# 2015 MTAP Reviewer for Grade 9 Solutions(6-10)

Previously, we solved the first 5 problems in 2015 Eliminations of MMC for grade 9. This time, let’s move on to the other 5 problems. You may utilize this site if you are not a grade 9 student. We have a search button at the top menu to take you to the right page. All reviewers for grade 9 can be checked here.

Problem:

If $\dfrac{x}{9}=y^2$  and $\dfrac{x}{3}=3y$, what is x?

Solution:

We solve for y for the second equation, $\dfrac{x}{3}=3y$ $\dfrac{x}{9}=y$

Now, we substitute this y to the first equation, $\dfrac{x}{9}=y^2$ $\dfrac{x}{9}=(\dfrac{x}{9})^2$

we equate this to 0, $(\dfrac{x}{9})^2-\dfrac{x}{9}=0$

Observe that this is a quadratic equation in x/9, thus by factoring we have $\dfrac{x}{9}(\dfrac{x}{9}-1)=0$

Solving for x’s: $\dfrac{x}{9}=0$ $x=0$

Solving for another x: $\dfrac{x}{9}-1=0$ $\dfrac{x}{9}=1$ $\boxed{x=9}$

Since x>0 is the given restriction in the problem, the only solution is x=9.

Problem 7:

If $f(x)=x^2+6x+9$, what is f(x-3)?

Solution:
We can rewrite f(x) in factor form as follows, $f(x)=(x+3)^2$

To find f(x-3), we let x=x-3, $f(x)=(x+3)^2$ $f(x-3)=(x-3+3)^2$ $\boxed{f(x-3)=(x)^2}$

Problem 8:

If $f(x)=x^3-1$ and $g(x)=x-1$, for all real number x. For what real number does $f(g(-a))=g(f(-a))$

Solution: $f(g(-a))=g(f(-a))$

Solving for g(-a) and f(-a), $g(x)=x-1$ $g(-a)=-a-1$

For f(-a), $f(x)=x^3-1$ $f(-a)=(-a)^3-1$ $f(-a)=-a^3-1$

Going back to the desired equation, $f(g(-a))=g(f(-a))$ $f(-a-1)=g(-a^3-1)$ $(-a-1)^3-1=-a^3-1-1$ $(-a-1)^3=-a^3-1$ $-1(a+1)^3=-(a^3+1)$ $(a+1)^3=a^3+1$ $a^3+3a^2+3a+1=a^3+1$ $3a^2+3a=0$ $3a(a+1)=0$

Solving for a: $3a=0$ $a=0$

Solving for another value of a: $a+1=0$ $a=-1$

Therefore, the answers are 0 and -1.

Problem 9:

Solve for x: $\sqrt{3+\sqrt{x}}=3$

Solution:

First step is to square both sides of equation to remove the outermost radical. $(\sqrt{3+\sqrt{x}})^2=3^2$ $3+\sqrt{x}=9$ $\sqrt{x}=9-3$ $\sqrt{x}=6$

Squaring both sides again, $(\sqrt{x})^2=6^2$ $\boxed{X=36}$

Problem 10:

Solve for x: $\dfrac{2x}{x+2}+\dfrac{x+2}{2x}=2$

Solution:

By cross multiplying the left side of equation we have, $\dfrac{2x}{x+2}+\dfrac{x+2}{2x}=2$ $\dfrac{(2x)^2+(x+2)^2}{(x+2)2x}=2$

let A=2x, B=x+2, $\dfrac{(A)^2+(B)^2}{AB}=2$

Simplifying, $A^2+B^2=2AB$ $A^2-2AB+B^2=0$

By factoring, $(A-B)^2=0$

Plugging in the real values of A and B we have, $(2x-(x+2))^2=0$ $(2x-x-2)^2=0$ $(x-2)^2=0$

Solving for x: $\boxed{x=2}$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.