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Solving matrix -3x+4y+5z = 7 4x+3y+2z = 9 -5x+5y+3z = -10

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  • Anonymous Mar 13, 2014

    What is the y-intercept from line -3x-4y=15

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Go to your matrix button and enter a "3x4" matrix.

Then enter it as follows:

-3 4 5 7
4 3 2 9
-5 5 3 -10

Then exit out and go to "2nd->matrix->math->rref(". Then press enter.

Your screen should look like this:

rref(

Then go to matrix and select your 3x4 matrix, press enter and close it with a parathesis. Your screen should look like this:

rref([A])

Press enter and the screen should say this:

1 0 0 2
0 1 0 -3
0 0 1 5

So,
x=3
y=-3
z=5

Hope this cleared up the confusion!

SJ_Sharks

Posted on Mar 14, 2009

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1 Answer

2x plus 5y equals 7 3x plus 6y equals 3


1) 2x + 5y = 7
2) 3x + 6y = 3

I'm going to use the method of elimination to solve for x and y.
Multiply 1) by 3 and 2) by 2 to allow the x's to be eliminated.

1) 6x + 15y = 21
2) 6x + 12y = 6

Now subtract line 2 from line 1.

0x + 3y = 15
---- ----
3 3 divide both sides by 3 to get y by itself.
y =5.
Substitute into 1) to calculate x.
2x + 5(5) = 7
2x + 25 = 7
2x + 25 -25 = 7 - 25
2x = -18
---- ----- divide both sides by 2 to get x by itself
2 2

x = -9

Check by plugging in answer into the other equation, in this case 2)

3 (-9) + 6(5) = 3
-27 + 30 = 3
3 = 3
We did it correctly and checked to prove that we did it right.

Good luck.

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1 Answer

Find the equations of circles passing through (1,-1),touching the lines 4x+3y+5=0 and 3x-4y-10=0


First, I graphed the lines and the point using Desmos.com.

I noticed that the two lines are perpendicular to each other and the point (1,-1) appears to be on the right side of the circle, on a line parallel to 3x -4y-10=0. The equation of this line is y= 3/4x - 1.75. The y-intercept is -1.75. Now we have two points on the opposite sides of the circle, (1, -1) and (0,-1.75). The midpoint formula will give you the centre of the circle and the distance formula will provide the radius.

Let me know if you have any questions.

Good luck.

Paul
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1 Answer

Problems matrix


This is no linear system. You cannot solve it like that using the matrix techniques. Haven't you made a mistake in writing the equations?
If that is tryly the system you want to solve, I suggest that you make a change of variables as follows:
X=1/x , Y= 1/y, Z=1/z (it being understood that x, y, z cannot be equal to 0). You will have to exclude the values x=0, y=0, z=0
Not I am not being sloppy, X and x are different entities, same with Y and y, Z and z.
Your system becomes
2X+3Y-1Z=26
1X+3Y-2Z=36
2X+4Y-5Z=52

Now that is a linear system. Solve it using matrices or Cramer's rule, When you obtain X, Y, and Z, get x=1/X, y=1/y, z=1/Z
The actual implementation of the solution method will depend on the exact model of calculator you are using. Not knowing that, I cannot advise you how to do it.

If I have not made any mistakes, the results are X=-58/9,Y=106/9, Z=-32/9. And x, y, z are just the reciprocals of their namesake.

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2 Answers

Tom is 5 times as old as todd.In seven years Tom's age will be 6 years more than 3 times as old as Todd. How old is Todd


Todd is 4 years old.

Let Tom's age = X
Let Todd's age = Y

From the given facts:

X = 5Y

and

X + 7 + 6 = 3 ( Y + 7 )
X+13 = 3Y + 21
X = 3Y + 8

So if:
X=5Y
and
X=3Y+8

Then:
5Y=3Y+8
2Y=8
Y=4

I hope that helps.

Joe.

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1 Answer

Michael buys two bags of chips and three boxes of pretzels for 413. He then buys another bag of chips and two boxes of pretzels for 239. Find the cost of each bag of chips and each box of pretzels.


Price of bag of chips s X and price of box of pretzels is Y.
Now you can write following equations:

2X+3Y=413
X+2Y=239

From second equation you know that X is 239-2Y, and you put that in first equation:

2*(239-2Y)+3Y=413

or

-4Y+3Y=413-2*239

Finally we have for Y: Y=65 and X=239-2Y=109.

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1 Answer

4x-y-2z=-7 -4x+8y+5z=46 8x-3y+z=13


Type solve(exp1 and exp2 and exp3, {x,y,z})
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1 Answer

Variables


Hello,
This calculator cannot handle symbolic algebra.
Use the rules to simplify the expressions: Get rid of the parentheses using distributivity od multiplication with respect to addition; group similar terms (terms with x, terms without x).
Hope it helps.

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