Question

Help me pleaseeee. It my math hw you don’t have to do all of them at least do 4 please

Answer 1

#5 -
`x=-2`

#6 -
`x=-4`

#7 -
`x=2`

#8 -
`x=10`

Explanation:

Solve the following multi-step equations.

#5 -
`-10-7x=-3x-2`

#6 -
`-13-4x=x+7`

#7 -
`x-2=10-5x`

#8 -
`3x-1=4x-11`

`\hrulefill`

I will use the SCAM method to solve these multi-step equations.

`\mathbb{S}\text{implify each side of the equation} \\\\\\\mathbb{C}\text{ombine like terms, collect variables on one side }\\\\\\\mathbb{A}\text{dd and/or subtract}\\\\\\\mathbb{M}\text{ultiply and/or divide}`

What is our goal when solving equations?

Our goal is to isolate the variable.

Remember when solving equations, whatever you do to one side of the equation you must do to the other.
`\hrulefill`

Now solving #5,

`-10-7x=-3x-2`

Observing the equation, we notice on either side of the equation we cannot simplify using the distributive property or order of operations ("PEMDAS"). Well will skip to the "C" in SCAM.

Add "7x" to each side of the equation.

`\Longrightarrow -10-7x+7x=-3x-2+7x\\\\\\\Longrightarrow -10=4x-2`

The variable "x" now appears on one side of the equation. Now we are on the "A" in SCAM.

Add the value of "2" to each side of the equation.

`\Longrightarrow -10+2=4x-2+2\\\\\\\Longrightarrow -8=4x`

Now on the "M" in SCAM.

Divide both sides of the equation by the value of "4."

`\Longrightarrow (-8)/(4)=(4)/(4)x\\\\\\\Longrightarrow -2=x\\\\\\\therefore \boxed{x=-2}`

Thus, problem #5 is solved.

`\hrulefill`

For the rest of the problems I will simply show the operations I am doing.

Now solving #6,

`-13-4x=x+7\\\\\\\Longrightarrow -13-4x+4x=x+7+4x\\\\\\\Longrightarrow -13=5x+7\\\\\\\Longrightarrow -13-7=5x+7-7\\\\\\\Longrightarrow -20=5x\\\\\\\Longrightarrow (-20)/(5)= (5)/(5)x\\\\\\\Longrightarrow-4=x\\\\\\\therefore \boxed{x=-4}`

Thus, problem #6 is solved.

`\hrulefill`

Now solving #7,

`x-2=10-5x\\\\\\\Longrightarrow x-2+5x=10-5x+5x\\\\\\\Longrightarrow 6x-2=10\\\\\\\Longrightarrow 6x-2+2=10+2\\\\\\\Longrightarrow 6x=12\\\\\\\Longrightarrow (6)/(6)x=(12)/(6)\\\\\\\therefore \boxed{x=2}`

Thus, problem #7 is solved.

`\hrulefill`

Now solving #8,

`3x-1=4x-11\\\\\\\Longrightarrow 3x-1-3x=4x-11-3x\\\\\\\Longrightarrow -1=x-11\\\\\\\Longrightarrow -1+11=x-11+11\\\\\\\Longrightarrow 10=x\\\\\\\therefore \boxed{x=10}`