Full Length PSAT Math Practice Test

The correct answer is $0.05 \ x \ + \ 4,500$
$x$ is the number of all John’s sales per month and $5\%$ of it is:
$5\% \ × \ x=0.05 \ x$
John’s monthly revenue: $0.05 \ x \ + \ 4,500$

The correct answer is $37$
$x^2=121→x=11$ (positive value) Or $x=  - \ 11$ (negative value)
Since $x$ is positive, then:
$f(121)=f(11^2 )=3 \ (11) \ + \ 4=33 \ + \ 4=37$

The correct answer is $- \ 5$
$2 \ x^2 \ - \ 11 \ x \ + \ 8= \ - \ 3 \ x \ + \ 18→$
$2 \ x^2 \ - \ 11 \ x \ + \ 3 \ x \ + \ 8 \ - \ 18=0→$
$2 \ x^2 \ - \ 8 \ x \ - \ 10=0→$
$2 \ (x^2 \ - \ 4 \ x \ - \ 5)=0→$ Divide both sides by $2$.
Then:
$x^2 \ - \ 4 \ x \ - \ 5=0$, Find the factors of the quadratic equation.
$→(x \ - \ 5) \ (x \ + \ 1)=0→x=5$ or $x= \ - \ 1$
$a \ > \ b$, then: $a=5$ and $b= \ - \ 1$
$\frac{a}{b}=\frac{5}{- \ 1}= \ - \ 5$

The correct answer is $(9,3)$
First, find the equation of the line.
All lines through the origin are of the form $y=m \ x$, so the equation is $y=\frac{1}{3} \ x$.
Of the given choices, only choice C $(9,3)$, satisfies this equation:
$y=\frac{1}{3} \ x→3=\frac{1}{3} \ (9)=3$

The correct answer is $x \ < \ 3$
$2 \ x \ + \ 4 \ > \ 11 \ x \ - \ 12.5 \ - \ 3.5 \ x→$ Combine like terms:
$2 \ x \ + \ 4 \ > \ 7.5 \ x \ - \ 12.5→$ Subtract $2 \ x$ from both sides: $4>5.5x-12.5$
Add $12.5$ both sides of the inequality.
$16.5 \ > \ 5.5 \ x$, Divide both sides by $5.5$.
$\frac{16.5}{5.5} \ > \ x→x \ < \ 3$

The correct answer is $11.5$
$3 \ a=4 \ b→b=\frac{3 \ a}{4}$ and $3 \ a=5 \ c→c=\frac{3 \ a}{5}$
$a \ + \ 2 \ b \ + \ 15 \ c=a \ + \ (2 \ × \ \frac{3 \ a}{4}) \ + \ (15 \ × \ \frac{3 \ a}{5})=a \ + \ 1.5 \ a \ + \ 9 \ a=11.5 \ a$
The value of $a \ + \ 2 \ b \ + \ 15 \ c$ is $11.5$ times the value of $a$.

The correct answer is $\frac{(x \ − \ 5)(x \ + \ 4)}{(x \ − \ 5) \ + \ (x \ + \ 4)}$
To rewrite $\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4)}}$, first simplify $\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}$.
$\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}=$
$\frac{1(x \ + \ 4)}{(x \ - \ 5) \ (x \ + \ 4)} \ + \ \frac{1 \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}=$
$\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}$
Then:
$\frac{1}{\frac{1}{x \ - \ 5} \ + \ \frac{1}{x \ + \ 4}}=$
$\frac{1}{\frac{(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4) \ (x \ - \ 5)}}=$
$\frac{(x \ - \ 5) \ (x \ + \ 4)}{(x \ - \ 5) \ + \ (x \ + \ 4)}$. (Remember, $\frac{1}{\frac{1}{x}}=x$)
This result is equivalent to the expression in choice A.

