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Recent problems solved by 'tishy'

tishy answered: 4 problems
Expressions-with-variables/11390: The problem is as follows-
Solve by useing Substitution:
x+y=20}
5x+10y=200}
I made it so the equation read y=20-x
so then i substituted that for the equatioin 5x+10y=200. It eventualy came out to -5x=0. But just today my algebra2 teacher said that is a wrong outcome for any problem like this. Ive done it over and over but I'm still geting that answer. Can you please help me?
1 solutions

Answer 5910 by tishy(4) About Me  on 2005-09-16 21:30:15 (Show Source):
You can put this solution on YOUR website!
x+y=20.....equation 1
5x+10y=200....equation 2
multiply equation 1 by negative 5
therefore you get (x) x -5, y x-5 and 20 x -5 (you must multiply everything my negative 5)
so your new equation would be -5x-5y= -100
so you work with your new equation -5x-5y= -100
5x + 10y= 200
NOTICE I made my x into a a negative 5 so that it may cross out ot contradict with the 5 in equation 2. so when i work out my problem those two will be elimianted watch closely
-5x-5y = -100
5x +10y = 200
therefore when i work it out the negative 5 has cancelled my 5 in equa. 2 so i am left with
-5y = -100
+10y = 200 (so i work it ot from here on)
you ALWAYS ADD AFTER YOU HAVE DONE YOUR CANCELATION(the cancelling i did with teh 5's) so now
-5y + 10y = 5
ans -100 + 200 (o i FORGOT YOU MUST ALWAYS ADD THE 2 NUMBERS AT THE END OF THE EQUATION ALSO)
so i am left with 5 = 100 ( now what i do with this is divide) so my answer would be 100/5y = 20
so therefore in my 1st part i have discovered that y = 20
so now i have to subsitute to ensure that that is correct
with substituion you can use any equation i will choose to use equation 2
so
5x + 10y = 200 now becomes:
5x +10(x20) = 200 (because i substitued)
5x + 200 = 200 ( so what i do next is carry the 200 on the left over to the right becasue there is a rule that states ONLY VARIABLES which are the letters can remain on the left)
so now my problem looks like this
5x = 200 - 200 which works out to
5x = 0
HEY you did a Great job at trying to do it you only got mix up with your negative sign i hope this is right bye !!!!


Geometry_Word_Problems/11404: Find the perimeter and area of a rectangle if its length is 8x-2, and its width is 4x.
1 solutions

Answer 5905 by tishy(4) About Me  on 2005-09-16 21:10:22 (Show Source):
You can put this solution on YOUR website!
Find the perimeter and area of a rectangle if its length is 8x-2, and its width is 4x.

8x-2= -16(length)
therefore perimeter is when you add all the sides together so therefore
perimeter = -16+-16+4+4
= -32 + 8
= -24
area is L (length) times W (width)
area = LxW
-16 x 4
= -64
answer :-)


Square-cubic-other-roots/11406: An 8-inch-square picture is to be enlarged so it covers an area of 100 inches.How much longer will each side of the enlarged picture be?
1 solutions

Answer 5903 by tishy(4) About Me  on 2005-09-16 21:04:00 (Show Source):
You can put this solution on YOUR website!
An 8-inch-square picture is to be enlarged so it covers an area of 100 inches.How much longer will each side of the enlarged picture be?
therefore 8x 4(no. of sides) = 32
so of 32 inches is taken up already then you ,msut see how much more is left there fore
100-32= 68
to see how much each side is enlarged by divide 68 into 4
you get an answer of 17
so if u add the 17 to the original 8 u get 25inches for each side and
25x4 = 100
answer!!! :-)


Polynomials-and-rational-expressions/11410: 2a+3b+c/2d, when a = 2, b = 0, c = 3, and d = -5
2(2)+3(0)+3/2(-5)
4+0+(3/-10)
4+0+ -3.33
=4+ -3.33
=0.67
1 solutions

Answer 5902 by tishy(4) About Me  on 2005-09-16 20:58:03 (Show Source):
You can put this solution on YOUR website!
2a+3b+c/2d, when a = 2, b = 0, c = 3, and d = -5
2(2)+3(0)+3/2(-5)
4+0+(3/-10)
4+0+ -3.33
=4+ -3.33
=0.67