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

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

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

$\displaystyle x = 2$

Literal Equations

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

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

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

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

No Solution

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

All Real Numbers

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

No Solution

5.) Solve for y:

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

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

Inequalities

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

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

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

No Solution

Absolute Value Equations

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

$\displaystyle x = \frac{8}{3}$
$\displaystyle x = -4$

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

No Solution

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

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

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

$\displaystyle x = -6$
$\displaystyle x = 18$

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

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

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

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

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

No Solution

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

All Real Numbers

Order Of Operations

1.) Simplify:

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

$\displaystyle 47$

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

$\displaystyle -21$

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

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

$\displaystyle \frac{11}{8}$

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

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.

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

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

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.

$\displaystyle x = 2$

Literal Equations

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

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

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

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

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

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

No Solution

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

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

All Real Numbers

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

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

No Solution

5.) Solve for y:

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

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 $\displaystyle \frac{3x - 8}{2} = \frac{3x}{2} - \frac{8}{2} = \frac{3}{2}x - 4$.

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

Inequalities

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

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

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

No Solution

Absolute Value Equations

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

$\displaystyle x = \frac{8}{3}$
$\displaystyle x = -4$

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

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

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

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

$\displaystyle x = -6$
$\displaystyle x = 18$

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

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

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

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

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

No Solution

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

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:

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

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

$\displaystyle 47$

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

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

$\displaystyle -21$

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

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

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

$\displaystyle \frac{11}{8}$