Solving linear equations

Worked solution

1. 1

   Multiply both sides by $2$.

   $3\left(x-2\right)=18$

   The denominator divides the whole numerator. Multiplying by $2$ undoes that division.

2. 2

   Divide both sides by $3$, then add $2$.

   $\begin{array}{rl}x-2& =6\\ x& =8\end{array}$

   The factor $3$ multiplies the entire bracket, so division removes it without expansion.

3. 3

   Substitute $x=8$ into the original equation.

   $\frac{3\left(8-2\right)}{2}=\frac{18}{2}=9$

   The left side equals the original right side, so the solution passes the check.

Worked solution

1. 1

   Expand the bracket and collect the numbers on the left.

   $\begin{array}{rl}7-2x+6& =x+1\\ 13-2x& =x+1\end{array}$

   The product $\left(-2\right)\left(-3\right)$ is $+6$. The negative factor applies to both terms.

2. 2

   Add $2x$ to both sides.

   $\begin{array}{rl}13-2x+2x& =x+1+2x\\ 13& =3x+1\end{array}$

   This removes the unknown term on the left and leaves all the unknown terms on the right.

3. 3

   Subtract $1$ from both sides, then divide by $3$.

   $\begin{array}{rl}12& =3x\\ 4& =x\end{array}$

   The statement $4=x$ means the same as $x=4$.

4. 4

   Check both original sides at $x=4$.

   $\begin{array}{rl}\text{Left}& =7-2\left(4-3\right)=5\\ \text{Right}& =4+1=5\end{array}$

   Both sides give $5$, including the original negative bracket.

Worked solution

1. 1

   Let $h$ be the number of hours and equate the costs.

   $6+3h=14+2h$

   Each cost is its fixed charge plus its hourly charge multiplied by the number of hours.

2. 2

   Subtract $2h$ from both sides, then subtract $6$.

   $\begin{array}{rl}6+h& =14\\ h& =8\end{array}$

   The same operations on both sides preserve equality while isolating the number of hours.

3. 3

   Check the two costs at $h=8$.

   $\begin{array}{rl}6+3\left(8\right)& =30\\ 14+2\left(8\right)& =30\end{array}$

   Each plan costs $30$ dollars, so the equal-cost condition holds.