Help with math algebra questions! T^T
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Help with math algebra questions! T^T

[From: ] [author: ] [Date: 12-06-16] [Hit: ]
Or just do the first one like Im doing the second one.Now for part b, the trick is to figure out what would be x and what would be a from that equation.Compare (x-a)^2 - x(x-2a)to12 335^2 - 12 340 * 12 330.The second part is easiest to unravel.Compare - x(x-2a)to - 12 340 * 12 330.......
I'm having trouble with algebra Q-Q

1. a) Simplify:
(i) (x-3)^2 - x(x-6),
(ii) (x-a)^2 - x(x-2a)

b) Use your answers from part (a) to write down the value of 12 335^2 - 12 340 * 12 330.


2. a) Given that x^2 - y^2 = 28 and x-y = 8 find the value of 2x + 2y.
b) If x^2 + y^2 = 64 and xy = -18, find the value of (2x-2y)^2.

3. Given that x = 2a+3 and y = a^2-1, express x-6/4y-5 in terms of a, giving your answer in its simplest form.

Thanks so much in advance!

-
1 a) You could use the second one to figure out the first one, too. Or just do the first one like I'm doing the second one.
(ii) (x-a)^2 - x(x-2a)
(x-a)(x-a) - x(x-2a)
( x^2 - 2ax + a^2 ) - (x^2 - 2ax)
( x^2 - 2ax + a^2 ) - x^2 + 2ax
x^2 - x^2 - 2ax + 2ax + a^2
a^2
Now for part 'b', the trick is to figure out what would be "x" and what would be "a" from that equation.

Compare (x-a)^2 - x(x-2a) to 12 335^2 - 12 340 * 12 330. The second part is easiest to unravel.
Compare - x(x-2a) to - 12 340 * 12 330. Can you see that x = 12340? and so, if x-2a = 12330, then 2a = 10, and a = 5.
So, if (x-a)^2 - x(x-2a) = a^2, and x = 12340 and a = 5, then 12 335^2 - 12 340 * 12 330 = a^2 = 5^2 = 25. :-)


2a. If x^2 - y^2 = 28, then x^2 = y^2 + 28. Look for two numbers that are 8 apart, whose squares are 28 apart. Ah, this has to get into negative numbers, because the smallest positive numbers, 9 and 1 (even 8 and 0) have the square values too high. But, stuff like 5 and -3 may work...
4 and -4 gives a difference of the squares of 0.
5 and -3? 16.
6 and -2? 32.
7 and -1? 48.
This isn't working, let's do it the mathematical way. You could change x-y = 8 to x = 8 + y and substitute it in elsewhere and see what that does:
( 8+y)^2 - y^2 = 28
64 + 16y + y^2 - y^2 = 28
16y = 28 - 64
y = -36 / 16
y = -9 / 4
x -y = 8, so x - (- 9/4) = 8 x + 9/4 = 8 x = 32/4 - 9/4 x = 23/4

b) x = -18/y, so, (-18/y)^2 + y^2 = 64
(-18/y)^2 = 64 - y^2
324/y^2 = 64 - y^2
How about multiplying both sides by y^2? And moving everything to one side, too.
y^2 * (y^2 - 64 + 324/y^2) = 0
y^2 * y^2 - 64y^2 + 324 = 0
Sorry, I'm not making headway on this problem, am I... :-(


3. Substitute in the values given for "x = " and "y =" into the x-6/(4y-5) term, as in:
(2a+3-6)/[4*(a^2-1) -5]
(2a-3) / ( 4a^2 - 4 - 5)
(2a-3) / ( 4a^2 - 9 )
( 2a-3 ) / [ ( 2a-3 ) * (2a+3 ) ]
1 / (2a+3)
1
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