The correct answer is $\frac{11}{30}$
Of the $30$ employees, there are $5$ females under age $45$ and $6$ males age $45$ or older.
Therefore, the probability that the person selected will be either a female under age $45$ or a male age $45$ or older is: $\frac{5}{30} \ + \ \frac{6}{30}=\frac{11}{30}$

The correct answer is $4$
Plug in the values of x and y of the point $(2, 12)$ in the equation of the parabola. Then:
$12=a \ (2)^2 \ + \ 5 \ (2) \ + \ 10→$
$12=4 \ a \ + \ 10 \ + \ 10→$
$12=4 \ a \ + \ 20$
$→4 \ a=12 \ - \ 20=- \ 8→$
$a=\frac{- \ 8}{4}=- \ 2→$
$a^2=(- \ 2)^2=4$

The correct answer is $10$
The input value is $5$.
Then: $x=5$
$f(x)=x^2 \ - \ 3 \ x→$
$f(5)=5^2 \ - \ 3 \ (5)=25 \ - \ 15=10$

The correct answer is The $y−$intercept represents the starting height of $6$ inches
To solve this problem, first recall the equation of a line:
$y=m \ x \ + \ b$
Where, $m=$ slope and $y=y-$intercept
Remember that slope is the rate of change that occurs in a function and that the $y-$intercept is the $y$ value corresponding to $x=0$.
Since the height of John’s plant is $6$ inches tall when he gets it. Time (or $x$) is zero.
The plant grows $4$ inches per year.
Therefore, the rate of change of the plant’s height is $4$.
The $y-$intercept represents the starting height of the plant which is $6$ inches.

The correct answer is $x=22, y=48$
$\begin{cases}\frac{ - \ x}{2} \ + \ \frac{y}{4} = 1 \\ \frac{ - \ 5 \ y}{6} \ + \ 2 \ x =4 \end{cases} \rightarrow$ Multiply the top equation by $4$. Then,
$\begin{cases}- \ 2 \ x \ + \ y =4\\\frac{- \ 5 \ y}{6} \ + \ 2 \ x =4\end{cases} \rightarrow$ Add two equations.
$\frac{1}{6} \ y=8→y=48$, plug in the value of $y$ into the first equation $→x=22$

The correct answer is $4.8$
Two triangles $\triangle$BAE and $\triangle$BCD are similar. Then:
$\frac{AE}{CD}=\frac{AB}{BC}→\frac{4}{6}=\frac{x}{12}→$
$48 \ - \ 4 \ x=6 \ x→$
$10 \ x=48→x=4.8$

The correct answer is $\frac{25}{16}$
First, simplify the numerator and the denominator.
$\frac{(10 \ (x) \ (y^2)^2}{(8 \ x \ y^2 )^2} =\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}$
Remove $x^2 \ y^4$ from both numerator and denominator.
$\frac{100 \ x^2 \ y^4}{64 \ x^2 \ y^4 )}=\frac{100}{64}=\frac{25}{16}$

The correct answer is $50$
$\frac{y}{5}=x \ - \ \frac{2}{5} \ x \ + \ 10$, Multiply both sides of the equation by $5$. Then:
$5 \ × \ \frac{y}{5}=5 \ × \ (x \ - \ \frac{2}{5} \ x \ + \ 10)→$
$y=5 \ x \ - \ 2 \ x \ + \ 50→y=3 \ x \ + \ 50$
Now, subtract $3 \ x$ from both sides of the equation. Then:
$y \ - \ 3 \ x=50$

The correct answer is $\frac{10}{3}$
First, factorize the numerator and simplify.
$\frac{x^2 \ - \ 9}{x \ + \ 3} \ + \ 2 \ (x \ + \ 4)=15→$
$\frac{(x \ - \ 3) \ (x \ + \ 3)}{(x \ + \ 3)} \ + \ 2 \ x \ + \ 8=15$
Divide both sides of the fraction by $(x \ + \ 3)$. Then:
$x \ - \ 3 \ + \ 2 \ x \ + \ 8=15→3 \ x \ + \ 5=15$
Subtract $5$ from both sides of the equation. Then:
$→3 \ x=15 \ - \ 5=10→x=\frac{10}{3}$

The correct answer is $\frac{10}{3}$
Let $L$ be the length of the rectangular and $W$ be the with of the rectangular. Then,
$L=4 \ W \ + \ 3$
The perimeter of the rectangle is $36$ meters. Therefore:
$2 \ L \ + \ 2 \ W=36$
$L \ + \ W=18$
Replace the value of $L$ from the first equation into the second equation and solve for $W$:
$(4 \ W \ + \ 3) \ + \ W=18→5 \ W \ + \ 3=18→5 \ W=15→W=3$
The width of the rectangle is $3$ meters and its length is:
$L=4 \ W \ + \ 3=4 \ (3) \ + \ 3=15$
The area of the rectangle is: length $×$ width $= 3 \ × \ 15 = 45$

