please help: Solve -3(x + 2) - 7 = x/2 + 5 for x

1) Solve for x: -3(x + 2) - 7 = x/2 + 5

Here's what I've done so far:

. . .-3x - 6 - 7 = x/2 + 5
. . .-3x - 13 = x/2 + 5
. . . . . . .-5. . . . . . .-5
. . .-----------------------
. . .-3x - 18 = x/2

What do I do next?

2) Find two solutions for x - 4y - 12 = 0

1) Try multiplying through by 2, to get:

. . . . .-6x - 36 = x

Then add 6x to both sides, etc.

2) Pick x-values. Plug them in. Solve for the corresponding y-values. The ordered pair, (x, y), is a solution to the equation.

Eliz.

-3x - 18 = x/2

What do I do next?

Click to expand...

multiply both sides by 2 to clear the fraction ...

-6x - 36 = x

add 6x to both sides ...

-36 = 7x

can you finish up?

Find two solutions for x - 4y - 12 = 0

Click to expand...

pick any value for x, then solve for the y-value that makes the equation true

I'll do one for you ... let x = 0

0 - 4y - 12 = 0
-4y = 12
y = -3

so, the ordered pair (0,-3) is a solution for the equation

now ... you pick a different value for x and find another ordered pair solution.

simple, huh?

Re: please help with these problems

Hello, nikki26!

An error in your first step . . .

Edit: And I copied the problem wrong! . . . (corrected now)

By the way, notice how it's easier to read with spaces.

Solve for: \(\displaystyle \,-3(x\,+\,2)\,-\,7\:=\:\frac{x}{2}\,+\,5\)

Click to expand...

We have: \(\displaystyle \,-3x\,-\,6\,-\,7 \:= \:\frac{x}{2}\,+\,5\)

. . . . . . . . . .\(\displaystyle \,-3x\,-\,13\:=\:\frac{x}{2}\,+\,5\)
. . . . . . . . . . . . .\(\displaystyle +13\) . . . . \(\displaystyle +13\)

. . . . . . . . . . . . .\(\displaystyle \,-3x\:=\:\frac{x}{2}\,+\,18\)


Now get the \(\displaystyle x\)'s on one side:

. . . \(\displaystyle \,-3x\,-\,\frac{x}{2}\:=\:18\)

. . . . . . \(\displaystyle \,-\frac{7}{2}x\:=\:18\)


Multiply both sides by\(\displaystyle \,-\frac{2}{7}:\)

. . .\(\displaystyle \left(-\frac{2}{7}\right)\left(-\frac{7}{2}x\right) \:=\:\left(-\frac{2}{7}\right)(18)\)

Therefore: \(\displaystyle \,x\:=\:-\frac{36}{7}\)

Find two solutions for: \(\displaystyle x\,-\,4y\,-\,12\:=\:0\)

Click to expand...

Here's a cute trick . . . Write it as: \(\displaystyle \,x\,-\,4y\:=\:12\)

Let \(\displaystyle x\,=\,0\;\;\Rightarrow\;\;0 \,-\,4y\:=\:12\;\;\Rightarrow\;\;-4y\:=\:12\;\;\Rightarrow\;\; y\,=\,-3\)
\(\displaystyle \;\;\)Hence, we have: \(\displaystyle \,(x,y)\,=\,(0,\,-3)\)

Let \(\displaystyle y\,=\,0\;\;\Rightarrow\;\;x\,-\,0\:=\:12\;\;\Rightarrow\;\; x\,=\,12\)
\(\displaystyle \;\;\) Hence, we have: \(\displaystyle \,(x,y)\,=\,(12,\,0)\)


Thanks for the heads-up, skeeter!