## Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2

This is the second part (questions 11-20) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. The first part can be read here.

Although reasonable care has been given to make the solution accurate as possible, the solver is also human. Please comment below if you see any errors.

11.) If $x \neq 1$, solve for $x$ in $2 \sqrt{x} + \frac{3}{\sqrt{x}} = 5$.

Solution
Multiplying both sides by $\sqrt{x}$, we obtain

$2x + 3 = 5\sqrt{x}$

Squaring both sides of the equation, we get

$4x^2 + 12x + 9 = 25x$

$4x^2 - 13x + 9 = 0$.

Factoring, we have

$(4x - 9)(x - 1) = 0$

So, $x = \frac{9}{4}$ or $x = 1$.

But from the given above, $x \neq 1$, so the only solution is $x = \frac{9}{4}$

## Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 1

This is the first part (questions 1-10) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. Although much care has been given to make the solution as accurate as possible, the solver is also human. Please comment below if you see any errors.

Also available: Questions 11-20

1.) Simplify $\sqrt{\frac{3}{2}} - \sqrt{\frac{2}{3}}$

Solution

$\sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{2}{3}} = \dfrac {\sqrt {3}}{\sqrt {2}} - \dfrac {\sqrt {2}}{\sqrt {3}} = \dfrac {(\sqrt {3})^2-(\sqrt {2})^2}{\sqrt {2}\sqrt {3}} = \dfrac {1}{\sqrt {6}}$.

Rationalizing the denominator,

$\dfrac {1}{\sqrt {6}} \times \dfrac {\sqrt {6}}{\sqrt {6}} = \dfrac {\sqrt {6}}{6}$.

Answer: $\sqrt{6}/6$

2.) Evaluate $\dfrac{2^0 + 2{^-1}}{2^{-2} + 2^{-3}}$

Solution

$\dfrac {2^0 + 2^{-1}}{2^{-2} + 2^{-3}} = \dfrac {1 + \frac {1}{2}}{\frac {1}{4} + \frac {1}{8}} = \dfrac {\frac {3}{2}}{\frac {3}{8}} = \dfrac {3}{2} \times \dfrac {8}{3} = 4$

3.) Simplify $\sqrt{ \dfrac{1}{9} + \dfrac{1}{16}}$.
$\sqrt{\dfrac{1}{9} + \dfrac {1}{16}} = \sqrt {\dfrac{16 + 9}{(9)(16)}} = \sqrt{\dfrac{25}{(9)(16)}} =\dfrac{5}{(3)(4)} = \dfrac{5}{12}$.