The correct answer is $\frac{10}{3}$
Let $L$ be the length of the rectangular and $W$ be the with of the rectangular. Then,
$L=4 \ W \ + \ 3$
The perimeter of the rectangle is $36$ meters. Therefore:
$2 \ L \ + \ 2 \ W=36$
$L \ + \ W=18$
Replace the value of $L$ from the first equation into the second equation and solve for $W$:
$(4 \ W \ + \ 3) \ + \ W=18→5 \ W \ + \ 3=18→5 \ W=15→W=3$
The width of the rectangle is $3$ meters and its length is:
$L=4 \ W \ + \ 3=4 \ (3) \ + \ 3=15$
The area of the rectangle is: length $×$ width $= 3 \ × \ 15 = 45$

The correct answer is $84$
The description $8 \ + \ 2 \ x$ is $16$ more than $20$ can be written as the equation $8 \ + \ 2 \ x=16 \ + \ 20$, which is equivalent to $8 \ + \ 2 \ x=36$.
Subtracting $8$ from each side of $8 \ + \ 2 \ x=36$ gives
$2 \ x=28$.
Since $6 \ x$ is $3$ times $2 \ x$, multiplying both sides of $2 \ x=28$ by $3$ gives $6 \ x=84$

The correct answer is $10$
$\frac{2}{5} \ × \ 25=\frac{50}{5}=10$

The correct answer is $y= x$
The slop of line A is: $m=\frac{y_{2} \ - \ y_{1}}{x_{2} \ - \ x_{1}}=\frac{3 \ - \ 2}{4 \ - \ 3}=1$
Parallel lines have the same slope and only choice D $(y=x)$ has slope of $1$.

The correct answer is $22$
Substituting $6$ for $x$ and $14$ for $y$ in $y = n \ x \ + \ 2$ gives $14=(n) \ (6) \ + \ 2$,
which gives $n=2$.
Hence, $y=2 \ x \ + \ 2$.
Therefore, when $x = 10$, the value of $y$ is:
$y=(2) \ (10) \ + \ 2 = 22$

The correct answer is $35$
Choices A, C and D are incorrect because $80\%$ of each of the numbers is a non-whole number.
A. $49$,      $80\%$ of $49 = 0.80 \ × \ 49=39.2$
B. $35$,      $80\%$ of $35=0.80 \ × \ 35=28$
C. $12$,      $80\%$ of $12=0.80 \ × \ 12=9.6$
D. $32$,      $80\%$ of $32=0.80 \ × \ 32=25.6$
Only choice B gives a whole number.

The correct answer is $25$
The capacity of a red box is $20\%$ bigger than the capacity of a blue box and it can hold $30$ books.
Therefore, we want to find a number that $20\%$ bigger than that number is $30$.
Let $x$ be that number. Then:
$1.20 \ × \ x=30$, Divide both sides of the equation by $1.2$. Then:
$x=\frac{30}{1.20}=25$

The correct answer is $− \ 5$
The smallest number is $- \ 15$.
To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them.
Let $x$ be the largest number. Then:
$- \ 70=(- \ 15) \ + \ (- \ 14) \ + \ (- \ 13) \ + \ (- \ 12) \ +(- \ 11) \ + \ x→- \ 70=- \ 65 \ + \ x$
$→x=- \ 70 \ + \ 65=- \ 5$

The correct answer is $x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1$
Let $x$ be equal to $0.5$, then: $x=0.5$
$\sqrt{x^2 \ + \ 1}=\sqrt{0.5^2 \ + \ 1}=\sqrt{1.25}≈1.12$
$\sqrt{x^2 } \ + \ 1=\sqrt{0.5^2} \ + \ 1=0.5 \ + \ 1=1.5$
Then, option A is correct.
$x \ < \ \sqrt{x^2 \ + \ 1} \ < \ \sqrt{x^2} \ + \ 1$

