It is helpful to think of this inequality as asking you to find all the values for x, including negative numbers, such that when you add three you will get a number less than 5.

[latex]\displaystyle \begin{array}{l}x+3<\,\,\,\,5\\\underline{\,\,\,\,\,-3\,\,\,\,-3}\\x\,\,\,\,\,\,\,\,<\,\,\,\,2\,\,\end{array}[/latex]

Isolate the variable by subtracting 3 from both sides of the inequality.

Answer

Inequality:  [latex]x<2[/latex] Interval:  [latex]\left(-\infty, 2\right)[/latex] Graph:

Isolate the variable by adding 10 to both sides of the inequality.

[latex]\displaystyle \begin{array}{r}x-10\le -12\\\underline{\,\,\,+10\,\,\,\,\,+10}\\x\,\,\,\,\,\,\,\,\,\,\le \,\,\,-2\end{array}[/latex]

Answer

Inequality: [latex]x\leq-2[/latex]
Interval: [latex]\left(-\infty,-2\right][/latex]
Graph: Notice that a closed circle is used because the inequality is “less than or equal to” [latex]\left(\leq\right)[/latex]. The blue arrow is drawn to the left of the point [latex]−2[/latex] because these are the values that are less than [latex]−2[/latex].

Substitute the end point [latex]−2[/latex] into the related equation  [latex]x-10=−12[/latex]

[latex]\displaystyle \begin{array}{r}x-10=-12\,\,\,\\\text{Does}\,\,\,-2-10=-12?\\-12=-12\,\,\,\end{array}[/latex]

Pick a value less than [latex]−2[/latex], such as [latex]−5[/latex], to check in the inequality. (This value will be on the shaded part of the graph.)

[latex]\displaystyle \begin{array}{r}x-10\le -12\,\,\,\\\text{ }\,\text{ Is}\,\,-5-10\le -12?\\-15\le -12\,\,\,\\\text{It}\,\text{checks!}\end{array}[/latex]

[latex]x\leq -2[/latex] is the solution to [latex]x-10\leq -12[/latex]

Divide both sides by 3 to isolate the variable.

[latex]\displaystyle \begin{array}{r}\underline{3x}>\underline{12}\\3\,\,\,\,\,\,\,\,\,\,\,\,3\\x>4\,\,\,\end{array}[/latex]

Check your solution by first checking the end point 4, and then checking another solution for the inequality.

[latex]\begin{array}{r}3\cdot4=12\\12=12\\3\cdot10>12\\30>12\\\text{It checks!}\end{array}[/latex]

Answer

Inequality: [latex]\displaystyle x>4[/latex]

Interval: [latex]\left(4,\infty\right)[/latex]

Graph:

Divide each side of the inequality by [latex]−2[/latex] to isolate the variable, and change the direction of the inequality sign because of the division by a negative number.

[latex]\displaystyle \begin{array}{r}\underline{-2x}<\underline{\,6\,}\\-2\,\,\,\,-2\,\\x<-3\end{array}[/latex]

Check your solution by first checking the end point [latex]−3[/latex], and then checking another solution for the inequality.

[latex]\begin{array}{r}-2\left(-3\right)=6 \\6=6\\ -2\left(-6\right)>6 \\ 12>6\end{array}[/latex]

It checks!

Answer

Inequality: [latex]\displaystyle x<-3[/latex] Interval: [latex]\left(-\infty, -3\right)[/latex] Graph:
Because both sides of the inequality were divided by a negative number, [latex]−2[/latex], the inequality symbol was switched from > to <.

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]