost $1.00 per pound, so the cost of the raisins will be

The total cost of the mixture is $13.00, so
(2)
The mixture calls for twice as much peanuts than cashews, so
(3)
Now I have 3 equations and 3 unknowns, so it should be solvable
Subtract (1) from (2)

Substitute (3) for



And, going back to (3),



Use (1) to find




The amounts of the ingredients are 2 pounds of cashews,
4 pounds of peanuts, and 3 pounds of raisins
Check answer:
(2)



OK

You can put this solution on YOUR website!
let the number of peanuts, cashews, and raisens be p,c,and r respectively
:
p+c+r=9........eq 1----->r=9-p-c
1.5p+2c+1r=13..eq 2
p=2c...........eq 3
:
lets plug p's value in eq 3 into eq 1 and 2 and re write them
:
2c+c+r=9.........eq (4)
1.5(2c)+2c+r=13..eq (5)
:
3c+r=9.... eq (4)
5c+r=13... eq (5)
:
now lets subtract eq 5 from eq 4
:
-2c=-4
:
pounds of cashews
:
pounds of peanuts
:
pounds of raisens

You can put this solution on YOUR website!
5.15-4.25=.85
:
rate of change from 1991 to 1197 is .85/5.15=.165 or 16.5%
:
on an annulized basis this would be %

You can put this solution on YOUR website!


1)
2(1-x)+3(x+4)=5(x+3)Complete multiplication using distributive property a(b + c) = ab + ac
2-2x+3x+12=5x+15 Group like terms on each side of the equal sign.
x+14=5x+15 Subtract x from both sides of the equation
x-x+14=5x-x+15
14=4x+15 Subtract 15 from both sides
14-15=4x+15-15
-1=4x divide both sides by 4
-1/4=x
2)
x(2x+3)-4(x-1)=2x(x+1) Complete multiplication
2x^2+3x-4x+4=2x^2+2x Group like terms on each side
2x^2-x+4=2x^2+2x subtract 2x^2 from both sides
2x^2-2x^2-x+4=2x^2-2x^2+2x
-x+4=2x add x to both sides
-x+x+4=2x+x
4=3x divide both sides by 3
x=4/3

You can put this solution on YOUR website!
2(1-x)+3(x+4)=5(x+3)
2-2x + 3x+12 = 5x+15
x + 14 = 5x + 15
4x = -1
x = -1/4
------------------------
x(2x+3)-4(x-1)=2x(x+1)
2x^2+3x -4x+4 = 2x^2+2x
Simplify:
2x^2 -x + 4 = 2x^2 + 2x
3x = 4
x = 4/3
============
Cheers,
Stan H.

You can put this solution on YOUR website!
250 - 115 = 9x
To write the equation, say it in words and write the math expressions that say it instead.
How's that?

You can put this solution on YOUR website!
If the camera costs $250 and you have $115, you figure out how much more you need to earn by subtracting 115 from 250. 250-115 = 135.

So, you need to earn $135. If you earn $9 per hour, you figure out how many hours it will take to earn the money by dividing 135 by 9 to see how many hours you need. 135/9 = 15.
You need to work 15 hours at $9 per hour to earn the $135 you need.

You can put this solution on YOUR website!
Do you have a question?

You can put this solution on YOUR website!
Same Problem172371
See Answer#127404. Done
Thank you

You can put this solution on YOUR website!
Let =Anthony's Age
Then, =Jacob' Age
Putting into eqn:

, divide both terms by 2
, Anthony's Age
, Jacob's Age
Check,


Thank you,
Jojo

You can put this solution on YOUR website!
let Jacob's and Anthony's ages be j and a respectively
:
j=a+8
a+j=38
:
now substitute j's value in first equation into 2nd equation
:
a+(a+8)=38
:
2a=30
:
Anthony's age
:
Jacob's age

You can put this solution on YOUR website!
Need help here, please with systems of linear equations in 3 variables!
I don't understand it at all. It's very confusing to me.
3x^2+4y+z=7
2y+z=3
-5x+3y+8z=-31
------------
You say linear, but the 1st eqn is 2nd order. Is that a typo?
PS I thought I just posted this.

