Practice questions

First isolate the absolute value.

$|y+2|=10\color{blue}+1$

$|y+2|=11$

Solution for the above equation is

$y+2=-11$ or $y+2=11$

Solving for $y$

$y=-11\color{red}-2$ or $y=11\color{red}-2$

$y=-13$ or $y=9$

The solution set is $\{-13, 9\}$

The absolute value, $|2x-1|$ is greater than or equal to a positive number, $7$.

The solution is

$2x-1\le -7$  or  $2x-1\ge 7$

Next, you need to solve for $x$.

Add $1$:

$2x\le -7\color{red}+1$  or  $2x\ge 7\color{red}+1$

$2x\le -6$  or  $2x\ge 8$

Divide $2$:

$x\le -6/2$  or  $x\ge 8/2$

$x\le -3$  or  $x\ge 4$

The solution in interval notation:

$(-\infty, -3]\cup [4, \infty)$

Isolate the absolute value:

$|x+5|\le -1\color{red}+6$

$|x+5|\le 5$

An absolute value is less than or equal to a positive number $5$.

The solution is

$-5\le x+5 \le 5$

Add $-5$

$-5-5 \le x+5-5 \le 5-5$

$-10 \le x \le 0$

In interval notation, the above solution is

$[-10, 0]$

Solve for $x$ in each inequality:

$-3x\ge 10\color{red}-4$ or $5x\gt 13\color{red}+2$

$-3x\ge 6$ or $5x\gt 15$

$\dfrac{-3x}{-3}\le \dfrac{6}{-3}$ or $\dfrac{5x}{5}\gt \dfrac{15}{5}$

$x\le -2$ or $x\gt 3$

In interval notation:

$(-\infty, -2]\cup(3, \infty)$

Solve for $x$ in each inequality:

$7x\ge -17\color{red}-4$ and $6x\ge -7 \color{red}-5$

$7x\ge -21$ and $6x\ge -12$

$\dfrac{7x}{7}\ge \dfrac{-21}{7}$ and $\dfrac{6x}{6}\ge \dfrac{-12}{6}$

$x\ge -3$ and $x\ge -2$

Draw the graph on the number line and take the overlapping region.

in interval notation,

$[-2, \infty)$

First find the slope from the equation of the line. For that write the equation in slope intercept form:

$y=-\dfrac{5}{2}x-\dfrac{27}{2}$

Slope is the coefficient of $x$:

$m=-\dfrac{5}{2}$

Slope of the perpendicular line is negative of the reciprocal of $m$.

$\dfrac{2}{5}$

Equation of the perpendicular line is

$y=\dfrac{2}{5}x+b$

Now, you need to find $b$.

Use the point, $(-5,-2)$ to find $b$. Substituting $x=-5$ and $y=-2$ in the above equation, you get

$-2=\dfrac{2}{5}(-5)+b$

$-2=-2+b$

$b=0$

Substitute the $b$ back into the equation, you get the answer:

$y=\dfrac{2}{5}x$

You are given two points, $(x_1,y_1)=(-3, 7)$ and $(x_2,y_2)=(2, -1)$.

First find the slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-1-7}{2-(-3)}=-\dfrac{8}{5}$

Use the following formula to find the equation:

$(y-y_1)=m(x-x_1)$

$y-7=-\dfrac{8}{5}(x-(-3))$

$y-7=-\dfrac{8}{5}(x+3)$

Now, you need to write this in standard form:

Multiply $5$ on both sides:

$5(y-7)=-8(x+3)$

Remove the parentheses by multiplication

$5y-35=-8x-24$

Move everything on the right to the left hand side:

$5y-35+8x+24=0$

or

$8x+5y=11$

Use the elimination method to solve:

Multiply the first equation by $4$ and the second equation by $-9$

$36x+20y=-76$

$-36x+27y=-18$

Add the equations that will eliminate $x$ terms:

$47y=-94$

Dividing $47$:

$y=-2$

Substitute $y=-2$ in one of the original equations and find $y$. Substituting $y=-2$ in the first equation:

$9x+5\cdot (-2)=-19$

$9x-10=-19$

Solving for $x$:

$x=-1$

So, the solutions are $x=-1$ and $y=-2$. Solution as ordered pair, $(-1, -2)$.