The correct answer is $𝑎=𝑐$
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: $6, 3, 4, 9, 8$
average (mean) $= \frac{sum \ of \ terms}{number \ of \ terms}= \frac{6 \ + \ 3 \ + \ 4 \ + \ 9 \ + \ 8}{5}=\frac{30}{5}=6$
Median is the number in the middle.
To find median, first list numbers in order from smallest to largest.
$3, 4, 6, 8, 9$
Median of the data is $6$.
Mode is the number which appears most often in a set of numbers.
Therefore, there is no mode in the set of numbers.
Median $=$ Mean, then, $𝑎=𝑐$

The correct answer is $60\%, 40\%, 90\%$
Percent of cities in the type of pollution A: $\frac{6}{10} \ × \ 100=60\%$
Percent of cities in the type of pollution C: $\frac{4}{10} \ × \ 100=40\%$
Percent of cities in the type of pollution E: $\frac{9}{10} \ × \ 100=90\%$

The correct answer is $2$
Let $x$ be the number of cities need to be added to type of pollutions B. Then:
$\frac{x \ + \ 3}{8}=0.625→x \ + \ 3=8 \ × \ 0.625→x \ + \ 3=5→x=2$

The correct answer is $12$
The ratio of boy to girls is $4:7$.
Therefore, there are $4$ boys out of $11$ students.
To find the answer, first divide the total number of students by $11$, then multiply the result by $4$.
$44 \ ÷ \ 11 = 4 ⇒ 4 \ × \ 4 = 16$
There are $16$ boys and $28 \ (44 \ – \ 16)$ girls.
So, $12$ more boys should be enrolled to make the ratio $1:1$

The correct answer is $\frac{1}{2}$
AB $=5$ And BC $=12$
AC $=\sqrt{12^2 \ + \ 5^2 }=\sqrt{144 \ + \ 25}=\sqrt{169}=13$
Perimeter $=5 \ + \ 12 \ + \ 13=30$
Area $=\frac{5 \ × \ 12}{2}=5 \ × \ 6=30$
In this case, the ratio of the perimeter of the triangle to its area is: $\frac{30}{30}=1$
If the sides AB and BC become twice longer, then:
AB $=24$ And AC $=10$
BC $=\sqrt{24^2 \ + \ 10^2}=\sqrt{576 \ + \ 100}=\sqrt{676}=26$
Perimeter $=26 \ + \ 24 \ + \ 10=60$
Area $=\frac{10 \ × \ 24}{2}=10 \ × \ 12=120$
In this case the ratio of the perimeter of the triangle to its area is: $\frac{60}{120}=\frac{1}{2}$

The correct answer is $- \ \frac{8}{7}$
Since $f(x)$ is linear function with a negative slop, then when $x=- \ 2, \ f(x)$ is maximum and when $x=3, \ f(x)$ is minimum.
Then the ratio of the minimum value to the maximum value of the function is: $\frac{f(3)}{f(- \ 2)}=\frac{- \ 3 \ (3) \ + \ 1}{- \ 3 \ (- \ 2) \ + \ 1}=\frac{- \ 8}{7}=- \ \frac{8}{7}$

The correct answer is $0.97$
Ratio of women to men in city A: $\frac{570}{600}=0.95$
Ratio of women to men in city B: $\frac{291}{300}=0.97$
Ratio of women to men in city C: $\frac{665}{700}=0.95$
Ratio of women to men in city D: $\frac{528}{550}=0.96$
Choice B provides the maximum ratio of women to men in the four cities.

The correct answer is $1.05$
Percentage of men in city A $= \frac{600}{1170} \ × \ 100=51.28\%$
Percentage of women in city C $= \frac{665}{1365} \ × \ 100=48.72\%$
Percentage of men in city A to percentage of women in city C $= \frac{51.28}{48.72}=1.05$

The correct answer is $132$
Let the number of women should be added to city D be $x$, then:
$\frac{528 \ + \ x}{550}=1.2→$
$528 \ + \ x=550 \ × \ 1.2=660→$
$x=132$

The correct answer is $5, 10$
The perimeter of the rectangle is: $2 \ x \ + \ 2 \ y=30→$
$x \ + \ y=15→$
$x=15 \ - \ y$
The area of the rectangle is: $x \ × \ y=50→$
$(15 \ - \ y) \ (y)=50→$
$y^2 \ - \ 15 \ y \ + \ 50=0$
Solve the quadratic equation by factoring method.
$(y \ - \ 5) \ (y \ - \ 10)=0→y=5$ (Unacceptable, because $y$ must be greater than $5$) or $y=10$
If $y=10 →x \ × \ y=50→x \ × \ 10=50→x=5$

