Simplify as far as possible, sovle the equation, sovle the quadratic equations

**Steven777** · April 22nd 2009, 04:05 AM

Simlify as far as possible

(i) 10y-5x+7x-3y-x-2y

(ii) 4(6x-7)-5(2x-3)

Sovle the equation

(i) 8x-9 = 4x-21

(ii) 2(2x-5)=4(2x-1)

(iii) 5-3x
----------- = 5 - x
4
Sovle the quadratic equations

(i) 3x² - x - 2 = 0

(ii) x² - 32 - 4x

I will appreciate if someone can help

thank you

**stapel** · April 22nd 2009, 04:28 AM

Originally Posted by Steven777

Simlify as far as possible

(i) 10y-5x+7x-3y-x-2y

(ii) 4(6x-7)-5(2x-3)

To learn how to add polynomials and combine "like" terms, try here and here. To learn how to work with parentheses, try here.

Originally Posted by Steven777

Sovle the equation

(i) 8x-9 = 4x-21

(ii) 2(2x-5)=4(2x-1)

(iii) 5-3x
----------- = 5 - x
4

To learn how to solve linear equations, try here.

Originally Posted by Steven777

Sovle the quadratic equations

(i) 3x² - x - 2 = 0

(ii) x² - 32 - 4x

To learn how to solve quadratic equations, try here.

Once you have learned the basic terms and techniques, please attempt at least one exercise from each of the three sets above. If you get stuck, you will then be able to reply with a clear listing of your work and reasoning so far.

Thank you!

**wik_chick88** · April 22nd 2009, 11:43 PM

Originally Posted by Steven777

Simplify as far as possible

(i) 10y-5x+7x-3y-x-2y

(ii) 4(6x-7)-5(2x-3)

(i)
$= 10y - 3y - 2y - 5x + 7x - x$
$= 5y + x$

(ii)
$= 4 \times 6x - 4 \times 7 - 5 \times 2x - 5 \times -3$
$= 24x - 28 - 10x +15$
$= 24x - 10x - 28 + 15$
$= 14x - 13$

Solve the equation

(i) 8x-9 = 4x-21

(ii) 2(2x-5)=4(2x-1)

(iii) 5-3x
----------- = 5 - x
4

(i)
$8x - 4x = -21 + 9$
$4x = -12$
$x = -3$

(ii)
$2 \times 2x + 2 \times -5 = 4 \times 2x + 4 \times -1$
$4x -10 = 8x -4$
$-10 + 4 = 8x - 4x$
$-6 = 4x$
$x = -1.5$

(iii) dont understand what you have written for this question

Solve the quadratic equations

(i) 3x² - x - 2 = 0

(ii) x² - 32 - 4x

(i)ok the way i was taught to solve these quadratic equations...

first you write it in the form $ax^2 + bx + c$ , thus for this question, $a = 3, b = -1, c = -2$ .

Then you work out $ac$ , which is $ac = 3 \times -2 = -6$ .

Now you have to find two numbers that multiply to give an answer of $-6$ and add to give an answer of $-1$ (which is $b$ in this question).

These two numbers are $-3$ and $2$ ( $-3 \times 2 = -6$ and $-3 + 2 = -1$ ).

So $3x^2 - x - 2 = 0$ can be rewritten as:
$3x^2 - 3x + 2x - 2 = 0$
$3x(x - 1) + 2(x - 1) = 0$
$(3x + 2) (x - 1) = 0$

This means that either $3x + 2$ or $x - 1$ have to be equal to 0 (or both)

Therefore,
$3x + 2 = 0$
$3x = -2$
$x = -\frac{2}{3}$

OR

$x - 1 = 0$
$x = 1$

(ii) have a go at this one by yourself now that I have shown you how to solve quadratic equations

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