The system is consistent and the equations are independent.

Use the elimination method to solve:

Multiply the first equation by $3$:

$6x-21y=-60$

Add this equation with the second equation that will eliminate the $x$ term and you get,

$0=-45$

This is not true.

So, the system has no solution. And the system is inconsistent. The equations are independent.

The linear function is

$C(t)=16+40t$

To find the amount of time for the total cost to reach $\$560$, take $C(t)=560$ and solve for $t$.

$16+40t =560$

Solving for $t$,

$t=13.6$ months

Let $x$ is the speed of Rosanna and $y$ the speed of Simone.

Rosanne walks 2 mph slower, therefore,

$x=y-2$

Use, $time =\dfrac{distance}{speed}$

Both takes same amount of time:

$\dfrac{5}{x}=\dfrac{8}{y}$

Cross multiplying:

$5y=8x$

Substitute, $y-2$ for $x$ from the first equation:

$5y=8(y-2)$

$5y=8y-16$

Solving for y:

$y=\dfrac{16}{3}=5.33$

Substitute, this $y$ in the first equation to find $x$:

$x=5.33-2=3.33$

Thus, the speed of Rosanna is 3.33 mph and that of Simone is 5.33 mph.

Use the following formula:

$\dfrac{1}{t}=\dfrac{1}{t_1}+\dfrac{1}{t_2}$

where $t$ is the time when both (well and spring) work together.

$t_1$ and $t_2$ are the time for each work alone. That is,

Let $t_1$ is the time to fill the fool , if the well work alone.

And, $t_2$ is the time to fill the fool, if the spring work alone.

It is given, $t=3$ hours. And the well takes, 8 hours less time than the spring. Therefore,

$t_1=t_2-8$

Substituting the values of $t$ and $t_1$ in the equation:

$\dfrac{1}{3}=\dfrac{1}{8}+\dfrac{1}{t_2}$

Isolate the $t_2$ term on the right:

$\dfrac{1}{3}-\dfrac{1}{8}=\dfrac{1}{t_2}$

$0.3333-0.125=\dfrac{1}{t_2}$

$0.2083=\dfrac{1}{t_2}$

Take the reciprocal:

$\dfrac{1}{0.2083}=t_2$

$t_2=\dfrac{1}{0.2083}=4.8$

So, the spring will take 4.8 hours to fill the pool alone.

Take the expression inside the radical sign and set that equal to $\ge=0$:

$4-9x\ge 0$

Solve for $x$:

Subtract $-4$

$-9x\ge -4$

Divide $-9$, dividing a negative number will change the inequality sign:

$x\le \dfrac{-4}{-9}$

$x\le \dfrac{4}{9}$

Now, you can write the domain in interval notation or can graph on a number line.

In interval notation, $\left(-\infty, \dfrac{4}{9}\right]$

The function is a rational function. Take the denominator and set that equal to zero:

$x^2+12x+35=0$

Solve this quadratic equation by factoring:

$(x+7)(x+5)=0$

Using the zero-product rule:

$x+7=0$ or $x+5=0$

$x=-7$ or $x=-5$

These are the excluded values. So, the domain is all the real numbers except $-7$ and $-5$.

We can write the domain in interval notation as

$(-\infty, -7)\cup (-7, -5) \cup (-5, \infty)$

Note that parentheses exclude the two numbers.

Let $x$ and $y$ are the amount of 25% and 35% alcohol respectively mixed together.