The correct answer is $80 \ - \ 6 \ x$
The amount of petrol consumed after $x$ hours is: $6 \ × \ x=6 \ x$
Petrol remaining after $x$ hours driving: $80 \ - \ 6 \ x$

The correct answer is $86$
In the figure angle A is labeled $(3 \ x \ - \ 2)$ and it measures $37$.
Thus, $3 \ x \ - \ 2=37$ and $3 \ x=39$ or $x=13$.
That means that angle B, which is labeled $(5 \ x)$, must measure $5 \ × \ 13=65$.
Since the three angles of a triangle must add up to $180, \ 37 \ + \ 65 \ + \ y \ - \ 8=180$, then:
$y \ + \ 94=108→y=180 \ - \ 94=86$

The correct answer is $14.5$
average (mean) $= \frac{sum \ of \ terms}{number \ of \ terms}=$
$\frac{9 \ + \ 12 \ + \ 15 \ + \ 16 \ + \ 19 \ + \ 16 \ + \ 14.5}{7}=14.5$

The correct answer is $28 \ x \ + \ 6$
If $f(x) = 3 \ x \ + \ 4 \ (x \ + \ 1) \ + \ 2$, then find $f(4 \ x)$ by substituting $4 \ x$ for every $x$ in the function.
This gives:
$f(4 \ x) = 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2$,
It simplifies to:
$f(4 \ x)= 3 \ (4 \ x) \ + \ 4 \ (4 \ x \ + \ 1) \ + \ 2=12 \ x \ + \ 16 \ x \ + \ 4 \ + \ 2=28 \ x \ + \ 6$

The correct answer is $x=− \ 2, y=3$
$\begin{cases}x \ + \ 4 \ y =10\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow$ Multiply the top equation by $- \ 5$ then,
$\begin{cases}- \ 5 \ x \ - \ 20 \ y =- \ 50\\ 5 \ x \ + \ 10 \ y =20\end{cases}\rightarrow$ Add two equations
$- \ 10 \ y=- \ 30→y=3$, plug in the value of $y$ into the first equation
$x \ + \ 4 \ y=10→x \ + \ 4 \ (3)=10→x \ + \ 12=10$
Subtract $12$ from both sides of the equation. Then:
$x \ + \ 12=10→x=- \ 2$

The correct answer is $\frac{𝑎}{\sqrt{𝑎^2 \ + \ 𝑏^2}}$
Cos ⁡$β=\frac{Adjacent \ side}{hypotenuse}$
To find the hypotenuse, we need to use Pythagorean theorem.
$a^2 \ + \ b^2=c^2→c=\sqrt{a^2 \ + \ b^2 }$
cos $(β)=\frac{a}{c}=\frac{a}{\sqrt{a^2 \ + \ b^2}}$

The correct answer is $\frac{10}{3} \ < \ x \ < \ 10$
$|\frac{x}{2} \ - \ 2 \ x \ + \ 10| \ < \ 5→|- \ \frac{3}{2} \ x \ + \ 10| \ < \ 5→- \ 5< \ - \ \frac{3}{2} \ x \ + \ 10 \ < \ 5$
Subtract $10$ from all sides of the inequality.
$→- \ 5 \ - \ 10 \ < \ - \ \frac{3}{2} \ x \ + \ 10 \ - \ 10 \ < \ 5 \ - \ 10→- \ 15 \ < \ - \ \frac{3}{2} \ x \ < \ - \ 5$
Multiply all sides by $2$.
$→2 \ × \ (- \ 15) \ < \ 2 \ × \ (- \ \frac{3 \ x}{2}) \ < \ 2 \ × \ (- \ 5)→- \ 30 \ < \ - \ 3 \ x \ < \ - \ 10$
Divide all sides by $- \ 3$.
(Remember that when you divide all sides of an inequality by a negative number, the inequality sing will be swapped. $<$ becomes $>$)
$→\frac{- \ 30}{- \ 3} \ > \ \frac{- \ 3 \ x}{- \ 3} \ > \ \frac{- \ 10}{- \ 3}$
$→10 \ > \ x \ > \ \frac{10}{3}→\frac{10}{3} \ < \ x \ < \ 10$

