zacismyname

$$3y+5x=18$$

$$3y=18-5x$$

$$y=6-\frac{5x}{3}$$   or  $$y=\frac{18-5x}{3}$$

$$3y+5x=18$$

$$5x=18-3y$$

$$x=\frac{18-3y}{5}$$

Hmmmm I think this can also be factorised. . .

$$a^2+ab^2+ab+b^3$$

$$a(a+b^2)b(a+b^2)$$

$$(a+b)(a+b^2)$$

:)

4/22100 = 1/5525

Thanks :)

Yeah I considered that but it basically means the same thing anyway so I left it how it was. 

1.) 5/10x-60

$$\frac{5}{10}x-60$$

$$\frac{x}{2}-60$$

2.) 12/6x-15

$$\frac{12}{6}x-15$$

$$2x-15$$

3.) 2x/2x^2+6x

$$\frac{2x}{2x^2}+6x$$

$$\frac{1}{x}+\frac{6x}{1}$$

$$\frac{1}{x}+\frac{6x^2}{x}$$

$$\frac{1+6x^2}{x}$$ (i think . . . or do you leave it as 1/x + 6x?)

$$\frac{2}{3}\times{(3x-6)}=10$$ Now exapnd the brackets . . .

$$\frac{6x}{3}-\frac{12}{3}=10$$ So . . .

$$2x-4=10$$

$$2x=14$$

$$x=7$$

So x does not equal 14.

We can also check this by substituting in [x=7]

$$\frac{2}{3}\times{(3\times{7}-6)}=10$$

$$\frac{2}{3}\times{15}=10$$

$$2\times{15}=30$$

$$30=30$$

:)

Thanks :)

1/4+1.8 = 0.25+1.8 = 2.05 = or $$2\frac{1}{20}$$

-2/3x+4=1/3x-2

$$\frac{-2x}{3}+4=\frac{x}{3}-2$$ Now move the 4 over,

$$\frac{-2x}{3}=\frac{x}{3}-6$$ Move the $$\frac{x}{3}$$ over,

$$\frac{-3x}{3}=-6$$ Simplify,

$$-x=-6$$ Now multiply both sides by -1 to make x a positive.

$$x=6$$

(If we substitute [x=6] into the sum we find that -4+4=2-2, which it does)

:)

The depreciation of the car can be written as

Future value(FV) = Present Value(P)*(1-rate(i))^number of years(n), or

$$FV=P(1-i)^n$$

so for 4 years . . .

$$FV=120493\times(1-0.319)^4=25914.94$$

And for 9 years . . .

$$FV=120493\times(1-0.319)^9=3795.65$$