# Thread: absolute values

1. ## absolute values

I put off this far too long and am now finding myself struggling on a number of problems. I'd appreciate any help anyone could give, and for the love of god, spell it out, show your work, as if you were trying to explain to someone with a very, very poor understanding of the subject. In fact, that's what you're doing.

3. An appliance store is having a sale on small TV's and stereos. One day a salesperson sells 3 of the TVs and 4 stereos for 2530$The next day the salesperson sells 4 of the same TV's and 3 of the same stereos for 2510$. What are the prices of a TV and a stereo?

11. Find the solution to the equation
{3x+1} = -2

9.Solve each absolute value inquailty and graph the solution the {}'s represent absoulte value bars

A) {2x+3} < 15

B) {2x+5} - 7 > -6

C) {2x-3} + 4 < -10

10. Graph each linear inequality in a rectangular coordinate system

A) 3x - 4y > 12

B)
2x - y < 4
X + Y > 5

Thanks for your help guys - it probably means more than you realize.

2. Originally Posted by bonapartist
Y + X = 4
3x + 3y = -6 ......... in this one, it seems impossible to me to eliminate one of the variables without taking the other, so what am I supposed to do?
Multiply the first equation by 3 to get,
3x+3y=12
3x+3y=-6

Which is impossible

3. Originally Posted by ThePerfectHacker
Multiply the first equation by 3 to get,
3x+3y=12
3x+3y=-6

Which is impossible
Should I write "impossible"? "incosistent"?

4. Originally Posted by bonapartist
Should I write "impossible"? "incosistent"?
no. write "no solution: parallel lines" since that's what you have here, parallel lines. i hope you see why

5. Hello, bonapartist!

Here's some help . . .

2. Solve the system by substitution

. . $\Large\begin{array}{cccc}\frac{x}{8} + \frac{3y}{4} & = & \frac{19}{4} & [1] \\ \text{-}\frac{x}{2} + \frac{3y}{4} & = & \frac{1}{2} & [2]\end{array}$
First, get rid of the fractions.

. . $\begin{array}{ccccc}\text{Multiply [1] by 8:} & x + 6y & = & 38 & [3] \\ \text{Multiply [2] by 4:} & \text{-}2x + 3y & = & 2 & [4]\end{array}$

Solve [3] for $x\!:\;\;x \:=\:38-6y$

Substitute into [4]: . $\text{-}2(38-6y) + 3y \:=\:2\quad\Rightarrow\quad\text{-}76 + 12y + 3y \:=\:2$

. . $15y \,=\,78\quad\Rightarrow\quad\boxed{ y \,=\,\frac{26}{5}}$

Substitute into [3]: . $x + 6\left(\frac{26}{5}\right) \:=\:38$

. . $x \:=\:38 - \frac{156}{5}\quad\Rightarrow\quad\boxed{ x\,=\,\frac{34}{5}}$

4. Solve the system by Elimination

. . $\begin{array}{cccc}x + y & = & 4 & [1] \\
3x + 3y & = & \text{-}6 & [2]\end{array}$
What happened is not your fault . . .

$\begin{array}{cccc}\text{Divide [2] by -3:} & \text{-}x - y & = & 2 \\ \text{Add [1]:} & x + y & = & 4 \\
\text{and we get:} & 0 & = & 6 \end{array}$

We get a statement which is obviously not true.
This means that the sytem has no solution . . . it is inconsistent.

6. Originally Posted by Soroban
Hello, bonapartist!

Here's some help . . .

First, get rid of the fractions.

. . $\begin{array}{ccccc}\text{Multiply [1] by 8:} & x + 6y & = & 38 & [3] \\ \text{Multiply [2] by 4:} & \text{-}2x + 3y & = & 2 & [4]\end{array}$

Solve [3] for $x\!:\;\;x \:=\:38-6y$

Substitute into [4]: . $\text{-}2(38-6y) + 3y \:=\:2\quad\Rightarrow\quad\text{-}76 + 12y + 3y \:=\:2$

. . $15y \,=\,78\quad\Rightarrow\quad\boxed{ y \,=\,\frac{26}{5}}$

Substitute into [3]: . $x + 6\left(\frac{26}{5}\right) \:=\:38$

. . $x \:=\:38 - \frac{156}{5}\quad\Rightarrow\quad\boxed{ x\,=\,\frac{34}{5}}$

What happened is not your fault . . .

$\begin{array}{cccc}\text{Divide [2] by -3:} & \text{-}x - y & = & 2 \\ \text{Add [1]:} & x + y & = & 4 \\
\text{and we get:} & 0 & = & 6 \end{array}$

We get a statement which is obviously not true.
This means that the sytem has no solution . . . it is inconsistent.

Thanks alot! As soon as I saw the fractions in the problem my idea of how to solve the problem fell apart and I didn't know what to do, and now I understand. This helps alot. I appreciate this alot guys - you're talking to an English major with zero inclination toward math and I appreciate it.

