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: \(\displaystyle \:x\:=\:-\frac{1}{5}\)
