Begin to isolate the variable by subtracting 5 from both sides of the inequality.

[latex]\displaystyle \begin{array}{l}4p+5<\,\,\,29\\\underline{\,\,\,\,\,\,\,\,\,-5\,\,\,\,\,-5}\\4p\,\,\,\,\,\,\,\,\,<\,\,24\,\,\end{array}[/latex]

Divide both sides of the inequality by 4 to express the variable with a coefficient of 1.

[latex]\begin{array}{l}\underline{4p}\,<\,\,\underline{24}\,\,\\\,4\,\,\,\,<\,\,4\\\,\,\,\,\,p<6\end{array}[/latex]

Answer

Inequality: [latex]p<6[/latex] Interval: [latex]\left(-\infty,6\right)[/latex] Graph: Note the open circle at the end point 6 to show that solutions to the inequality do not include 6. The values where p is less than 6 are found all along the number line to the left of 6.

Check the end point 6 in the related equation.

[latex]\displaystyle \begin{array}{r}4p+5=29\,\,\,\\\text{Does}\,\,\,4(6)+5=29?\\24+5=29\,\,\,\\29=29\,\,\,\\\text{Yes!}\,\,\,\,\,\,\end{array}[/latex]

Try another value to check the inequality. Let’s use [latex]p=0[/latex].

[latex]\displaystyle \begin{array}{r}4p+5<29\,\,\,\\\text{Is}\,\,\,4(0)+5<29?\\0+5<29\,\,\,\\5<29\,\,\,\\\text{Yes!}\,\,\,\,\,\end{array}[/latex]

[latex]p<6[/latex] is the solution to [latex]4p+5<29[/latex]

Note how the variable is on the right hand side of the inequality, the method for solving does not change in this case.

Begin to isolate the variable by subtracting 14 from both sides of the inequality.

[latex]\displaystyle \begin{array}{l}−58\,\,>14−6p\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,-14\,\,\,\,\,\,\,-14}\\-72\,\,\,\,\,\,\,\,\,\,\,>-6p\end{array}[/latex]

Divide both sides of the inequality by [latex]−6[/latex] to express the variable with a coefficient of 1. Dividing by a negative number results in reversing the inequality sign.

[latex]\begin{array}{l}\underline{-72}>\underline{-6p}\\-6\,\,\,\,\,\,\,\,\,\,-6\\\,\,\,\,\,\,12\lt{p}\end{array}[/latex]

We can also write this as [latex]p>12[/latex].   Notice how the inequality sign is still opening up toward the variable p.

Answer

Inequality: [latex]p>12[/latex]
Interval: [latex]\left(12,\infty\right)[/latex]
Graph: The graph of the inequality p > 12 has an open circle at 12 with an arrow stretching to the right.

First, check the end point 12 in the related equation.

[latex]\begin{array}{r}-58=14-6p\\-58=14-6\left(12\right)\\-58=14-72\\-58=-58\end{array}[/latex]

Then, try another value to check the inequality. Try 100.

[latex]\begin{array}{r}-58>14-6p\\-58>14-6\left(100\right)\\-58>14-600\\-58>-586\end{array}[/latex]

Distribute to clear the parentheses.

[latex]\displaystyle \begin{array}{r}\,2(3x-5)\leq 4x+6\\\,\,\,\,6x-10\leq 4x+6\end{array}[/latex]

Subtract 4x from both sides to get the variable term on one side only.

[latex]\begin{array}{r}6x-10\le 4x+6\\\underline{-4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4x}\,\,\,\,\,\,\,\,\,\\\,\,\,2x-10\,\,\leq \,\,\,\,\,\,\,\,\,\,\,\,6\end{array}[/latex]

Add 10 to both sides to isolate the variable.

[latex]\begin{array}{r}\\\,\,\,2x-10\,\,\le \,\,\,\,\,\,\,\,6\,\,\,\\\underline{\,\,\,\,\,\,+10\,\,\,\,\,\,\,\,\,+10}\\\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\le \,\,\,\,\,16\,\,\,\end{array}[/latex]

Divide both sides by 2 to express the variable with a coefficient of 1.

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

Answer

Inequality: [latex]x\le8[/latex]
Interval: [latex]\left(-\infty,8\right][/latex]
Graph: The graph of this solution set includes 8 and everything left of 8 on the number line.

First, check the end point 8 in the related equation.

[latex]\displaystyle \begin{array}{r}2(3x-5)=4x+6\,\,\,\,\,\,\\2(3\,\cdot \,8-5)=4\,\cdot \,8+6\\\,\,\,\,\,\,\,\,\,\,\,2(24-5)=32+6\,\,\,\,\,\,\\2(19)=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\38=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Then, choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 0.

[latex]\displaystyle \begin{array}{l}2(3\,\cdot \,0-5)\le 4\,\cdot \,0+6?\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2(-5)\le 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-10\le 6\,\,\end{array}[/latex]

[latex]x\le8[/latex] is the solution to [latex]\left(-\infty,8\right][/latex]