Total mixture is $20$ gallon:

$x+y=20$   Equation (1)

Looking at the amount of salt in each solution mixed and in the total mixture of 20 L:

$0.25x+0.35y=0.32*20$

Multiplying $100$ to remove the decimal point:

$25x+35y=640$   Equation (2)

Solve this and the first equation by elimination method:

Multiply equation (1) with $-25$:

$-25x-25y=-500$

Add this and equation (2):

$10y=140$

$y=14$

Now, substitute this in eqn.(1) to find $x$:

$x+14=20$

Solving for $x$:

$x=6$

So, to get the desired mixture, you need to mix 6 gallons of 25% and 14 gallons of 35% alcohol solution.

Let $x$ and $y$ are the numbers of \$8.50 and $9.75 brushes respectively.

So,

$x+y=45$

Total price:

$8.5x+9.75y=398.75$

Solve these two equations by elimination method, you get

Let $x$ is the speed of the current and $y$ is the speed of the boat in still water.

Use the formula:

$speed =\dfrac{distance}{time}$

For upstream:

$x-y=\dfrac{72}{6}$

For downstream:

$x+y=\dfrac{72}{4}$

Solving the two equations:

$x=$

$=\dfrac{3^{-4}x^{5(-4)}y^{-3(-4)}}{9xy^2}$

$=\dfrac{3^{-4}x^{-20}y^{12}}{9xy^2}$

$=\dfrac{x^{-20}y^{12}}{3^4\cdot 9xy^2}$

$=\dfrac{y^{12-2}}{81\cdot 9x^{1+20}}$

$=\dfrac{y^{10}}{729x^{21}}$

Use the formula, $(a-b)^2=a^2-2ab+b^2$

$=(3x)^2-2\cdot 3x\cdot 7y+(7y)^2$

$=9x^2-42xy+49y^2$

$=\dfrac{(x+5)(x-2)}{x+5}$

$=x-2$

$=8x^2y^6(8x^7y^3+3)$

$=m^2(m+4)-6(m+4)$

$=(m+4)(m^2-6)$

Write the equation in standard form of a quadratic equation:

$2k^2-9k+9=0$

Factor:

$2k^2-6k-3k+9=0$

$2k(k-3)-3(k-3)=0$

$(k-3)(2k-3)=0$

By zero-product rule:

$k-3=0$ or $2k-3=0$

Solve for k:

$k=3$ or $k=3/2$

Solution:

$\{3, 3/2\}$

First you need to factor, whatever you can factor.

$\dfrac{3}{k+2}-\dfrac{2}{(k+2)(k-2)}=\dfrac{1}{k-2}$

Least common denominator is $(k+2)(k-2)$

Multiply this in each term:

$\dfrac{3}{k+2}\cdot (k+2)(k-2)-\dfrac{2}{(k+2)(k-2)}\cdot (k+2)(k-2)$

$=\dfrac{1}{k-2}\cdot (k+2)(k-2)$

Now, cancel out whatever you can cancel out:

$3(k-2)-2=k+3$

Now, solve for $k$:

$3k-6-2=k+3$

$2k=11$

$k=\dfrac{11}{2}$

You need to factor whatever you can factor:

$=\dfrac{(m+7)(m-7)}{m+1}\div \dfrac{7-m}{m}$

Replace the multiplication by taking the reciprocal of the rational expression that follows the division sign:

$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{7-m}$

$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{-(m-7)}$

$=-\dfrac{m(m+7)}{m+1}$

Factor the expressions in the denominators:

=$\dfrac{5x}{(x+2y)(x-y)}-\dfrac{3x}{(x+6y)(x-y)}$

LCD$=(x+2y)(x-y)(x+6y)$

Multiply and divide the LCD in each term:

=$\dfrac{5x}{(x+2y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$

$-\dfrac{3x}{(x+6y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$

Cancel out the common factors between the numerator and the denominator;

=$\dfrac{5x}{(x+2y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$

$-\dfrac{3x}{(x+6y)(x-y)}\cdot \dfrac{(x+2y)(x-y)(x+6y)}{(x+2y)(x-y)(x+6y)}$

First write the radicals in exponential notation:

