
Solve the following equations
a) $5x - 11 = 2x + 7$
b) $3(x - 8) - 2(x - 5) = 2 - 5x$
c) $2(3x + 1) - 7 = 2(6x - 7)$
Answer
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Hint: Since these are only one variable problem, bringing the variable on one side of the equation will help you solve the problem.
Complete step-by-step answer:
All the problems have only one variable. So each equation needs rearrangement with the terms to solve, i.e. find the value of $x$.
a) $5x - 11 = 2x + 7$
We will bring the variable terms on the left hand side of the equation and numbers on the other side. Bringing the variables on the left side of the equation is just a convention and is not related to the solution.
$\begin{gathered}
\Rightarrow 5x - 2x = 7 + 11 \\
\Rightarrow 3x = 18 \\
\end{gathered} $
Dividing the equation by 3 on both sides we get,
$ \Rightarrow x = 6$ This is the solution.
b) $3(x - 8) - 2(x - 5) = 2 - 5x$
Opening brackets first and bringing all the variables on one side, we get,
$\begin{gathered}
\Rightarrow 3x - 24 - 2x + 10 + 5x = 2 \\
\Rightarrow 6x = 2 + 24 - 10 \\
\Rightarrow 6x = 16 \\
\Rightarrow x = \dfrac{{16}}{6} = 2.\overline 6 \\
\end{gathered} $
This is the solution and we have got a fraction. Our variable is a representative of all types of numbers, rational, irrational. In this case we get a recurring digit 6 if we solve the fraction.
c) $2(3x + 1) - 7 = 2(6x - 7)$
Again, as we did with earlier problems, we will open the bracket and rearrange the terms by bringing variables on the left hand side and numbers on the right hand side.
$\begin{gathered}
\Rightarrow 6x + 2 = 12x - 14 + 7 \\
\Rightarrow 6x - 12x = - 7 - 2 \\
\Rightarrow - 6x = - 9 \\
\end{gathered} $
Dividing both sides by (-3), we get,
$\begin{gathered}
\Rightarrow 2x = 3 \\
\Rightarrow x = \dfrac{3}{2} = 1.5 \\
\end{gathered} $
This is the solution.
Note: These problems are single variables and simple rearrangement will do the magic but we need to be careful with signs while opening brackets and bringing terms on the other side of the equation.
Complete step-by-step answer:
All the problems have only one variable. So each equation needs rearrangement with the terms to solve, i.e. find the value of $x$.
a) $5x - 11 = 2x + 7$
We will bring the variable terms on the left hand side of the equation and numbers on the other side. Bringing the variables on the left side of the equation is just a convention and is not related to the solution.
$\begin{gathered}
\Rightarrow 5x - 2x = 7 + 11 \\
\Rightarrow 3x = 18 \\
\end{gathered} $
Dividing the equation by 3 on both sides we get,
$ \Rightarrow x = 6$ This is the solution.
b) $3(x - 8) - 2(x - 5) = 2 - 5x$
Opening brackets first and bringing all the variables on one side, we get,
$\begin{gathered}
\Rightarrow 3x - 24 - 2x + 10 + 5x = 2 \\
\Rightarrow 6x = 2 + 24 - 10 \\
\Rightarrow 6x = 16 \\
\Rightarrow x = \dfrac{{16}}{6} = 2.\overline 6 \\
\end{gathered} $
This is the solution and we have got a fraction. Our variable is a representative of all types of numbers, rational, irrational. In this case we get a recurring digit 6 if we solve the fraction.
c) $2(3x + 1) - 7 = 2(6x - 7)$
Again, as we did with earlier problems, we will open the bracket and rearrange the terms by bringing variables on the left hand side and numbers on the right hand side.
$\begin{gathered}
\Rightarrow 6x + 2 = 12x - 14 + 7 \\
\Rightarrow 6x - 12x = - 7 - 2 \\
\Rightarrow - 6x = - 9 \\
\end{gathered} $
Dividing both sides by (-3), we get,
$\begin{gathered}
\Rightarrow 2x = 3 \\
\Rightarrow x = \dfrac{3}{2} = 1.5 \\
\end{gathered} $
This is the solution.
Note: These problems are single variables and simple rearrangement will do the magic but we need to be careful with signs while opening brackets and bringing terms on the other side of the equation.
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