7. Originally Posted by bonapartist
5. Solve the system by A) Elimination and B) Cramers rule or Matrices

X + 2y + 3z = -5
2x + Y + Z = 1
X + Y - Z = 8
A) Elimination
First we'll eliminate $z$:
Multiply the second equation by -3 and add with the first:
$-5x-y=-8$ (1)
Add the second and the third equations:
$3x+2y=9$ (2)
Now, we'll eliminate $y$:
Multiply (1) by -2 and add with (2)
$-7x=-7\Rightarrow x=1$
Replace $x=1$ in (1) $\Rightarrow y=3$.
Replace $x$ and $y$ in an equation of the initial system $\Rightarrow z=-4$
So, the solution is $x=1,y=3,z=-4$

B) Cramer's rule
Let $A=\begin{pmatrix}1 & 2 & 3\\2 & 1 & 1\\1 & 1 & -1\end{pmatrix}$ be the matrix of system.
We have $d=\det A=7$.
$d_x=\begin{vmatrix}-5 & 2 & 3\\1 & 1 & 1\\8 & 1 & -1\end{vmatrix}=7$, $d_y=\begin{vmatrix}1 & -5 & 3\\2 & 1 & 1\\1 & 8 & -1\end{vmatrix}=21$, $d_z=\begin{vmatrix}1 & 2 & -5\\2 & 1 & 1\\1 & 1 & 8\end{vmatrix}=-28$

Now, $\displaystyle x=\frac{d_x}{d}=\frac{7}{7}=1$, $\displaystyle y=\frac{d_y}{d}=\frac{21}{7}=3$, $\displaystyle z=\frac{d_z}{d}=-\frac{28}{7}=-4$

C) Matrix method
The system can be written as $A\cdot X=B$, where $A$ is the matrix of the system (see B)), $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$, $B=\begin{pmatrix}-5\\1\\8\end{pmatrix}$

$\det A=7\Rightarrow A$ is invertible.
Multiplying both members of the matrix equation, to the left, with $A^{-1}$ we have $X=A^{-1}\cdot B=\begin{pmatrix}1\\3\\-4\end{pmatrix}$

(I hope you know how the find $A^{-1}$).

8. 6. Solve the linear inequality

3(2x-1) - 2(x-4) > 7 + 2(3+4x)
$3(2x-1)-2(x-4)>7+2(3+4x)\Leftrightarrow 6x-3-2x+8>7+6+8x\Leftrightarrow$
$\Leftrightarrow -4x>8\Leftrightarrow x<-2\Leftrightarrow x\in(-\infty ,-2)$

7. Solve each compound inequality

A) 2x-5 > -1 or 3x < 3

B) 5x + 3 < 18 and 2x - 7 < 5
A) $2x-5>-1\Rightarrow x>2\Rightarrow x\in (2,\infty )$
$3x<3\Rightarrow x<1\Rightarrow x\in (-\infty ,1)$
Then $x\in (-\infty ,1)\cup (2,\infty )$

B) $5x+3<18\Rightarrow x<3\Rightarrow x\in (-\infty ,3)$
$2x-7<5\Rightarrow x<6\Rightarrow x\in (-\infty ,6)$
Then $x\in (-\infty, 3)\cap (-\infty ,6)=(-\infty ,3)$

8. Find the solution set of this equation: 4x-3 = 7x + 9 (both {4x-3} and {7x + 9} are absolute values but I haven't the slightest how to make that on a keyboard.)
$|4x-3|=|7x+9|$
We have $|a|=|b|\Leftrightarrow a=\pm b$
If $4x-3=7x+9\Rightarrow x=-4$
If $4x-3=-(7x+9)\Rightarrow\displaystyle x=-\frac{6}{11}$

9. 10. Graph each linear inequality in a rectangular coordinate system

A) 3x - 4y > 12

B)
2x - y < 4
X + Y > 5
Here are the graphs.

10. I have no idea how to solve

3. An appliance store is having a sale on small TV's and stereos. One day a salesperson sells 3 of the TVs and 4 stereos for 2530$The next day the salesperson sells 4 of the same TV's and 3 of the same stereos for 2510$. What are the prices of a TV and a stereo?

Can anyone help?

11. Originally Posted by bonapartist
I have no idea how to solve

3. An appliance store is having a sale on small TV's and stereos. One day a salesperson sells 3 of the TVs and 4 stereos for 2530$The next day the salesperson sells 4 of the same TV's and 3 of the same stereos for 2510$. What are the prices of a TV and a stereo?

Can anyone help?
Let $t$ be the price of the TV
Let $s$ be the price of the stereo

he sold 3 TV's and 4 stereos for 2530, so we have:

$3t + 4s = 2530$

then he sold 4 TV's and 3 stereos for 2510, so we have:

$4t + 3s = 2510$

thus all we need to do to find the prices is solve the system:

$3t + 4s = 2530$ ................(1)
$4t + 3s = 2510$ ................(2)

i leave solving it to you

12. Originally Posted by Jhevon
Let $t$ be the price of the TV
Let $s$ be the price of the stereo

he sold 3 TV's and 4 stereos for 2530, so we have:

$3t + 4s = 2530$

then he sold 4 TV's and 3 stereos for 2510, so we have:

$4t + 3s = 2510$

thus all we need to do to find the prices is solve the system:

$3t + 4s = 2530$ ................(1)
$4t + 3s = 2510$ ................(2)

i leave solving it to you
I hate to say it, but I have no idea what to do from here.

13. Originally Posted by bonapartist
I hate to say it, but I have no idea what to do from here.
As this is a take home test I expect you to learn from this, NOT simply copy it. (In other words, see the solution and rederive it yourself!!) Presumably you should already know how to do this, else you wouldn't have been tested on it.

$3t + 4s = 2530$
$4t + 3s = 2510$

There are many solution methods, this one is called the "substitution method."

Solve one equation (say, the top one) for one of the unknowns (say, t):
$3t + 4s = 2530$

$3t = 2530 - 4s$

$t = \frac{2530 - 4s}{3}$

Now insert this value into the other equation and solve for the remaining variable:
$4 \left ( \frac{2530 - 4s}{3} \right ) + 3s = 2510$ <-- Multiply both sides by 3

$4(2530 - 4s) + 9s = 7530$

$10120 - 16s + 9s = 7530$

$10120 -7s = 7530$

$-7s = -2590$

$s = 370$

Now plug this value of s into the (solved) t equation:
$t = \frac{2530 - 4(370)}{3} = 350$

So the solution to the system of equations is
$s = 370$ and $t = 350$

-Dan