The correct answer is $5$
$x$ is directly proportional to the square of $y$. Then:
$x=c \ y^2$
$12=c \ (2)^2→12=4 \ c→c=\frac{12}{4}=3$
The relationship between $x$ and $y$ is:
$x=3 \ y^2$
$x=75$
$75=3 \ y^2→y^2=\frac{75}{3}=25→y=5$

The correct answer is $\frac{𝑎}{𝑏}=\frac{21}{11}$
The equation $\frac{a \ - \ b}{b}=\frac{10}{11}$ can be rewritten as $\frac{a}{b} \ - \ \frac{b}{b}=\frac{10}{11}$, from which it follows that $\frac{a}{b \ - \ 1}=\frac{10}{11}$, or $\frac{a}{b}=\frac{10}{11} \ + \ 1=\frac{21}{11}$.

The correct answer is $1$
The intersection of two functions is the point with $2$ for $x$.
Then:
$f(2)=g(2)$ and $g(2)=(2 \ × \ (2)) \ - \ 3=4 \ - \ 3=1$
Then, $f(2)=1→a \ (2)^2 \ + \ b \ (2) \ + \ c=1→4 \ a \ + \ 2 \ b \ + \ c=1$ (i)
The value of $x$ in the vertex of the parabola is: $x=- \ \frac{b}{2 \ a}→- \ 2=- \ \frac{b}{2 \ a}→b=4 \ a$ (ii)
In the point $(- \ 2, 5)$, the value of the $f(x)$ is $5$.
$f(- \ 2)=5→a \ (- \ 2)^2 \ + \ b \ (- \ 2) \ + \ c=5→4 \ a \ - \ 2 \ b \ + \ c=5$ (iii)
Using the first two equations:
$\begin{cases}4 \ a \ + \ 2 \ b \ + \ c=1\\4 \ a \ - \ 2 \ b \ + \ c=5\end{cases}$
Equation $1$ minus equation $2$ is:
(i) $-$ (iii) $→4 \ b=- \ 4→b=- \ 1$ (iv)
Plug in the value of b in the second equation:
$b=4 \ a →a=\frac{b}{4}=- \ \frac{1}{4}$
Plug in the values of a and be in the first equation. Then:
$→4(- \ \frac{1}{4}) \ + \ 2 \ (- \ 1) \ + \ c=1→- \ 1 \ - \ 2 \ + \ c=1→c=1 \ + \ 3→c=4$
The product of $a, \ b$ and $c=(- \ \frac{1}{4}) \ × \ (- \ 1) \ × \ 4=1$

The correct answer is $90$
The relationship among all sides of special right triangle
$30^\circ \ - \ 60^\circ \ - \ 90^\circ$ is provided in this triangle:
In this triangle, the opposite side of $30^\circ$ angle is half of the hypotenuse.
Draw the shape of this question.
The latter is the hypotenuse.
Therefore, the latter is $90$ feet.

The correct answer is $2$
Let $x$ be the length of an edge of cube, then the volume of a cube is: $V=x^3$
The surface area of cube is: $SA=6 \ x^2$
The volume of cube $A$ is $\frac{1}{3}$ of its surface area. Then:
$x^3=\frac{6 \ x^2}{3}→x^3=2 \ x^2$, divide both side of the equation by $x^2$. Then:
$\frac{x^3}{x^2} =\frac{3 \ x^2}{x^2} →x=2$

The correct answer is $5$
$3 \ x \ + \ 6 \ y=\frac{- \ 3 \ y^2 \ + \ 15}{x}$, Multiply both sides by $x$.
$x \ × \ (3 \ x \ + \ 6 \ y)=x \ × \ (\frac{- \ 3 \ y^2 \ + \ 15}{x})→3 \ x^2 \ + \ 6 \ x \ y=- \ 3 \ y^2 \ + \ 15$
$→3 \ x^2 \ + \ 6 \ x \ y \ + \ 3 \ y^2=15→3 \ × \ (x^2 \ + \ 2 \ x \ y \ + \ y^2 )=15→x^2 \ + \ 2 \ x \ y \ + \ y^2=\frac{15}{3}$
$x^2 \ + \ 2 \ x \ y \ + \ y^2=(x \ + \ y)^2$, Then:
$→(x \ + \ y)^2=5$