I need to find "x" is the algebraic equation: What is x in 5-2x+9-3=3x+2

I submitted this question earlier, but I did not get an answer because I accidentally left off the equal sign. Thank you.

I need to find "x" is the algebraic equation: What is x in 5-2x+9-3=3x+2

I submitted this question earlier, but I did not get an answer because I accidentally left off the equal sign. Thank you.

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5-2x+9-3=3x+2

Because you are solving for x, you want all the x's on one side, so your final equation is x = #. You do this by getting all the x's on one side of the equation, and all the numbers on the other. You can move a number to the other side of the equation by performing the opposite operation (add <--> subtract; multiply <--> divide)

**remember: whatever you do to one side of the equation, you MUST do to the other side, so it stays equal on both sides.

Let's start by adding 2x to each side, so that all the x's are on the right side. We now have:

5+9-3=3x+2+2x

Next subract 2 on both sides, so the right side is only in terms of x:

5+9-3-2=3x+2x

We can now perform the operations on the left, and then the right:

Left: 5+9-3-2= 9

Right: 3x+2x=5x

9=5x

We now divide by 5 to get x all by itself:

9/5 = x

5-2x+9-3. You can use the associative property of addition to simplify it and add 5+9 and subtract 3, which gives you 11. Now you are left with -2x+11=3x+2. From there, you need to try to get both the x's to the same side of the equation so that you can solve for x. You can do this by adding 2x to both sides, resulting in 11=5x+2. Now you can simply solve for x using the basic rules for solving a two step equation.

My first step I would do is move (-2x) to the right side of the equals sign so I have a positive value for X. I can move it over by addition. Now that you have all the values of x on one side of the problem you should move all non values of x to the side of the equals sign.

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