# Thread: #1 Graph ellipse. #2 Subtracting polynomials.

1. ## #1 Graph ellipse. #2 Subtracting polynomials.

Stuck on two questions that was in a review section.

1) How can I graph the ellipse: x^2/4 + y^2/36 = 1?

2) How do I solve for x: 3x/2 - 3/4 = x/8?

Here's what I've done so far:

3x(4)/2 - 3(2)/4 = x/8

12x/2 - 6/4 = x/8

48x/8 - 12/8 = 4x/32.........this is where I got stuck when trying to get all the denominators the same.

2. Originally Posted by Mulya66
2) How do I solve for x: 3x/2 - 3/4 = 8/x?

Here's what I've done so far:

3x(4)/2 - 3(2)/4 = x/8

12x/2 - 6/4 = x/8

48x/8 - 12/8 = 4x/32.........this is where I got stuck when trying to get all the denominators the same.

Hint: Multiply everything by 8x. (See what happens).

3. Originally Posted by Mulya66
Stuck on two questions ...

3x/2 - 3/4 = 8/x

12x/2 - 6/4 = x/8
Hello,

you unfortunately turned over the fraction...

Use the TPH's hint...

EB

4. Hello, Mulya66!

Graph the ellipse: .x²/4 + y²/36 .= .1

First of all, the ellipse is centered at the origin.

Recall that the denominators give you the "dimensions" of the ellipse.

Since a² = 4, a = ±2.
. . The ellipse extends 2 units to the left and right of center.

Since b² = 36, b = ±6.
. . The ellipse extends 6 units up and down from the center.

2) Solve: .3x/2 - 3/4 .= .8/x

Multiply through by the LCM, 4x

. . . . 3x . - . .3 . . . . . 8
. . 4x·--- - 4x·-- .= .4x·-- . . 6x² - 3x .= .32
. . . . .2 . - . .4 . - . - . x

5. Originally Posted by Soroban
Hello, Mulya66!

First of all, the ellipse is centered at the origin.

Recall that the denominators give you the "dimensions" of the ellipse.

Since a² = 4, a = ±2.
. . The ellipse extends 2 units to the left and right of center.

Since b² = 36, b = ±6.
. . The ellipse extends 6 units up and down from the center.

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Oh okay, so a^2 is for left and right and b^2 is up and down.

#2 was actually x/8, sorry. But last night, I got x = 6/11 ~ 0.5455.

Thanks a lot, everyone. =)