Question

Solve for x in the proportion 3 over the quantity x plus 2 equals 6 over the quantity x minus 1.

x = −5

x = 1

x = 4

x = 5

Answer 1

first step is to write the equation \[\huge \frac 3{x+2} = \frac 6{x-1}\]

Answer 2

then cross multiply. what do you get?

Answer 3

12 and 3

Answer 4

\[3(x - 1) = ?\]
\[6(x + 2) =?\]

Answer 5

-3 and 12?

Answer 6

Don't forget about the 'x' part

When you multiply 3 times (x - 1)

you multiply 3 by x....and then 3 by -1

so

\[3(x - 1) = 3x - 3\]

Right?

Answer 7

So use that same logic with

\[6(x + 2) = ?\]

Answer 8

Did that make sense..??

Answer 9

So 6x + 12

Answer 10

Yeah but how do i get the answeR?

Answer 11

Yes @ahoward79 6x + 12

so combining what I showed you and what you got...now you have

\[3x - 3 = 6x + 12\]

can you solve for 'x' from here?

Answer 12

Uhh, can you show me :/

Answer 13

Sure...don't get discouraged! :)...but let me ask you...do you understand everything up until now..?

Answer 14

I'll break it down just in case

Step 1: Write the equation

\[\frac{ 3 }{ x + 2 } = \frac{ 6 }{ x - 1 }\]

Step 2...when you have 2 fractions like this equal to each other...you cross multiply

Multiply

\[3(x - 1) = 6(x + 2)\]

*see how it crossed?

Step 3 Distribute

\[3(x - 1) = (3 \times x)+(3 \times -1) = 3x - 3\]
and
\[6(x + 2) = (6 \times x)+(6 \times 2) = 6x + 12\]

so now you have

\[3x - 3 = 6x + 12\]

Step 4: Combining like terms...

you can only combine numbers with an 'x' and numbers that are constants...so

lets get all the numbers with an 'x' on 1 side....and the other numbers on the other

Lets start by adding 3 to both sides

\[3x -3 + 3 = 6x + 12 + 3\]
becomes
\[3x = 6x + 15\]

Now lets subtract 6x from both sides

\[3x - 6x = 6x - 6x + 15\]
becomes

\[-3x = 15\]

can you solve for 'x' from here?

Answer 15

LOL