Tuesday
May 3, 2016

# Homework Help: Algebra II

Posted by Jessica on Saturday, August 18, 2007 at 8:49pm.

I need some help with some problems as well as need some checked. Could you please help me? Thanks!

1. How do you graph y=1/x
2. How do I the x and y intecepts using a graph?

Solve and check the linear equation.

9. (-4x-2)+7=-3(x+3) I got 14

10. -2[7x-7-6(x+1)]=2x+5 I got 21/4

Solve the equation.

13. (x+7)/4=2-(x-1)/6 I am not too sure how to do this one.

Find all the values of x satisfying the given conditions.

16. y1= (x+6)/3, y2=(x+8)/6, and y1=y2 This one I dont understand.

First write the value(s) that make the denominator(s) zero. Then solve the equation.

19. (x-8)/2x +5= (x+6)/x This one I also dont understand.

Determine whether the equation is an identity, a conditional equation, or inconsistent equation.

24. -2(x+7)+52=4x-6(x+3) I got Inconsistent

25. (3x+2)/4 +2= -7x/2 I got inconsistent

I need some help with some problems as well as need some checked. Could you please help me? Thanks!

1. How do you graph y=1/x make a table of values, x vs y. Plot the values
2. How do I the x and y intecepts using a graph? The question makes no sense as typed

Solve and check the linear equation.

9. (-4x-2)+7=-3(x+3) I got 14
correct
10. -2[7x-7-6(x+1)]=2x+5 I got 21/4
correct
Solve the equation.

13. (x+7)/4=2-(x-1)/6 I am not too sure how to do this one.
multiply both sides by 12, then reduce the equation
Find all the values of x satisfying the given conditions.

16. y1= (x+6)/3, y2=(x+8)/6, and y1=y2 This one I dont understand. If y1=12, then (x-6)/3=(x+8)/6 . multiply both sides of the equation by six, then gather terms.

Sorry, I meant to ask how do I find the x and y intercepts using a graph.

Hi, I need some help with some inequalitites. Thanks!!

Use graphs to find the set.

1.(-9,0) intersection [-4,] I got [-4,0)

2. (-9,0) U [-4,10] I got (-9,10]

Solve the linear inequality. Other than (empty set, use interval notation to express the solution set and graph the solution set on a number line.

4. 21x -21>3(6x-2) I got x<5, graph would be <------)5

5. 5(4x+7)-4x<4(8+4x)-6 I got 0>9, but I dont know how the graph would be.

6. 6(x+4)> or equal to 5(x-3)+x I got 0 > greater than or equal to -39, but dont know how the graph would be.

Solve the compound inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

10. -24 < or equal to -5x+1 < -9 I am not too sure how to solve this.

14. 3 < or equal to (8/5x)-5<11 one this one am I suppose to multiply each side by 5 to undo the fraction.

Solve the absolute value inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

20. |x+9| -4 < or equal to I got [-5 -13] and the graph would be <----)-13 -5---->

21. |10y+30/3| < 10 This one I dont understand what to.

Hi, I need some help with some inequalitites. Thanks!!

Use graphs to find the set.

1.(-9,0) intersection [-4,] I got [-4,0)

2. (-9,0) U [-4,10] I got (-9,10]

Solve the linear inequality. Other than (empty set, use interval notation to express the solution set and graph the solution set on a number line.

4. 21x -21>3(6x-2) I got x<5, graph would be <------)5

5. 5(4x+7)-4x<4(8+4x)-6 I got 0>9, but I dont know how the graph would be.

6. 6(x+4)> or equal to 5(x-3)+x I got 0 > greater than or equal to -39, but dont know how the graph would be.

Solve the compound inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

10. -24 < or equal to -5x+1 < -9 I am not too sure how to solve this.

14. 3 < or equal to (8/5x)-5<11 one this one am I suppose to multiply each side by 5 to undo the fraction.

Solve the absolute value inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

20. |x+9| -4 < or equal to I got [-5 -13] and the graph would be <----)-13 -5---->

21. |10y+30/3| < 10 This one I dont understand what to.

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