a. To solve an equation our goal is to isolate the variable. That is, get an equivalent equation of the form variable [latex]=[/latex] number. In the equation [latex]x+1=7[/latex] we need to have the variable x alone on one side of the equation. To remove the constant term [latex]1[/latex] on the left side, we use the additive property of equality to add the additive inverse of [latex]1[/latex] to each side of the equation.

[latex]x+1=7[/latex]

[latex]x+1+(-1)=7+(-1)[/latex]

[latex]x+0=6[/latex]

[latex]x=6[/latex]

Since adding [latex]-1[/latex] is the same as subtracting [latex]1[/latex], we often say we subtract [latex]1[/latex] from each side of the equation. We can check our solution in the original equation.

[latex]6+1=7[/latex]

b. To solve the equation [latex]3x=45[/latex] we need to remove the constant factor [latex]3[/latex] from the left side. We do this by multiplying each side by its multiplicative inverse (reciprocal), [latex]\frac{1}{3}[/latex].

[latex]3x=45[/latex]

[latex]\frac{1}{3} \cdot 3x = \frac{1}{3} \cdot 45[/latex]

[latex]1 \cdot x = \frac{45}{3}[/latex]

[latex]x=15[/latex]

Since multiplying by [latex]\frac{1}{3}[/latex] produces the same result as dividing by [latex]3[/latex], we often say we divide each side of the equation by [latex]3[/latex]. We can check our solution in the original equation.

[latex]3 \cdot 15 = 45[/latex]

Apply the distributive property to expand [latex]4\left(2a+3\right)[/latex] to [latex]8a+12[/latex].

[latex]\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}[/latex]

Subtract [latex]12[/latex] from both sides to isolate the variable term.

[latex]\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}[/latex]

Divide both terms by [latex]8[/latex] to get a coefficient of [latex]1[/latex].

[latex]\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\\a\,=\,\,2\end{array}[/latex]

Answer

[latex]a=2[/latex]

Apply the distributive property to expand [latex]2\left(4t-5\right)[/latex] to [latex]8t-10[/latex] and [latex]-3\left(2t+1\right)[/latex] to[latex]-6t-3[/latex]. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.

[latex]\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Add [latex]-6t[/latex] to both sides to begin combining like terms.

[latex]\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}[/latex]

Add [latex]10[/latex] to both sides of the equation to isolate t.

[latex]\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}[/latex]

The last step is to divide both sides by [latex]14[/latex] to completely isolate t.

[latex]\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}[/latex]

Answer

[latex]t=\frac{1}{2}[/latex]

We simplified the fraction [latex]\frac{7}{14}[/latex] into [latex]\frac{1}{2}[/latex].