$\sqrt[5]{s^3}\:\sqrt[4]{s}$

$=s^{\frac{3}{5}}\:s^{\frac{1}{4}}$

$=s^{\frac{3}{5}+\frac{1}{4}}$

$=s^{\frac{3}{5}+\frac{1}{4}}$

$=s^{\frac{17}{20}}$

Write in radical form:

$=\sqrt[20]{s^{17}}$

Factoring what is inside the radical sign:

$\sqrt{108}$

=$\sqrt{36\cdot 3}$

=$6\sqrt{3}$

$-\sqrt[3]{-125a^6b^9c^{12}}$

=$-(-5)a^2b^3c^4$

=$5a^2b^3c^4$

$3x\sqrt[3]{xy^2}-2\sqrt[3]{8x^4y^2}$

$=3x\sqrt[3]{xy^2}-2\cdot 2 x\sqrt[3]{xy^2}$

$=3x\sqrt[3]{xy^2}-4x\sqrt[3]{xy^2}$

$=-x\sqrt[3]{xy^2}$

$\sqrt{x^2-4x+4}$;

Factor the expression inside the radical sign:

=$\sqrt{(x-2)^2}$

=$|x-2|$

Note that, I put the absolute value symbol as the square root radical sign represents a positive number (principal square root). It is given $x\ge 0$, so, $x$ includes $1$ (for example), so if you put $x=1$ in $x-2$, you get $1-2=-1$, which is negative. So, you must put the absolute value symbol to keep the square root positive.

$\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}}$

Multiply and divide the conjugate of the denominator:

$=\dfrac{(\sqrt{2}-\sqrt{3})}{(\sqrt{6}-\sqrt{5})} \cdot \dfrac{(\sqrt{6}+\sqrt{5})}{(\sqrt{6}+\sqrt{5})}$

The denominator, you can write as a difference of squares. And multiply the factors in the numerator using the distributive property.

$=\dfrac{\sqrt{2}\, \sqrt{6} + \sqrt{2}\,\sqrt{5}-\sqrt{3}\,\sqrt{6}-\sqrt{3}\,\sqrt{5}}{6-5}$

$=\dfrac{\sqrt{12} + \sqrt{10}-\sqrt{18}-\sqrt{15}}{1}$

  We can simplify, $\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}$

   and $\sqrt{18}=\sqrt{9\cdot 2}=3\sqrt{2}$

$=2\sqrt{3} + \sqrt{10}-3\sqrt{2}-\sqrt{15}$

It is given, the quadratic equation has only one solution. So, the discriminant must be zero.

Discriminant,

$b^2-4ac=0$

Multiply and divide the complex conjugate:

$=\dfrac{(4+3i)(5-3i)}{(5+3i)(5-3i)}$

$=\dfrac{20-12i+15i-9i^2}{5^2+9^2}$

$=\dfrac{20-12i+15i-9(-1)}{25+81}$

$=\dfrac{29+3i}{106}$

$=\dfrac{29}{106}+\dfrac{3i}{106}$

$=2i(-(4+i))^2$

$=2i(4+i)^2$

Expanding using the forumla $(a+b)^2 = a^2+2ab+b^2$:

$=2i(4^2+8i+i^2)$

$=2i(16+8i-1)$

$=2i(15+8i)$

$=30i+16i^2$

$=30i+16(-1)$

$=-16+30i$

$i^{42}=i^2=-1$

Note that 2 is the remainder after dividing 42 by 4.

$i^{17}=i^1=i$

Note that 1 is the remainder after dividing 17 by 4.

