We still need some help with solving linear equations

[greenwood]

Just looking for someone knowledgeable enough to walk us through
the steps of how to get the answer and solving equations like these:
Thank u in advance for stepping up to the newcomers and helping us out.......
Much appreciated~

3x = 5 + 2(3 + 4x) OR 10 + 3x = 2(3 + 4x) + 5

 

[soroban]

Re: We still need some assistance....8th grade math- algebra

Hello, greenwood!

$\displaystyle 3x \;= \;5\,+\,2(3\,+\,4x)$

Use the Distributive Property to clear the parentheses:$\displaystyle \:3x\;=\;5\,+\,6\,+\,8x$

Combine terms (if possible): $\displaystyle \:3x\;=\;11\,+\,8x$

Get the x-terms on one side; the constants on the other side.

Subtract $\displaystyle 8x$ from both sides: $\displaystyle \:3x\,-\,8x\;=\;11\,+\,8x\,-\,8x$

$\displaystyle \;\;$And we have: $\displaystyle \:-5x\;=\;11$

Divide both sides by -5: $\displaystyle \,\frac{-5x}{-5}\;=\;\frac{11}{-5}$

Therefore: $\displaystyle \;x\;=\;-\frac{11}{5}$

$\displaystyle 10\,+\,3x\;=\;2(3\,+\,4x)\,+\,5$

It's the same routine.

Clear parentheses: $\displaystyle \:10\,+\,3x\;=\;6\,+\,8x\,+\,5$

Combine terms: $\displaystyle \:10\,+\,3x\;=\;8x\,+\,11$

Get the x-terms on one side: subtract $\displaystyle 8x$ from both sides:
\(\displaystyle \;\;10\,+\,3x\,-\,8x\;=\;8x\.+\,11\,-\,8x\)
. . . . . . .$\displaystyle 10\,-\,5x\;=\;11$

Get the constants on the other side: subtract 10 from both sides:
$\displaystyle \;\;10\,-\,5x\,-\,10\;=\;11\,-\,10$
. . . . . . . . . $\displaystyle \,-5x\;=\;1$

Divide both sides by -5: $\displaystyle \:\frac{-5x}{-5}\;=\;\frac{1}{-5}$

Therefore: $\displaystyle \:x\:=\:-\frac{1}{5}$

 

[greenwood]

THANK YOU!!!

Thanks this helped out a bunch.....