- Thu Sep 10, 2009 2:34 am
- Forum: Intermediate Algebra
- Topic: solving x + 2y = 6 for y, and then graphing
- Replies:
**3** - Views:
**1559**

I know the y-intercept is 3, but when I try to graph even numbers, it gets confusing.

- Sat Sep 05, 2009 2:24 am
- Forum: Intermediate Algebra
- Topic: solving x + 2y = 6 for y, and then graphing
- Replies:
**3** - Views:
**1559**

solving x + 2y = 6 for y, and then graphing

I tried to solve for y, and got y= -x/2 + 3. This is hard to graph.

I tried to solve for y, and got y= -x/2 + 3. This is hard to graph.

- Fri Sep 04, 2009 12:36 am
- Forum: News
- Topic: New Lessons?
- Replies:
**10** - Views:
**15764**

You definitely need Geometry for all the lost 8th Graders out there.

- Wed Aug 05, 2009 6:10 pm
- Forum: Intermediate Algebra
- Topic: Connections: 5X 1st of 3 consec. even integers is 14 more
- Replies:
**7** - Views:
**2388**

So it's 5x = 3(x + 2) +14?

X = 10

X = 10

- Tue Aug 04, 2009 5:34 pm
- Forum: Intermediate Algebra
- Topic: Connections: 5X 1st of 3 consec. even integers is 14 more
- Replies:
**7** - Views:
**2388**

- Mon Aug 03, 2009 11:16 pm
- Forum: Intermediate Algebra
- Topic: Connections: 5X 1st of 3 consec. even integers is 14 more
- Replies:
**7** - Views:
**2388**

1: X

2: Stuck here. I don't understand.

2: Stuck here. I don't understand.

- Mon Aug 03, 2009 6:11 pm
- Forum: Intermediate Algebra
- Topic: Connections: 5X 1st of 3 consec. even integers is 14 more
- Replies:
**7** - Views:
**2388**

This problem baffles me.

Do I solve for X? Or do I solve for Y? Or something completely different?

The equation I have so far is 5x(first integer) = 3y(second integer) + 4.5 times the first of 3 consecutive even integers is 14 more than 3 times the second. What is the third consecutive integer?

Do I solve for X? Or do I solve for Y? Or something completely different?

- Mon Aug 03, 2009 3:38 am
- Forum: Intermediate Algebra
- Topic: [SPLIT] length of a rectangle is 3 more than twice its width
- Replies:
**6** - Views:
**3096**

- Sun Aug 02, 2009 7:27 pm
- Forum: Intermediate Algebra
- Topic: [SPLIT] length of a rectangle is 3 more than twice its width
- Replies:
**6** - Views:
**3096**

1. W

2. 2w + 3

3. P = 2(w) + 2L

4. P = 2(w) + 2(2w + 3)

5. 2W + 4W + 3 = 48

6W + 3 = 48

48 - 3 = 45

6w = 45

/6 /6

w = 7.5

2. 2w + 3

3. P = 2(w) + 2L

4. P = 2(w) + 2(2w + 3)

5. 2W + 4W + 3 = 48

6W + 3 = 48

48 - 3 = 45

6w = 45

/6 /6

w = 7.5

- Sat Aug 01, 2009 6:52 pm
- Forum: Intermediate Algebra
- Topic: [SPLIT] length of a rectangle is 3 more than twice its width
- Replies:
**6** - Views:
**3096**

I tried again, and came up with 2w + 3 + 2w = 48

Solve for W = 11.25

Solve for W = 11.25