First, rewrite the equation so that both have same exponents:

$(4^2)^{2x+1}=(4^3)^{x+3}$

$(4)^{2(2x+1)}=(4)^{3(x+3)}$

Now, the bases are the same, so the exponents should be the same:

$2(2x+1)=3(x+3)$

Now, solve for $x$:

$4x+4=3x+9$

Solving for $x$:

$x=5$

Write the logarithmic equation in exponential form:

$4^3=2x+4$

$64=2x+4$

Solving for $x$:

$x=30$

You cannot write this equation with same base on both sides. So take log on both sides:

$\log(2^{x+3})=\log(5^x)$

$(x+3)\log(2)=x\log(5)$

$x\log(2)+3\log(2)=x\log(5)$

$x(\log(2)-\log(5))=2\log(2)$

Divide $(\log(2)-\log(5))$ to get $x$:

$x=\dfrac{2\log(2)}{\log(2)-\log(5)}$

After using the calculator:

$x=-1.513$

First combine the two logarithms on the left:

$\log_2(x(x-7))=3$

Now, write in exponential form:

$2^3=x(x-7)$

$8=x^2-7x$

This is a quadratic equation, write in standard form to solve:

$0=x^2-7x-8$

i.e., $x^2-7x-8=0$

Factoring:

$(x-7)(x+1)=0$

Applying zero-product rule:

$x-7=0$ or $x+1=0$

$x=7$ or $x=-1$

So, the solutions are

$x=-1, 7$

Write the coefficients of log as exponents using the formula $\log_a{x^p}=p\log x$

$=\log_p(x^3)+\log_p(y^{1/2})-\log_p(z^{3/2})$

Combine them:

$=\log_p\left(\dfrac{x^3y^{1/2}}{z^{3/2}}\right)$

You can also write this as

$=\log_p\left(\dfrac{x^3\sqrt{y}}{\sqrt{z^3}}\right)$

Use the change of base formula. Change the base to $e$ as there is an $e$ in the log. (You can also change the base to 10).

$\log_\pi(e)=\dfrac{\ln(e)}{\ln (\pi)}$

 $=\dfrac{1}{\ln (\pi)}$

 $=\dfrac{1}{1.1447}$

 $=0.874$, after rounding to the thousandth.

For the interest compounded continuously, you should use the following formula:

$A=P e^{rt}$

$t$ is the number of years, $A$ is the amount after $t$ years, $P$ is the Principal (initial deposit or loan if money is borrowed) and $r$ is the interest rate.

You are given:

$P=4000\:\$$; $r=3.5\%=3.5/100=0.035$; and $t=6$ years. You need to find the amount $A$ after $6$ years. Substituting, $P$,$r$ and $t$ in the equation:

$A=4000 e^{0.035\cdot 6}$

$A=4000 e^{0.035\cdot 6}$

$ =4000 e^{0.21}$

$ =4000 \cdot 1.2337$

$ =4934.71 \:\$$

This is one of the answer.

Next, you need to find how long, $(t)$ does it take to double the amount.

Amount is doubled, so, $A=2P$. Substituting this, $A$ and $r$ in the formula:

$2P=Pe^{0.035t}$

Divide out $P$, and solve for $t$:

$2=e^{0.035t}$

To solve, take $\ln$ on both sides:

$ln (2)= ln(e^{0.035t})$

$ln (2)= 0.035t$

$t=ln(2)/0.035=19.8$ years

In this problem, the interest is compounded annually, that is periodically. So the formula to be used is

$A=P\left(1+\dfrac{r}{n}\right)^{nt}$

where $n$ is the number of times per year, the interest is compounded.

Since the interest is compunded annually, $n=1$.

Interest rate is, $r=3\%=3/100=0.03$

You need to find the time $t$ to reach the amount, $35,000$.

Substitute, $A=35,000$ and solve for $t$:

$35000=29000(1.03)^t$

To solve this exponential equation, take log on both sides:

$\log(35000)=\log(29000(1.03)t)$

$\log(35000)=1.03t\log(29000)$

$t=\dfrac{\log(35000)}{1.03\log(29000)}=0.989$ years