1. ## Algebraic Fractions

Solve the equation algebraically.

1. $\displaystyle \frac{2x} {x+1} = \frac{x+1} {2}$

2. $\displaystyle 6.5 = \frac {52} {sqrt[z]}$

3. 4.3 = sqrt 18 divided by 7

2. For the first one:
Given: $\displaystyle \frac{2x} {x+1} = \frac{x+1} {2}$

Cross Multiply: $\displaystyle 2x \cdot 2 = (x+1)\cdot(x+1)$

Simply/expand: $\displaystyle 4x = x^2 + 2x + 1$

Subtract 4x from both sides:$\displaystyle 0 = x^2 - 2x + 1$

Factor: $\displaystyle 0 = (x-1)(x-1) = (x-1)^2$

x = 1

Can you rewrite the second question?

3. 2. $\displaystyle 6.5 = \frac {52} {sqrt[z]}$
the 6.5 = is correct.
the z is a squareroot.
52 is on top.
sorry, still learning the coding for it.
don't know how to put z in squareroot.

4. It's better that you write \sqrt{z} instead, not sqrt[z]. If you wanted cube root though, you should write sqrt[3]{z}. The nth root is $\displaystyle \sqrt[n]{x}$ which is \sqrt[n]{x}.

Given: $\displaystyle 6.5 = \frac{52}{\sqrt{z}}$
Multiply by the multiplicative inverse of 6.5 on both sides: $\displaystyle 1 = \frac{52}{6.5\sqrt{z}}$
Multiply by $\displaystyle \sqrt{z}$ on both sides: $\displaystyle \sqrt{z} = \frac{52}{6.5}$
Square both sides (principal of powers): $\displaystyle z = \left(\frac{52}{6.5}\right)^2$

5. Originally Posted by Chop Suey
It's better that you write \sqrt{z} instead, not sqrt[z]. If you wanted cube root though, you should write sqrt[3]{z}. The nth root is $\displaystyle \sqrt[n]{x}$ which is \sqrt[n]{x}.

Given: $\displaystyle 6.5 = \frac{52}{\sqrt{z}}$
Multiply by the multiplicative inverse of 6.5 on both sides: $\displaystyle 1 = \frac{52}{6.5\sqrt{z}}$
Multiply by $\displaystyle \sqrt{z}$ on both sides: $\displaystyle \sqrt{z} = \frac{52}{6.5}$
Square both sides (principal of powers): $\displaystyle z = \left(\frac{52}{6.5}\right)^2$
hehe, thanks. = me be scare of you.
what do you mean by "multiplicative inverse"?

6. If x is a real number (not equal to 0), then the multiplicative inverse of x is $\displaystyle \frac{1}{x}$

Don't be scared of me! I'm not really the cookie-monster, hehe.