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}

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