You can put this solution on YOUR website!
there is no question on this problem so I will write this in slope intercept form and hope your looking for slope and intercept
:
2y=3x+4---divide by 2----y=(3/2)x+2....so slope is 3/2 and y intecept is 2
:
and a graph (3/2)x+2----notice how the line passes through y-axis at (0,2)

You can put this solution on YOUR website!
The problem description contains a lot of extraneous data that will not be needed in solving this problem, so, beware of this on your test.
You'll need to sort out the wheat from the chaff, as the saying goes.
The fact that the Auto Center has ten fifty-five gallon drums of 90% and eight fifty-five gallon drums of 40% ethylene glycol is interesting but not required to solve the problem.
Let x = the number of gallons of 90% ethylene glycol solution needed ((90%)x).
Then the manager will need to mix (100-x) gallons of the 40% ethylene glycol solution ((40%(100-x)) with that to get the 100 gallons of 55% ethylene glycol solution (55%(100)).
After changing the percentages to their decimal equivalents: (90% = 0.9, 40% = 0.4, and 55% = 0.55), you can express this as follows:
0.9x+0.4(100-x) = 0.55(100) Simplify and solve for x.
0.9x+40-0.4x = 55 Combine like-terms.
0.5x+40 = 55 Subtract 40 from both sides.
0.5x = 15 Divide both sides by 0.5
x = 30 gallons of 90% ethylene glycol solution.
Suffice it to say that the manager has enough ethylene glycol solution on hand to do the job.

You can put this solution on YOUR website!
3 square root 81B^5
= 3 sqrt[81B^4*B]
= 3 sqrt[81B^4]* sqrt[B]
= 3 [9B^2]*sqrt[B]
= 27B^2*sqrt[B]
=======================
Cheers,
Stan H.

You can put this solution on YOUR website!
L=2W-1
L*W=55
(2W-1)W=55
2W^2-W-55=0
(2W-11)(W+5)=0
2W=11
W=11/2
W=5.5 FT. ANS.
L=2*5.5-1
L=11-1
L=10 FT. ANS.
PROOF:
10*5.5=55
55=55
THUS THE PERIMETERIS:
2L+2W=PERIMETER.
2*10+2*5.5=20+11=31 FT. IS THE PERIMETER.

You can put this solution on YOUR website!
L=2W-1
L*W=55
(2W-1)W=55
2W^2-W-55=0
(2W-11)(W+5)=0
2W=11
W=11/2
W=5.5 FT. ANS.
L=2*5.5-1
L=11-1
L=10 FT. ANS.
PROOF:
10*5.5=55
55=55

You can put this solution on YOUR website!
X+Y=20 OR X=20-Y
X^2-Y^2=120
(20-Y)^2-Y^2=120
400-40Y+Y^2=120
-40Y=120-400
-40Y=-280
Y=-280/-40
Y=7 ans.
X+7=20
X=20-7
X=13 ans.
Proof:
13^2-7^2=120
169-49=120
120=120

You can put this solution on YOUR website!
can u help me our e-mail me an website to solve linear equations
----------
Enter linear algebra on Google.

You can put this solution on YOUR website!

Remember, the idea is to have the SAME variable
missing in two equations, not just to eliminate
any variable you can.

You should start either of the following two 
ways, but not the way you started.  

1. Take advantage of the fact that y is already 
missing in the 1st equation. To do so, use the other 
two equations, 2nd and 3rd, to get another 
equation with y missing.  Then solve that with
the 1st.

or

2. Take advantage of the fact that x is already 
missing in the 2nd equation. To do so, use the other 
two equations, 1st and 3rd, to get another 
equation with x missing.  Then solve that with
the 2nd.

I'll arbitrarily choose the first way.

Since y is already eliminated in the first equation,

let's put it aside for now, and use only the other two
equations to eliminate y, the same variable that is
missing in the first.



We'll write  as 



Remember we want to eliminate y, so we multiply
the upper equation through by  and add
it to the lower equation:



We add term by term and get



Now we go back and get the very first original
equation which we put aside, 

We put it together with the one we just found
and now we have a system of two equations with 
the SAME missing variable, y. 

 

Now we eliminate  by multiplying the
upper equation through by -2, and adding:



and get





Substitute this into 







Now substitute these values in either one of the 
original equations which contains . The 
simpler one is 







So the solution is , , 

Edwin

You can put this solution on YOUR website!
Help me solve each system by the substitution method:
y=x^2-4x-10
y=-x^2-2x+14
------------
These are quadratics (2nd order eqns). I don't think substution applies.
y=x^2-4x-10

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=56 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 5.74165738677394, -1.74165738677394. Here's your graph:

It's
2 + sqrt(14)
2 - sqrt(14)
----------------
y=-x^2-2x+14

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . The discriminant -52 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -52 is + or - . The solution is , or Here's your graph:

This one is:
1 + sqrt(-13)
1 - sqrt(-13)

You can put this solution on YOUR website!
I don't have that book.