# Math Help - linear equations

1. ## linear equations

I need some help with the following problems. If you could, please show me how to find the correct answers, since I know what the correct answers are already. I'm taking my Midterm on Tuesday. So, I would like to see some replies by tomorrow. I will also post some new problems tomorrow and maybe even on Monday. Thanks in advance for those who can help me.

Linear Equations

1.) $
\frac{1}{2}x -\frac{7}{4} = -\frac{3}{4}x + 5
$

$
x = \frac{27}{5}
$

2.) $
\frac{2}{3}x -\frac{11}{6} = 1 -\frac{3}{4}x
$

$
x = 2
$

Literal Equations

1.) The surface area of a cylinder is given by the following equation. Solve it for h:

$
S = 2(pi)rh + 2(pi)r^2
$

$
h = \frac{S - 2(pi)r^2}{2(pi)r}
$

2.) $
|5x - 1| = -4
$

No Solution

3.) $
|3x - 2| \geq -2
$

All Real Numbers

4.) $
|5x - 2| < -4
$

No Solution

5.) Solve for y:

$
x = \frac{2}{3}(y + 4)
$

$
y = \frac{3}{2}x - 4
$

Inequalities

1.) $
2x + 1 \leq 6x - 1
$
OR $
4x - 7 \leq -11
$

$
x \geq \frac{1}{2}
$
OR $
x \leq -1
$

2.) $
-1 \leq -2x + 1 < 5
$

No Solution

Absolute Value Equations

1.) $
|3x + 2| = 10
$

$
x = \frac{8}{3}
$

$
x = -4
$

2.) $
2|5x - 1| + 12 = 4
$

No Solution

3.) $
2|3x - 4| + 4 = 8
$

$
x = \frac{2}{3}
$

$
x = 2
$

4.) $
|\frac{1}{2}x - 3| + 5 = 11
$

$
x = -6
$

$
x = 18
$

5.) $
|3x + 2| < 8
$

$
x < 2
$
AND $
x > \frac{-10}{3}
$

6.) $
2|x - 3| + 5 \geq 17
$

$
x \geq 9
$
OR $
x \leq -3
$

7.) $
|5x - 7| < 5 -3
$

No Solution

7.) $
4|3x + 5| + 9 > 1
$

All Real Numbers

Order Of Operations

1.) Simplify:

$
-3^2 + [4(2^2 - 1) + 6]3 -(-2)
$

$
47
$

2.) Evaluate $x^3 + x^2 + x$ when x = -3:

$
-21
$

3.) Evaluate the following when a = 2, b = -3, and c = -4:

$
\frac{a^3 + b^2 + c^2}{|abc|} =
$

$
\frac{11}{8}
$

2. Thats' quite a mountain of work, sport! I'll give a few tips for the first few.
Originally Posted by Spoonman
[snip]
Linear Equations

1.) $
\frac{1}{2}x -\frac{7}{4} = -\frac{3}{4}x + 5
$

Mr F says: Multiply the whole of the equation through by 4 and solve the resulting much simpler linear equation. Note: 4 is the lowest common multiple of 2 and 4, the numbers in the denominators of the fractions.

$
x = \frac{27}{5}
$

2.) $
\frac{2}{3}x -\frac{11}{6} = 1 -\frac{3}{4}x
$

Mr F says: Multiply the whole of the equation through by 12 and solve the resulting much simpler linear equation. Note: 12 is the lowest common multiple of 3, 6 and 4, the numbers in the denominators of the fractions.

$
x = 2
$

Literal Equations

1.) The surface area of a cylinder is given by the following equation. Solve it for h:

$
S = 2(pi)rh + 2(pi)r^2
$

Mr F says: Step 1: Subtract $2 \pi r^2$ from both sides. Step 2: Divide both sides by $2 \pi r$.

$
h = \frac{S - 2(pi)r^2}{2(pi)r}
$

2.) $
|5x - 1| = -4
$

Mr F says: Can a magnitude ever be negative? Therefore .....

No Solution

3.) $
|3x - 2| \geq -2
$

Mr F says: A magnitude is always gretare than or equal to 0. Therefore ......

All Real Numbers

4.) $
|5x - 2| < -4
$

Mr F says: Can a magnitude ever be negative? Therefore .....

No Solution

5.) Solve for y:

$
x = \frac{2}{3}(y + 4)
$

Mr F says: Step 1: Multiply the whole equation through by 3. Step2: Expand the right hand side. Step 3: Subtract 8 from both sides. Step 4: Make y the subject. Note that $\frac{3x - 8}{2} = \frac{3x}{2} - \frac{8}{2} = \frac{3}{2}x - 4$.

$
y = \frac{3}{2}x - 4
$

Inequalities

1.) $
2x + 1 \leq 6x - 1
$
OR $
4x - 7 \leq -11
$

$
x \geq \frac{1}{2}
$
OR $
x \leq -1
$

2.) $
-1 \leq -2x + 1 < 5
$

No Solution

Absolute Value Equations

1.) $
|3x + 2| = 10
$

$
x = \frac{8}{3}
$

$
x = -4
$

2.) $
2|5x - 1| + 12 = 4
$

Mr F says: Step 1: Subtract 12 from both sides. Step 2: Divide both sides by 2. Step 3: Can a magnitude ever be negative? Therefore .....

No Solution

3.) $
2|3x - 4| + 4 = 8
$

$
x = \frac{2}{3}
$

$
x = 2
$

4.) $
|\frac{1}{2}x - 3| + 5 = 11
$

$
x = -6
$

$
x = 18
$

5.) $
|3x + 2| < 8
$

$
x < 2
$
AND $
x > \frac{-10}{3}
$

6.) $
2|x - 3| + 5 \geq 17
$

$
x \geq 9
$
OR $
x \leq -3
$

7.) $
|5x - 7| < 5 -3
$

No Solution

7.) $
4|3x + 5| + 9 > 1
$

Mr F says: Step 1: Subtract 9 from both sides. Step 2: Divide both sides by 4. Step 3: A magnitude is always gretare than or equal to zero. Therefore .....

All Real Numbers

Order Of Operations

1.) Simplify:

$
-3^2 + [4(2^2 - 1) + 6]3 -(-2)
$

Mr F says: [tex]= -9 + [4(4 - 1) + 6]3 + 2 = ....

$
47
$

2.) Evaluate $x^3 + x^2 + x$ when x = -3:

Mr F says: [tex]= (-3)^3 + (-3)^2 + (-3) = -27 + 9 - 3 = ....

$
-21
$

3.) Evaluate the following when a = 2, b = -3, and c = -4:

$
\frac{a^3 + b^2 + c^2}{|abc|} =
$

Mr F says: Substitute the given values. Note that $2^3 = 8, (-3)^2 = 9, (-4)^2 = 16, |(2)(-3)(-4)| = |24| = 24$ ......

$