as maths problems concerning algebraWatch

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Thread starter 14 years ago
#1
i am having a bit of trouble solving this simulateous equations:

(a) y=x-4, xy=12 (b) y=x+1, y= 6/x (c) y=x+3, x2+2y2=6

(the 2 beside the x and y are squared on question c)

any help will be much appreciated 0
14 years ago
#2
(a) your unknowns are x and y. One of your equations tells you what y is in terms of x, so if you substitute that into the other equation, you end up with only x and numbers in the equation, telling you what x is (take care: two answers). Then you can find y because one equation tells you what y is when you know x

(b) you have two expressions for y, so they must be equal to one another. Once you put them as equal to one another, you can find x by rearranging.

(c) you know what y is in terms of x, so you know what y^2 is in terms of x. you can then substitute that into your other equation, and leave only x terms. you then rearrange for x, and then find y.

NB they all have two answers (an answer is both an x value and the y value that goes with it) each
0
Thread starter 14 years ago
#3
Thanks, i got the answer for the first one and it's (2,6)(6,2), is that right 0
14 years ago
#4
y=x-4, xy=12

x(x-4)=12 (Substitute x -4 for y so you eliminate the y term)
x² - 4x - 12 = 0
(x + 2)(x - 6) = 0 (Factorise and solve)
x = -2 or 6

y = 4 - 2 OR 6 + 2 (Substitute x values into y = x + 4)

So... x = -2, y = 4
or
x = 6, y = 8

I can give you solutions to the other two if you need them.
0
14 years ago
#5
ekk....looks horrible!
0
14 years ago
#6
Simultaneous equations are quite nice actually If you sat / do sit A Level Maths, by the time you get to C4 you'll be praying for SEs, LOL.
0
14 years ago
#7
Some solutions ahead, if you want them!

Solutions for b) y=x+1, y= 6/x

x + 1 = 6 / x (Substitute x + 1 for y)
x² + x - 6 = 0
(x + 3) (x - 2) = 0
x = -3
x = 2

Substituting x gives
y = 1 - 3 = -2
y = 2 + 1 = 3

SO...
x = -3, y = -2
x = 2, y = 3

c) y=x+3, x2+2y2 = 6

x² + 2(x+3)² - 6 = 0
x² + 2x² + 18 - 6 + 12x = 0
3x²+ 12x + 12 = 0
x² + 4x + 3 = 0
(x + 3)(x + 1) = 0
x = -3
x = -1

Substituting x into y = x + 3 gives
y = 3 - 3 = 0
y = 3 - 1 = 2

SO....
x = -3, y = 0
x = -1, y = 2
0
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